Update:: I think that 228 integral was a mistake, I found another project file from Dr. Crawford which listed 406 unique integrals for the same system ($H_2O$ with 7 molecular orbitals). In case if anyone wants to work on that project, they can use this data file.

-----------------------------------------------------------------------
      PSI4: An Open-Source Ab Initio Electronic Structure Package
                          PSI 4.0.0-alpha Driver
T. D. Crawford, C. D. Sherrill, E. F. Valeev, J. T. Fermann, R. A. King,
M. L. Leininger, S. T. Brown, C. L. Janssen, E. T. Seidl, J. P. Kenny,
and W. D. Allen, J. Comput. Chem. 28, 1610-1616 (2007)

                     Additional Contributions by
Francesco Evangelista, Andrew Simmonett, Justin Turney, Jeremiah Wilke
-----------------------------------------------------------------------

Using LocalCommunicator (Number of processes = 1)

Memory level set to 256.000 MB

./plugin_aointegrals.so loaded.
Calling plugin plugin_aointegrals.so.
    Molecular point group: c2v

Geometry (in Angstrom), charge = 0, multiplicity = 1:

   Center              X                  Y                   Z
------------   -----------------  -----------------  -----------------
       O          0.000000000000     0.000000000000    -0.071151380605
       H          0.000000000000     0.757939245855     0.564612021746
       H          0.000000000000    -0.757939245855     0.564612021746

Number of unique atoms: 2

Atoms equivalency:
   unique atom 0: 0
   unique atom 1: 1 2

Nuclear repulsion energy:       8.90770810

## Overlap ##

             1            2            3            4            5            6            7

1    1.0000000    0.2367039    0.0000000    0.0000000   -0.0000000    0.0500137    0.0500137
2    0.2367039    1.0000000    0.0000000    0.0000000   -0.0000000    0.4539953    0.4539953
3    0.0000000    0.0000000    1.0000000    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000    1.0000000    0.0000000    0.2927386   -0.2927386
5   -0.0000000   -0.0000000    0.0000000    0.0000000    1.0000000    0.2455507    0.2455507
6    0.0500137    0.4539953    0.0000000    0.2927386    0.2455507    1.0000000    0.2510021
7    0.0500137    0.4539953    0.0000000   -0.2927386    0.2455507    0.2510021    1.0000000

## Kinetic ##

             1            2            3            4            5            6            7

1   29.0031999   -0.1680109    0.0000000    0.0000000   -0.0000000   -0.0045400   -0.0045400
2   -0.1680109    0.8081280    0.0000000    0.0000000   -0.0000000    0.1137475    0.1137475
3    0.0000000    0.0000000    2.5287312    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000    2.5287312    0.0000000    0.1993355   -0.1993355
5   -0.0000000   -0.0000000    0.0000000    0.0000000    2.5287312    0.1672036    0.1672036
6   -0.0045400    0.1137475    0.0000000    0.1993355    0.1672036    0.7600319    0.0083249
7   -0.0045400    0.1137475    0.0000000   -0.1993355    0.1672036    0.0083249    0.7600319

## Potential ##

             1            2            3            4            5            6            7

1  -61.6882822   -7.4363118    0.0000000    0.0000000   -0.0186797   -1.6150635   -1.6150635
2   -7.4363118  -10.1101908    0.0000000    0.0000000   -0.2221598   -3.6569582   -3.6569582
3    0.0000000    0.0000000   -9.9595668    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000  -10.0957534    0.0000000   -2.0901916    2.0901916
5   -0.0186797   -0.2221598    0.0000000    0.0000000  -10.0553869   -1.8259929   -1.8259929
6   -1.6150635   -3.6569582    0.0000000   -2.0901916   -1.8259929   -5.7165220   -1.5685885
7   -1.6150635   -3.6569582    0.0000000    2.0901916   -1.8259929   -1.5685885   -5.7165220

## One Electron Ints ##

             1            2            3            4            5            6            7

1  -32.6850823   -7.6043227    0.0000000    0.0000000   -0.0186797   -1.6196036   -1.6196036
2   -7.6043227   -9.3020628    0.0000000    0.0000000   -0.2221598   -3.5432106   -3.5432106
3    0.0000000    0.0000000   -7.4308356    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000   -7.5670222    0.0000000   -1.8908561    1.8908561
5   -0.0186797   -0.2221598    0.0000000    0.0000000   -7.5266557   -1.6587893   -1.6587893
6   -1.6196036   -3.5432106    0.0000000   -1.8908561   -1.6587893   -4.9564901   -1.5602636
7   -1.6196036   -3.5432106    0.0000000    1.8908561   -1.6587893   -1.5602636   -4.9564901

Two-electron Integrals

( 0  0 |  0  0) =    4.785065404705502
( 1  0 |  0  0) =    0.741380351973408
( 1  1 |  0  0) =    1.118946866342470
( 1  0 |  1  0) =    0.136873385354388
( 1  1 |  1  0) =    0.256633394730974
( 1  1 |  1  1) =    0.817206321526058
( 2  0 |  0  0) =    0.000000000000000
( 3  0 |  0  0) =    0.000000000000000
( 4  0 |  0  0) =   -0.000000000000000
( 2  1 |  0  0) =    0.000000000000000
( 3  1 |  0  0) =    0.000000000000000
( 4  1 |  0  0) =   -0.000000000000000
( 2  0 |  1  0) =    0.000000000000000
( 3  0 |  1  0) =    0.000000000000000
( 4  0 |  1  0) =   -0.000000000000000
( 2  1 |  1  0) =    0.000000000000000
( 3  1 |  1  0) =    0.000000000000000
( 4  1 |  1  0) =   -0.000000000000000
( 2  0 |  1  1) =    0.000000000000000
( 3  0 |  1  1) =    0.000000000000000
( 4  0 |  1  1) =   -0.000000000000000
( 2  1 |  1  1) =    0.000000000000000
( 3  1 |  1  1) =    0.000000000000000
( 4  1 |  1  1) =    0.000000000000000
( 2  2 |  0  0) =    1.115813812152427
( 3  2 |  0  0) =    0.000000000000000
( 3  3 |  0  0) =    1.115813812152427
( 4  2 |  0  0) =    0.000000000000000
( 4  3 |  0  0) =    0.000000000000000
( 4  4 |  0  0) =    1.115813812152427
( 2  0 |  2  0) =    0.024477412258099
( 3  0 |  2  0) =    0.000000000000000
( 3  0 |  3  0) =    0.024477412258099
( 4  0 |  2  0) =    0.000000000000000
( 4  0 |  3  0) =    0.000000000000000
( 4  0 |  4  0) =    0.024477412258099
( 2  2 |  1  0) =    0.256683985810103
( 3  2 |  1  0) =    0.000000000000000
( 3  3 |  1  0) =    0.256683985810103
( 4  2 |  1  0) =    0.000000000000000
( 4  3 |  1  0) =    0.000000000000000
( 4  4 |  1  0) =    0.256683985810103
( 2  1 |  2  0) =    0.037808607416361
( 3  0 |  2  1) =    0.000000000000000
( 4  0 |  2  1) =    0.000000000000000
( 3  1 |  2  0) =    0.000000000000000
( 3  1 |  3  0) =    0.037808607416361
( 4  0 |  3  1) =    0.000000000000000
( 4  1 |  2  0) =    0.000000000000000
( 4  1 |  3  0) =    0.000000000000000
( 4  1 |  4  0) =    0.037808607416361
( 2  2 |  1  1) =    0.817022605320914
( 3  2 |  1  1) =    0.000000000000000
( 3  3 |  1  1) =    0.817022605320914
( 4  2 |  1  1) =    0.000000000000000
( 4  3 |  1  1) =    0.000000000000000
( 4  4 |  1  1) =    0.817022605320914
( 2  1 |  2  1) =    0.180518392104632
( 3  1 |  2  1) =    0.000000000000000
( 3  1 |  3  1) =    0.180518392104632
( 4  1 |  2  1) =    0.000000000000000
( 4  1 |  3  1) =    0.000000000000000
( 4  1 |  4  1) =    0.180518392104632
( 2  2 |  2  0) =    0.000000000000000
( 3  2 |  2  0) =    0.000000000000000
( 3  3 |  2  0) =    0.000000000000000
( 4  2 |  2  0) =    0.000000000000000
( 4  3 |  2  0) =    0.000000000000000
( 4  4 |  2  0) =    0.000000000000000
( 3  0 |  2  2) =    0.000000000000000
( 3  2 |  3  0) =    0.000000000000000
( 3  3 |  3  0) =    0.000000000000000
( 4  2 |  3  0) =    0.000000000000000
( 4  3 |  3  0) =    0.000000000000000
( 4  4 |  3  0) =    0.000000000000000
( 4  0 |  2  2) =   -0.000000000000000
( 4  0 |  3  2) =    0.000000000000000
( 4  0 |  3  3) =   -0.000000000000000
( 4  2 |  4  0) =    0.000000000000000
( 4  3 |  4  0) =    0.000000000000000
( 4  4 |  4  0) =   -0.000000000000000
( 2  2 |  2  1) =    0.000000000000000
( 3  2 |  2  1) =    0.000000000000000
( 3  3 |  2  1) =    0.000000000000000
( 4  2 |  2  1) =    0.000000000000000
( 4  3 |  2  1) =    0.000000000000000
( 4  4 |  2  1) =    0.000000000000000
( 3  1 |  2  2) =    0.000000000000000
( 3  2 |  3  1) =    0.000000000000000
( 3  3 |  3  1) =    0.000000000000000
( 4  2 |  3  1) =    0.000000000000000
( 4  3 |  3  1) =    0.000000000000000
( 4  4 |  3  1) =    0.000000000000000
( 4  1 |  2  2) =    0.000000000000000
( 4  1 |  3  2) =    0.000000000000000
( 4  1 |  3  3) =    0.000000000000000
( 4  2 |  4  1) =    0.000000000000000
( 4  3 |  4  1) =    0.000000000000000
( 4  4 |  4  1) =    0.000000000000000
( 2  2 |  2  2) =    0.880159093375045
( 3  2 |  2  2) =    0.000000000000000
( 3  2 |  3  2) =    0.047444445118384
( 3  3 |  2  2) =    0.785270203138277
( 3  3 |  3  2) =    0.000000000000000
( 3  3 |  3  3) =    0.880159093375045
( 4  2 |  2  2) =    0.000000000000000
( 4  2 |  3  2) =    0.000000000000000
( 4  2 |  3  3) =    0.000000000000000
( 4  2 |  4  2) =    0.047444445118384
( 4  3 |  2  2) =    0.000000000000000
( 4  3 |  3  2) =    0.000000000000000
( 4  3 |  3  3) =    0.000000000000000
( 4  3 |  4  2) =    0.000000000000000
( 4  3 |  4  3) =    0.047444445118384
( 4  4 |  2  2) =    0.785270203138277
( 4  4 |  3  2) =    0.000000000000000
( 4  4 |  3  3) =    0.785270203138277
( 4  4 |  4  2) =    0.000000000000000
( 4  4 |  4  3) =    0.000000000000000
( 4  4 |  4  4) =    0.880159093375045
( 5  0 |  0  0) =    0.159028953813045
( 5  1 |  0  0) =    0.381788700431517
( 5  0 |  1  0) =    0.029107255214247
( 5  1 |  1  0) =    0.088639046604491
( 5  0 |  1  1) =    0.054208207239172
( 5  1 |  1  1) =    0.311911628346485
( 5  2 |  0  0) =    0.000000000000000
( 5  3 |  0  0) =    0.208975170450133
( 5  4 |  0  0) =    0.175289465612119
( 5  0 |  2  0) =    0.000000000000000
( 5  0 |  3  0) =    0.001028596856434
( 5  0 |  4  0) =    0.000862792421254
( 5  2 |  1  0) =    0.000000000000000
( 5  3 |  1  0) =    0.049151047350638
( 5  4 |  1  0) =    0.041228155506758
( 5  1 |  2  0) =    0.000000000000000
( 5  1 |  3  0) =    0.005091750233326
( 5  1 |  4  0) =    0.004270986718219
( 5  0 |  2  1) =    0.000000000000000
( 5  0 |  3  1) =    0.001923027401919
( 5  0 |  4  1) =    0.001613045439387
( 5  2 |  1  1) =    0.000000000000000
( 5  3 |  1  1) =    0.184701361441018
( 5  4 |  1  1) =    0.154928467698285
( 5  1 |  2  1) =    0.000000000000000
( 5  1 |  3  1) =    0.038683534029509
( 5  1 |  4  1) =    0.032447950603007
( 5  2 |  2  0) =    0.009336058122476
( 5  2 |  3  0) =    0.000000000000000
( 5  2 |  4  0) =    0.000000000000000
( 5  3 |  2  0) =    0.000000000000000
( 5  3 |  3  0) =    0.011813922734513
( 5  3 |  4  0) =    0.002078445792233
( 5  4 |  2  0) =    0.000000000000000
( 5  4 |  3  0) =    0.002078445792233
( 5  4 |  4  0) =    0.011079469320272
( 5  0 |  2  2) =    0.054104219459936
( 5  0 |  3  2) =    0.000000000000000
( 5  0 |  3  3) =    0.054291571211162
( 5  0 |  4  2) =    0.000000000000000
( 5  0 |  4  3) =    0.000157151628507
( 5  0 |  4  4) =    0.054236039067604
( 5  2 |  2  1) =    0.054125156914812
( 5  2 |  3  1) =    0.000000000000000
( 5  2 |  4  1) =    0.000000000000000
( 5  3 |  2  1) =    0.000000000000000
( 5  3 |  3  1) =    0.075855681518665
( 5  3 |  4  1) =    0.018227677656995
( 5  4 |  2  1) =    0.000000000000000
( 5  4 |  3  1) =    0.018227677656995
( 5  4 |  4  1) =    0.069414628237947
( 5  1 |  2  2) =    0.306655599347507
( 5  1 |  3  2) =    0.000000000000000
( 5  1 |  3  3) =    0.315738293535417
( 5  1 |  4  2) =    0.000000000000000
( 5  1 |  4  3) =    0.007618611374202
( 5  1 |  4  4) =    0.313046130402530
( 5  2 |  2  2) =    0.000000000000000
( 5  2 |  3  2) =    0.008064485483452
( 5  2 |  3  3) =    0.000000000000000
( 5  2 |  4  2) =    0.006764532589140
( 5  2 |  4  3) =    0.000000000000000
( 5  2 |  4  4) =    0.000000000000000
( 5  3 |  2  2) =    0.176295759644369
( 5  3 |  3  2) =    0.000000000000000
( 5  3 |  3  3) =    0.197638924560966
( 5  3 |  4  2) =    0.000000000000000
( 5  3 |  4  3) =    0.011138225736852
( 5  3 |  4  4) =    0.179964436324773
( 5  4 |  2  2) =    0.147877804962089
( 5  4 |  3  2) =    0.000000000000000
( 5  4 |  3  3) =    0.152251498109802
( 5  4 |  4  2) =    0.000000000000000
( 5  4 |  4  3) =    0.011733162163856
( 5  4 |  4  4) =    0.164484175321776
( 5  5 |  0  0) =    0.517125360977304
( 5  0 |  5  0) =    0.006270522616106
( 5  5 |  1  0) =    0.122013556886528
( 5  1 |  5  0) =    0.019208705902818
( 5  5 |  1  1) =    0.491873027324534
( 5  1 |  5  1) =    0.146642827474200
( 5  5 |  2  0) =    0.000000000000000
( 5  5 |  3  0) =    0.009188483135799
( 5  5 |  4  0) =    0.007707347696814
( 5  2 |  5  0) =    0.000000000000000
( 5  3 |  5  0) =    0.011128525245716
( 5  4 |  5  0) =    0.009334665162233
( 5  5 |  2  1) =    0.000000000000000
( 5  5 |  3  1) =    0.095810960370799
( 5  5 |  4  1) =    0.080366734512074
( 5  2 |  5  1) =    0.000000000000000
( 5  3 |  5  1) =    0.105474819021288
( 5  4 |  5  1) =    0.088472829676139
( 5  5 |  2  2) =    0.471913115835107
( 5  5 |  3  2) =    0.000000000000000
( 5  5 |  3  3) =    0.506704669305872
( 5  5 |  4  2) =    0.000000000000000
( 5  5 |  4  3) =    0.029183336960891
( 5  5 |  4  4) =    0.496392251464638
( 5  2 |  5  2) =    0.021048752318672
( 5  3 |  5  2) =    0.000000000000000
( 5  3 |  5  3) =    0.096048971720199
( 5  4 |  5  2) =    0.000000000000000
( 5  4 |  5  3) =    0.062910576176908
( 5  4 |  5  4) =    0.073818472547383
( 5  5 |  5  0) =    0.026686622886860
( 5  5 |  5  1) =    0.289333860384054
( 5  5 |  5  2) =    0.000000000000000
( 5  5 |  5  3) =    0.237637271677700
( 5  5 |  5  4) =    0.199331385983227
( 5  5 |  5  5) =    0.774605943919898
( 6  0 |  0  0) =    0.159028953813045
( 6  1 |  0  0) =    0.381788700431517
( 6  0 |  1  0) =    0.029107255214247
( 6  1 |  1  0) =    0.088639046604491
( 6  0 |  1  1) =    0.054208207239172
( 6  1 |  1  1) =    0.311911628346485
( 6  2 |  0  0) =    0.000000000000000
( 6  3 |  0  0) =   -0.208975170450133
( 6  4 |  0  0) =    0.175289465612119
( 6  0 |  2  0) =    0.000000000000000
( 6  0 |  3  0) =   -0.001028596856434
( 6  0 |  4  0) =    0.000862792421254
( 6  2 |  1  0) =    0.000000000000000
( 6  3 |  1  0) =   -0.049151047350638
( 6  4 |  1  0) =    0.041228155506758
( 6  1 |  2  0) =    0.000000000000000
( 6  1 |  3  0) =   -0.005091750233326
( 6  1 |  4  0) =    0.004270986718219
( 6  0 |  2  1) =    0.000000000000000
( 6  0 |  3  1) =   -0.001923027401919
( 6  0 |  4  1) =    0.001613045439387
( 6  2 |  1  1) =    0.000000000000000
( 6  3 |  1  1) =   -0.184701361441018
( 6  4 |  1  1) =    0.154928467698285
( 6  1 |  2  1) =    0.000000000000000
( 6  1 |  3  1) =   -0.038683534029509
( 6  1 |  4  1) =    0.032447950603007
( 6  2 |  2  0) =    0.009336058122476
( 6  2 |  3  0) =    0.000000000000000
( 6  2 |  4  0) =    0.000000000000000
( 6  3 |  2  0) =    0.000000000000000
( 6  3 |  3  0) =    0.011813922734513
( 6  3 |  4  0) =   -0.002078445792233
( 6  4 |  2  0) =    0.000000000000000
( 6  4 |  3  0) =   -0.002078445792233
( 6  4 |  4  0) =    0.011079469320272
( 6  0 |  2  2) =    0.054104219459936
( 6  0 |  3  2) =    0.000000000000000
( 6  0 |  3  3) =    0.054291571211162
( 6  0 |  4  2) =    0.000000000000000
( 6  0 |  4  3) =   -0.000157151628507
( 6  0 |  4  4) =    0.054236039067604
( 6  2 |  2  1) =    0.054125156914812
( 6  2 |  3  1) =    0.000000000000000
( 6  2 |  4  1) =    0.000000000000000
( 6  3 |  2  1) =    0.000000000000000
( 6  3 |  3  1) =    0.075855681518665
( 6  3 |  4  1) =   -0.018227677656995
( 6  4 |  2  1) =    0.000000000000000
( 6  4 |  3  1) =   -0.018227677656995
( 6  4 |  4  1) =    0.069414628237947
( 6  1 |  2  2) =    0.306655599347507
( 6  1 |  3  2) =    0.000000000000000
( 6  1 |  3  3) =    0.315738293535417
( 6  1 |  4  2) =    0.000000000000000
( 6  1 |  4  3) =   -0.007618611374202
( 6  1 |  4  4) =    0.313046130402530
( 6  2 |  2  2) =    0.000000000000000
( 6  2 |  3  2) =   -0.008064485483452
( 6  2 |  3  3) =    0.000000000000000
( 6  2 |  4  2) =    0.006764532589140
( 6  2 |  4  3) =    0.000000000000000
( 6  2 |  4  4) =    0.000000000000000
( 6  3 |  2  2) =   -0.176295759644369
( 6  3 |  3  2) =    0.000000000000000
( 6  3 |  3  3) =   -0.197638924560966
( 6  3 |  4  2) =    0.000000000000000
( 6  3 |  4  3) =    0.011138225736852
( 6  3 |  4  4) =   -0.179964436324773
( 6  4 |  2  2) =    0.147877804962089
( 6  4 |  3  2) =    0.000000000000000
( 6  4 |  3  3) =    0.152251498109802
( 6  4 |  4  2) =    0.000000000000000
( 6  4 |  4  3) =   -0.011733162163856
( 6  4 |  4  4) =    0.164484175321776
( 6  5 |  0  0) =    0.154983826095369
( 6  0 |  5  0) =    0.006178562861131
( 6  5 |  1  0) =    0.036324481156051
( 6  1 |  5  0) =    0.018637075964466
( 6  0 |  5  1) =    0.018637075964466
( 6  5 |  1  1) =    0.138899129503383
( 6  1 |  5  1) =    0.120424410513661
( 6  5 |  2  0) =    0.000000000000000
( 6  5 |  3  0) =    0.000000000000000
( 6  5 |  4  0) =    0.002735070642527
( 6  2 |  5  0) =    0.000000000000000
( 6  3 |  5  0) =   -0.009851358272803
( 6  4 |  5  0) =    0.009063088333113
( 6  0 |  5  2) =    0.000000000000000
( 6  0 |  5  3) =    0.009851358272803
( 6  0 |  5  4) =    0.009063088333113
( 6  5 |  2  1) =    0.000000000000000
( 6  5 |  3  1) =    0.000000000000000
( 6  5 |  4  1) =    0.024748716021254
( 6  2 |  5  1) =    0.000000000000000
( 6  3 |  5  1) =   -0.059473510026833
( 6  4 |  5  1) =    0.070735331310826
( 6  1 |  5  2) =    0.000000000000000
( 6  1 |  5  3) =    0.059473510026833
( 6  1 |  5  4) =    0.070735331310826
( 6  5 |  2  2) =    0.135540657848790
( 6  5 |  3  2) =    0.000000000000000
( 6  5 |  3  3) =    0.138097677983239
( 6  5 |  4  2) =    0.000000000000000
( 6  5 |  4  3) =    0.000000000000000
( 6  5 |  4  4) =    0.142908608378693
( 6  2 |  5  2) =    0.016764912164332
( 6  2 |  5  3) =    0.000000000000000
( 6  2 |  5  4) =    0.000000000000000
( 6  3 |  5  2) =    0.000000000000000
( 6  3 |  5  3) =   -0.013036508898613
( 6  3 |  5  4) =   -0.034131389240033
( 6  4 |  5  2) =    0.000000000000000
( 6  4 |  5  3) =    0.034131389240033
( 6  4 |  5  4) =    0.057682230376614
( 6  5 |  5  0) =    0.007800931718220
( 6  0 |  5  5) =    0.025610632874342
( 6  5 |  5  1) =    0.065308066611221
( 6  1 |  5  5) =    0.196819899854417
( 6  5 |  5  2) =    0.000000000000000
( 6  5 |  5  3) =    0.040452941573134
( 6  5 |  5  4) =    0.044287849234219
( 6  2 |  5  5) =    0.000000000000000
( 6  3 |  5  5) =   -0.085489074048849
( 6  4 |  5  5) =    0.126882792323867
( 6  5 |  5  5) =    0.133977082851046
( 6  6 |  0  0) =    0.517125360977304
( 6  0 |  6  0) =    0.006270522616106
( 6  6 |  1  0) =    0.122013556886528
( 6  1 |  6  0) =    0.019208705902818
( 6  6 |  1  1) =    0.491873027324534
( 6  1 |  6  1) =    0.146642827474200
( 6  6 |  2  0) =    0.000000000000000
( 6  6 |  3  0) =   -0.009188483135799
( 6  6 |  4  0) =    0.007707347696814
( 6  2 |  6  0) =    0.000000000000000
( 6  3 |  6  0) =   -0.011128525245716
( 6  4 |  6  0) =    0.009334665162233
( 6  6 |  2  1) =    0.000000000000000
( 6  6 |  3  1) =   -0.095810960370799
( 6  6 |  4  1) =    0.080366734512074
( 6  2 |  6  1) =    0.000000000000000
( 6  3 |  6  1) =   -0.105474819021288
( 6  4 |  6  1) =    0.088472829676139
( 6  6 |  2  2) =    0.471913115835107
( 6  6 |  3  2) =    0.000000000000000
( 6  6 |  3  3) =    0.506704669305872
( 6  6 |  4  2) =    0.000000000000000
( 6  6 |  4  3) =   -0.029183336960891
( 6  6 |  4  4) =    0.496392251464638
( 6  2 |  6  2) =    0.021048752318672
( 6  3 |  6  2) =    0.000000000000000
( 6  3 |  6  3) =    0.096048971720199
( 6  4 |  6  2) =    0.000000000000000
( 6  4 |  6  3) =   -0.062910576176908
( 6  4 |  6  4) =    0.073818472547383
( 6  6 |  5  0) =    0.025610632874342
( 6  5 |  6  0) =    0.007800931718220
( 6  6 |  5  1) =    0.196819899854417
( 6  5 |  6  1) =    0.065308066611221
( 6  6 |  5  2) =    0.000000000000000
( 6  6 |  5  3) =    0.085489074048849
( 6  6 |  5  4) =    0.126882792323867
( 6  5 |  6  2) =    0.000000000000000
( 6  5 |  6  3) =   -0.040452941573134
( 6  5 |  6  4) =    0.044287849234219
( 6  6 |  5  5) =    0.342541413419711
( 6  5 |  6  5) =    0.035634211559360
( 6  6 |  6  0) =    0.026686622886860
( 6  6 |  6  1) =    0.289333860384054
( 6  6 |  6  2) =    0.000000000000000
( 6  6 |  6  3) =   -0.237637271677700
( 6  6 |  6  4) =    0.199331385983227
( 6  6 |  6  5) =    0.133977082851046
( 6  6 |  6  6) =    0.774605943919898

There are 406 unique integrals


Why does thallium hydroxide increase the yield of product in a Suzuki reaction?

The Suzuki coupling reaction (also called Suzuki-Miyaura coupling reactions; Ref.1) is the coupling of an aryl or vinyl boronic acid with an aryl or vinyl halide or triflate using a palladium(0) catalyst similar to Heck reaction and Negishi reactions in mechanistic aspects. In particular, Negishi reaction uses organozinc reagents instead of organoboronic acids. Both reactions are powerful cross-coupling methods that allows the synthesis of conjugated olefins, styrenes, and biphenyls. The mechanism of Suzuki coupling reaction is depicted below:

The palladium catalyzed mechanism begins with oxidative addition of the organohalide ($\ce{R^1 -X}$) to the $\ce{Pd(0)}$ (e.g., $\ce{Pd(0)L_n}$) to form a $\ce{Pd(II)}$ complex (denoted in mechanism as $\ce{R^1 -Pd(II)L2X}$). It is worth nothing that this first step in both of the Suzuki and Negishi cross-coupling reactions is identical to that of the Heck reaction. This complex undergoes halide ligand exchange with the base present in the reaction mixture.

This base is present to serve one or two possible roles: Either to react with the organoboron reagent (denoted as $\ce{R^2 -B(OH)2}$) to form a trialkoxyboronate (denoted as $\ce{^-B(OH)3R^2}$), which then attacks the palladium(II) halide complex (denoted as $\ce{R^1 -Pd(II)L2X}$), and/or to convert of the palladium(II) halide complex to a palladium(II) oxo complex (denoted as $\ce{R^1 -Pd(II)L2OH}$) that reacts with the neutral organoboron reagent ($\ce{R^2 -B(OH)2}$) (Ref.1 & 2).

Transmetalation with the organoboron reagent then follows where the $\ce{R^2}$ group of $\ce{R^2 -B(OH)2}$ replaces the hydroxide anion on the palladium(II) complex to give $\ce{R^1R^2 -Pd(II)L2}$ and boric acid. Keep in mind that organoboron compounds are highly covalent in character, and may not undergo transmetallation readily in the absence of a base. Therefore, the reaction of the base with the organoboron reagent, $\ce{R^2 -B(OH)2}$ to form $\ce{^-B(OH)3R^2}$ is very likely. Also note that transmetalation step is highly depend on the base availability and the types of ligand on the catalyst (Ref 1 & 3).

Reductive elimination then follows to give the final coupled product regenerating the catalyst, hence the catalytic cycle can begin again.

Based on this mechanism, availability of a base is very important to reaction to proceed. It is clear that AP's question does not indicate the solvent used in the reaction. However, mentioned reference has been using THF as the solvent. Since solubility of $\ce{KOH}$ in THF is very low and hence its availability to the reaction. Thallium(I) hydroxide is strong base compatible with alkali bases such as $\ce{KOH}$ and key to its capability of enhancing the reaction here should be its readily availability in such solvents. This is evidence that in recent publication (Ref.4) where $\ce{KOH}$ and $\ce{TlOH}$ are equally effective in $\ce{Pd(0)}$  catalyzed coupling reaction, yield of which increases when solvent has been changed from THF to DMF.

References:

Norio Miyaura, Akira Suzuki, "Palladium-Catalyzed Cross-Coupling Reactions of Organoboron Compounds," Chem. Rev. 1995, 95(7), 2457-2483 (https://doi.org/10.1021/cr00039a007).
Karl Matos, John A. Soderquist, "Alkylboranes in the Suzuki−Miyaura Coupling:  Stereochemical and Mechanistic Studies," J. Org. Chem. 1998, 63(3), 461-470 (https://doi.org/10.1021/jo971681s).
Norio Miyaura, Kinji Yamada, Hiroshi Suginome, Akira Suzuki, "Novel and convenient method for the stereo- and regiospecific synthesis of conjugated alkadienes and alkenynes via the palladium-catalyzed cross-coupling reaction of 1-alkenylboranes with bromoalkenes and bromoalkynes," J. Am. Chem. Soc. 1985, 107(4), 972-980 (https://doi.org/10.1021/ja00290a037).
Hiroko Koyama, Hisashi Doi, Masaaki Suzuki, "Evaluation of $\ce{TlOH}$ Effect for $\ce{Pd^0}$-Mediated Cross-Coupling of Methyl Iodide and Excess Boronic Acid Ester toward Fabrication of $\ce{[^{11}C]CH3}$-Incorporated PET Tracer," Intl. J. Org. Chem. 2013, 3(3), 220-223 (DOI: 10.4236/ijoc.2013.33030).


Configuration Interaction matrix composition from electronic configurations

Given those six configurations, the CI matrix with be a six-by-six matrix with rows and columns defined by the respective configuration. The matrix elements themselves will be the Hamiltonian expressed in those configurations.

For example, say column one corresponds to $\psi_0$, or $\left|0\right>$ in bra-ket notation. Similarly, row one corresponds to $\psi_0$, or $\left<0\right|$ in bra-ket notation. The CI matrix element where these coincide, namely $\left<0|H|0\right>$, where H is your Hamiltonian, is simply the ground state energy (the Hartree-Fock energy, if the configurations are the result of HF orbitals). The other matrix elements, like $\left<1|H|0\right>$ or $\left<0|H|2\right>$ have rules for determining them. The best reference is the book Modern Quantum Chemistry, by Szabo and Ostlund. Dr. David Sherrill has a good reference as well, which you can find here: http://vergil.chemistry.gatech.edu/notes/ci/ci.html

Okay, to answer the second part of your question. There are a few rules for determining CI matrix elements. One of the first ones is that singly excited configurations do not mix with the ground state. This is Brillouin's theorem. I don't see any singly excited configurations in the six you gave, so we won't have any of those. The other rule is that configurations that differ by more than two orbital occupations are zero. Since you truncated to doubles, you also won't have this case. The other thing you need to check is spin. If spin is not conserved, you'll have a zero matrix element. I think you'll have this case with the first configuration and the last one, which have different multiplicities.

As to your last question, I'm not real sure what you are asking. MP4(SDQ) is a perturbative method, which is different from CI, which is a variational method. In other words you do a CI by diagonalization, whereas MP4 is not. So I think CISDQ would be a better question...anyway, the point is that you are increasing your configuration space from singles and doubles (SD) to singles, doubles and quadruples (SDQ). The point I made in the preceding paragraph still holds, and expressing your Hamiltonian between ground state and quadruply excited configurations will be zero because their occupation differs by more than two. Same for singly excited configurations and quadruples.

EDIT: Just had the thought, you should also search for "Slater's rules" when trying to construct CI matrices. These are the rules I mentioned earlier.


VSEPR theory and hybridization in determining the shape of a molecule

The VSEPR theory can indeed be used to predict geometry a priori. You would count the number of substituents around a central atom, counting any lone pairs as substituents also, and then find an arrangement that maximises the distance between all substituents. This theory works well for elements of the first period and compounds in which the central element is in the highest oxidation state, but tends to fail very rapidly for all others, most notably because it cannot account for anything outside of the traditional 2-electron-2-centre bond and because it considers all lone pairs equal even though they are typically not.

The hybridisation theory can only be used to predict geometry in a very limited number of cases that involve carbon atoms, and it only works because carbon is extremely regular. However, using hybridisation to predict carbon geometry is basically circling back onto VSEPR: you check how many atoms are bound to a certain carbon and thereby deduce whether this carbon atom will be $\mathrm{sp,sp^2}$ or $\mathrm{sp^3}$. Carbocations and carbanions can also be accomodated for (by either counting the anionic lone pair or ignoring it — as in VSEPR).

In general, however, it is not possible to predict the geometry following a certain hybridisation. Rather, one determines the geometry by an independent mean (e.g. quantum chemical calculations, gas phase diffraction), deduces the bond angles (and bond lengths) from the geometry and then derives a hybridisation. In short: hybridisation follows geometry, not the other way around.


Is the decomposition reaction of nitrogen monoxide to dinitrogen and dioxygen possible?

You should look up enthalpy of formation and enthalpy of reaction in your textbook. Also have a look at (Gibbs) Free energy change on reaction. This will tell you  if the reaction is going to be spontaneous, as it is in this case.
The thermodynamic energies never tell you how fast the reaction will be as these relate only to the starting (reactant) and ending (product) energies. The activation energy, the barrier between reactants and product is unknown to thermodynamics.
Secondly, a catalyst will not change the thermodynamics of the reaction, the starting and ending states are the same, but it speeds up a reaction by allowing the reaction to proceed by an alternative pathway with a smaller activation barrier.


Why can Grignard's Reagent be stored in ether?

The free nucleophile $\ce{R-}$ never really exists in a Grignard reagent. That's why we write it $\ce{RMgX}$. Ethers are used as a solvent as the oxygen coordinates to the $\ce{Mg}$, improving solubility and stability.

Regarding your mentioned substitution reaction, it probably would happen if a Grignard reagent was left long enough. There's a reason you make Grignard reagents immediately before use. They don't last if left on the shelf!

But Grignard reagents are usually reacted with carbonyl compounds. These are much more electrophilic than ethers, because the oxygen atom is pulling electrons away from one carbón atom instead of spreading its pull between 2, and because a double $\ce{C=O}$ bond is inherently weaker than two $\ce{C-O}$ bonds. Hence the desired reaction happens much faster than the undesired one.


Do alkylhydroxylamines form N-nitrosamines?

Yes they can in acidic solutions according to the abstract of this paper$^{[1]}$:

  O-Alkylhydroxylamines react with nitrous acid with a rate law $v = k_2\ce{[R^1NH2^+OR^2][HNO2]}$; the mechanism involves electrophilic nitrosation at the nitrogen atom of the free base. At acid concentrations below $\pu{2M}$-per-chloric acid, hydroxylamine and N-methylhydroxylamine react with nitrous acid by an acid-catalysed mechanism with a rate law $v = k_3\ce{[RNH2^+OH][HNO2]}h_\circ$; this involves initial nitrosation at the oxygen atom of the conjugate acid, followed by migration of the nitrosonium ion to the nitrogen. At higher acidities the O-nitrosation can be reversed, and the migration of the nitrosonium ion becomes the rate-determining stage.

Reference

T. D. B. Morgan, G. Stedman, M. N. Hughes, "Kinetics and mechanism of the reaction between nitrous acid and hydroxylamine," J. Chem. Soc. B 1968, 344-349 (DOI: https://doi.org/10.1039/J29680000344).


Can running water carry an electrical current indefinitely, and for long distances?

I assume that the fundamental problem with this idea is that due to inherent resistance of the water, heat will be generated. As a result the water will evaporate. If there is no space left for the gaseous water vapor the "pipeline" will just explode.

That is basically why conducting wires are made of copper, it mirrors a remarkably low resistance. Silver and gold would be even better but more expensive on the other hand. Consequently copper is the standard material of wires.

If constant cooling of the system would be permitted, it could basically work.
water itself does not significantly conduct electricity, but with electrolytes dissolved in the mentioned (as suggested by the author of this question) it becomes conducting.
As long as the ions do not precipitate the liquid retains the property of conductance.


How does addition of impurities in a solid decrease the melting point?

one of the most prevalent (and simplest) model which explains this fact is colligative freezing point depression.

Freezing (or melting) point is temperature at which solid and liquid phase are in equilibrium, and their vapor pressures are equal. Adding (non-volatile) impurity in solid phase lowers the vapor pressure of solid phase (because you lower the "concentration" of your compound in solid phase), and phase equilibrium can be achieved at lower temperature.

Difference between melting points for pure and impure compounds can be quantified with following equation:
$$ \Delta T = i K_f b $$
where $\Delta T$ is melting point difference, $i$ is Van't Hoff factor which tells you on how many species does the impurity dissociate, $K_f$ is cryoscopic constant which tells you how great the effect is, and $b$ is molality of the impurity.

Basically, all impurities lower the melting point, if they do not interact strongly with the given compound. In some cases (i.e. urea/choline chloride) freezing point depression can be in order of 100 °C! Organic compounds in particular have pretty high cryoscopic constants, which can cause headaches in organic synthesis (in most cases, you end up with some kind of oil which has to be purified to obtain solid product).

Elevation of freezing point is possible only if pressure is decreased, but the effect is so minuscule that it is not practically useful.


What is the mechanism of formation of alkali metal superoxide?

Getting to know the superoxide anion radical
Hayyan et al. (2016) have studied and characterized the formation of the superoxide anion radical, $\ce{O2^.-}$. This chemical entity forms as a result of the reduction of diatomic oxygen
$\ce{O2 + e^- -> O2^.-}$
There is some conflicting data on the reactivity of this radical, however it can be isolated:

A few studies have reported that $\ce{O2^.-}$ is relatively reactive. In contrast, a previous study showed that $\ce{O2^.-}$ or hyperoxide, as some scientists name it, is not highly reactive despite being a free radical. The 17-electron $\ce{O2^.-}$ is one of the isolable paramagnetic main-group ions and one of the many intermediate ORR species.

This chemical species can be synthesized in a myriad of ways. It is actually an intermediary in the mitochondrial respiratory chain and it plays a really important role in the activity of respiratory and immune defense systems.

Superoxide can be generated not only enzimatically (as it has been previosuly mentioned), but also electrochemically, photochemically and photocatalitically. Usually, it is either hydrogen peroxide or the peroxy radical used as target-molecules for the synthesis of this species.
Carrying out the synthesis of superoxide photochemically - a mechanism based on solvated electrons
As mentioned earlier, crossing the boundary between diatomic oxygen and superoxide can be done in mutliple ways. Photochemically, this leads to the formation of solvated electrons in the reaction mixture. One electron can attack one oxygen molecule, forming superoxide. Holroyd et al. have concluded in 1978 that photochemical composition of water is the main process leading to the formation of $\ce{e^-(H2O)n}$. The hydroxyl anion $\ce{OH-}$ that forms after the primary dissociation can absorb energy and disproportionate into hydroxyl radical $\ce{OH^.}$ and a solvated electron.
Chemically generating the species
Superoxides of alkali metals have been heavily documented, as well as their synthetic pathways. These alkali cations (such as potassium, sodium and even rubidium and cesium are referred to as superoxide carriers in this context). Zhdanov et al. (2004) have studied the kinetics of synthesis of potassium superoxide. The chemical environment in which this reaction was carried out is alkaline solution of hydrogen peroxide.
$\ce{H2O2 + HO- -> HO2- + H2O}$
Apart from the hydroperoxide species (which involves proton exchange between hydrogen peroxide and the hydroxide species), there has been observed an exchange leading to the formation of an intermediate adduct:

Being a weak acid, hydrogen peroxide enters, under certain conditions, into an exchange reaction with potassium hydroxide to give an intermediate adduct $\ce{K2O2*H2O}$ (potassium peroxide diperoxohydrate).

$\ce{KOH + 3H2O2 -> K2O2*2H2O2 + 2H2O}$
It is this intermediary that disproprotionates upon mild heating into the superoxide species:
$\ce{K2O2*2H2O2 ->[40^o C] 2KO2 + 2H2O}$
This is the most widely used method for obtaining these salts. It is different from the photochemical/electrochemical methods, since the superoxide radical is only generated in situ (if, at all).

Recently, an in situ method was proposed for preparing superoxides by integrating $\ce{H2O2}$ with aqueous $\ce{KOH}$ or sodium hydroxide ($\ce{NaOH}$). $\ce{NaO2}$ was generated by mixing $\ce{NaOH}$ and $\ce{H2O2}$ in aqueous media under ambient conditions (eq 3). $\ce{NaO2}$ is generated by the rapid formation of alkali peroxide, followed by its decomposition in excess $\ce{H2O2}$. However, this method did not provide alkali superoxides in a powder form. By contrast, it was stated that, in this in situ method, $\ce{O2^.-}$ can be generated in solutions for diverse applications.

Conclusions
Two different mechanistic pathways are known for the synthesis of superoxide-based alkali metal salts. One of them involves directly the $\ce{O2^.-}$ species (obtained electrochemically, enzimatically or photochemically) which gets to react with the alkali metal's exposed surface.

Sodium metal, coated in a mixture of superoxide, peroxide, oxide, hydroxide and possibly carbonates
The other mechanism involves the disproportionation of the alkali peroxide-hydrogen peroxide adduct, freeing the superoxide species in the reaction mixture (apologies, but I have found no concrete mechanistic details for this rather bizarre disproportionation).
References

https://en.wikipedia.org/wiki/Sodium_superoxide
Hayyan m., Hashim M.A., Al Nashef I.M., Superoxide Ion: Generation and Chemical Implications, Chem. Rev., 2016, 116 (5), pp 3029–3085.
Sawyer, D. T. Superoxide Chemistry, McGraw-Hill,
D. V. Zhdanov, M. A. Ul’yanova & Yu. A. Ferapontov, A Study of the Kinetics of Synthesis of Potassium Superoxide from an Alkaline Solution of Hydrogen Peroxide, Russian Journal of Applied Chemistry
Richard A. Holroyd and Benon H. J. Bielski, Photochemical Generation of Superoxide Radicals in Aqueous Solutions, Contribution from the Chemistry Department, Brookhaven National Laboratory, Upton, New York 11 973, Received March 9, I978


Why does an acid exist when dissolved in water?

Your doubts are legitimates, but there are some aspects that you should take in account:
1-Not all acid are fully dissociated
Not all acids fully dissociate, so if you take acetic acid you have in fact an equilibrium with some acetic acid molecules not dissociated. So your question holds only for strong acid like sulfuric acid you mentioned.
2-You are using pseudo-Arrhenius definition, why don't you use Lewis acid definition?
If you think the acid as a 'substance that that releases $H^{+}$' this is still an acid even if it dissociate, and even here you should notice that acidity is linked with $H^{+}$. However if you want to dig deeper there are many definitions of acid if you use Lewis acid theory and so you think the acid as a specie that accepts electrons you can easily see that in fact that the acidity is linked mainly with $H^{+}$!


Well, I think, in principle, depending on the energy of electrons, different events can happen, but in the classical experiment electrons will diffract from the spacing within crystal lattice of gold in the foil.

P.S. The experiment is actually very well known in the history of science due to the following reason (quoted from here).

  It is interesting to recall that G.P. Thompson, who shared the 1937
  Nobel Prize with Davisson for these experiments which proved that
  electrons are waves, is the son of J.J. Thompson who received the
  Nobel Prize in 1906 for proving that cathode rays were actually
  particles - electrons! And the amazing thing is that they were both
  right.


Are there colorful metals besides gold and copper?

Osmium has a bluish-gray tint.
Well; slightly.
Cesium is silvery-golden!,
But don't wear it.


What is an example of a very low temperature endothermic reaction?

Yes there are. In Vogels famous textbook on practical organic chemistry there is a section about the cooling effect of certain salt (mixture) solutions in water.

  If ice is temporarily not available, advantage may be taken of the
  cooling effect attending the solution of certain salts or salt
  mixtures in water. Thus a mixture produced by dissolving 1 part of
  NH4C1 and 1 part of NaNO3 in 1-2 parts of water causes a reduction in
  temperature from 10 to — 15 °C to — 20 °C; 3 parts of NH4C1 in 10
  parts of water from 13 to -15°C; 11 parts of Na2S2O3.5H2O in 10 parts
  of water from 11 to - 8 °C; and 3 parts of NH4NO3 in 5 parts of water
  from 13 to — 13°C

(Taken from Vogels 5th edition, p. 70)


Purpose of electrolyte in a galvanic cell

Unless I'm badly misunderstanding your question, I think your confusion arises from an error in the balanced chemical equation you wrote.  As you have it, a salt is reacting with a strong acid to make another strong acid and a new salt. This reaction may occur spontaneously as written (I haven't figured its Gibb's free energy) , but it's not the oxidation-reduction reaction that drives a galvanic cell.  In fact, it's not an oxidation reduction at all, it's some sort of acid-base reaction, and all of the reactants and products remain in solution anyway.

The actual chemistry that occurs in this system is between the zinc anode and the acid.

$\ce{Zn + 2 HCl -> ZnCl2 + H2}$

From this it is much clearer why the net ionic equation is as you wrote.  The zinc sulfate could be replaced with zinc chloride with almost no effect on the battery's potential.  In fact, the zinc solution exists only to create standard conditions for the cell and to carry electrical current through the solution - if it were replaced by a nonreactive electrolyte solution the cell would still function, albeit at a different (higher) potential.

You might then ask what the purpose of the hydrogen cathode is.  Again, it's to create standard state.  If you did this in reality, you would be using some sort of inert electrode at the hydrogen end to carry the electrical current into the acid solution and likely would just vent the hydrogen to the surroundings.

In short, this galvanic cell does not function because of a reaction between its electrolytes, nor because of a reaction between its electrodes.  Instead it is ultimately a reaction between the anode and the cathode's electrolyte.  The anode's electrolyte and cathode are required for electrical continuity and are used in examples of this sort to put the cell in standard state.  Beyond that they don't serve much of a purpose.


Brønsted Acid has been used in Friedel-Crafts alkylation before (for example, Ref.1). According to this reference, when toluene use as the solvent, the F-C alkylation reaction of toluene is fruitful for tertiary and secondary alkyl bromides, tosylates, and alkenes (which give both carbocations upon protonation). However, the only primary bromide, which worked in this alkylation are benzyl and cinnamyl bromides. Other active primary bromides such as allyl and crotonyl bromides gave ditolyl products (e.g., crotyl group substituted first, then underwent double bond protonation followed by hydride shift to give secondary benzyl carbocation, which then reacted with the second toluene nucleus to give 1,1-ditolylbutane). Keep in mind that these reactions worked only with activated phenyl nucleus. For example, chlorobenzene gave only trace of products even with higher temperatures.
Later, this reaction is successfully used in Tandem-Friedel-Crafts alkylation reactions to synthesize hitherto unknown perylene derivatives (Ref.2):

References:

Mathew P. D. Mahindaratne, Kandatege Wimalasena, "Detailed Characterization of p-Toluenesulfonic Acid Monohydrate as a Convenient, Recoverable, Safe, and Selective Catalyst for Alkylation of the Aromatic Nucleus," J. Org. Chem. 1998, 63(9), 2858–2866 (https://doi.org/10.1021/jo971832r).
Mark A. Penick, Mathew P. D. Mahindaratne, Robert D. Gutierrez, Terrill D. Smith, Edward R. T. Tiekink, George R. Negrete, "Tandem Friedel–Crafts Annulation to Novel Perylene Analogues," J. Org. Chem. 2008, 73(16), 6378–6381 (https://doi.org/10.1021/jo800558c)(PDF).


How can one say pressure is constant in thermodynamics?

Sometimes in thermodynamics we wish to constrain pressure to be constant.  [Or we may wish to do a thermodynamic analysis for a system that is at constant pressure, which is a common laboratory condition.]   However, that is a choice.  Pressure is not generally constrained to be constant in thermodynamics.
For a closed system:
$$d\text{U} = \text{đ}q + \text{đ}w$$
[Note that I'm using a different sign convention from you for work, but the results come out the same.]
If the only type of work is $pV$- work, then  $\text{đ}w = -p_{ext}dV$, where $p_{ext}$ is the external pressure applied to the system by the surroundings.  Thus we have:
$$d\text{U} = \text{đ}q - p_{ext}dV,$$
where the pressure can change in any way you please during the process.
If the process is reversible, then (in that idealized case) the system is in continous equilibrium with its surroundings. As as a consequence, $p_{sys} =p* = p_{ext}$, and we can write:
$$d\text{U} = \text{đ}q - pdV$$
[*The convention is that if a subscript is not present, the variable is assumed to apply to the system.]


Work done by irreversible expansion of gas against piston of fixed pressure

In an irreversible expansion like this, the aveage pressure of the gas within the cylinder is greater than the pressure acting at the internal piston face $P_{ext} $ (where the work is being performed).  Of course, at the internal piston face, by Newton's 3rd law, the pressure exerted by the gas on the piston face is equal to the pressure exerted by the piston face on the gas.

The reason that, in an irreversible expansion, the average pressure of the gas (averaged over the volume) is greater than the value at the piston faces is that the gas pressure is non-uniform spatially within the cylinder.  This is the result both of mass times acceleration effects within the gas itself (yes, the gas has mass) as well as viscous stresses present within the gas that are proportional, not to the gas volume, but to the rate at which the gas is deforming locally within the cylinder.  So the mechanical behavior of a gas experiencing an irreversible expansion is very different from that of a gas experiencing a reversible (quasi-static) expansion.

Now let's reconsider the force balance on the piston.  If $P_{gas}(t)$ represents the instantaneous pressure exerted by the gas on the piston at time t, then the force balance on the piston becomes:
$$M\frac{dv}{dt}=P_{gas}(t)A-Mg$$If we multiply both sides of this equation by v=dx/dt and integrate with respect to t (from time = 0 when the piston is released), we obtain:  $$\int_0^{x(t)}{P_{gas}Adx}=W(t)=\frac{1}{2}Mv^2-\frac{Mg}{A}[V(t)-V(0)]$$where W(t) is the work done by the gas on the piston up to time t, v is the velocity of the piston at time t, and V(t) is the volume of the gas at time t.

Basically what will happen here at short times is that the piston will rise and gain velocity, but it will overshoot its equilibrium position (like a mass on a spring experiencing simple harmonic motion).  It will then slow down, until it begins moving downward.  But it will again overshoot the equilibrium position.

If the piston were frictionless and the gas were inviscid, this oscillation would continue forever.  However, even if the piston were frictionless, the gas is not inviscid.  So the viscosity of the gas will act as a damping factor, and the piston, rather than oscillating forever, will experience a damped oscillation until if finally comes to a stop at the equilibrium position (i.e., when v = 0).  So the equation for the work will become:  $$W(\infty)=\frac{Mg}{A}(V_{\infty}-V(0))$$
So, even for a piston with mass, the kinetic energy effect will die out and the work will be the same as just the change in potential energy of the piston.


How do we know the natural abundance of isotopes on Earth?

Because we can sample the environment with great precision
You seem to assume that we can't know the exact makeup of a sample unless we already know the components. This is wrong. Mass spectroscopy can reliably measure the exact mass of every isotope in a sample and the relative abundance of each. If, for example, a sample of sea-water is measured, we can reliably tell the exact proportion of all the hydrogen isotopes and all the oxygen isotopes in it.
Given that his is possible, estimating the true natural abundance is a matter of statistical sampling. Ocean and river water is fairly well mixed so this is easier than it sounds (at least when there are no processes that radically alter the isotopic composition). So we have a lot of samples, for example, from different oceans and different locations in those oceans. Given that, we can be fairly confident that the isotopic composition doesn't vary a lot and, therefore, we can get a good estimate of the "natural" abundance of each isotope.
But the techniques we use are precise enough to notice that there are some (small) variations in that composition. Some processes do (very slightly) alter the isotopic ratios. Evaporation tends to happen slightly faster for lighter isotopes, for example, as do some biological processes. Some of those processes are temperature dependent leading to the ability to identify the ambient temperature preset when some fossils were alive by measuring the ratio of oxygen isotopes in them.
The key point is that we do have tools that can measure isotope ratios very precisely. So we know the ratios in any sample very precisely. With a very large number of samples, we can generate a reliable picture of the typical abundance of each isotope which is what we refer to as the natural abundance.


How dangerous is cobalt(II) chloride used in silica gel balls?

Cobalt poisoning is apparently very rare. The only situation in which cobalt poisoning is a serious problem is when it is exposed directly to your bloodstream (Recently it has been an issue with people who have undergone certain hip implants -- see this page).
The amount of Cobalt chloride in silica gel balls would be absolutely harmless. According to most sources, you have to be exposed to significant amounts of cobalt (at least more than 20g for an average adult) for a long period of time for symptoms to appear.
Note: Cobalt is actually necessary for our body (in small amounts) in the form of Vitamin B12.


Are homemade batteries economically practical?

No: trying to catch up with 100+ years of technological optimizations in your kitchen won't be an economical enterprise.  Store-brought batteries are standardized and relatively inexpensive.  Power from an outlet is even less expensive (by one or two orders of magnitude).

On the other hand,
in many places you can save on electricity by going solar. Even if you can't install solar panels on your roof because you don't own a house, you could buy solar chargers for small appliances if you have a sunny corner in your apartment...


Ring formation via oxymercuration-demercuration

Observe that there is no nucleophile give in the mercuration step and recall that the three-membered mercury ring intermediate is very reactive. Hence, this is suggestive that the oxygen would attack the intermediate with its lone pair on the more substituted position as follows:

Edit: Oops! Missed a carbon on the last step as correctly pointed out by @Waylander. I have updated the answer with a new image that corrects the mistake.

For reference see  Addition reactions of Alkene's in David Kleins organic chemistry (around page-410 ) and for a quick reference see wiki


Calculation of silver nanoparticle size using Mie Theory

Plug it in to the math equation
Ok, so we have this equation: $γ(R)=γ_0+(Av_f)/R$
First, we substitute the values into the equation that we know.
$γ(R) = (5\cdot10^{12}) + (3/4\cdot1.39\cdot10^6)/R$. Simplified, this is:
$γ(R) = (5\cdot10^{12}) + (1.0425 \cdot 10^6)/R$
$γ(R)$ can be measured in your experiment. Then we can solve your equation for $R...$
$\large{R = \fbox{$\frac{1.0425 \cdot 10^6}{γ(R) - 5\cdot10^{12}}$}}$
Plug in your value of $γ(R)$, and you're there!*

*In other words, how should I use the λmax obtained through the experiment in the provided equation?
The equation only mentions $\gamma (R)$, I don't know how to help you there**

**Zhe: @JavaScriptCoder For these types of questions, the first thing to ask is "do you work in a lab?"
If not, why are you doing experiments outside of a lab, and if so, why is it that no one in your lab has any expertise on this?


How do water insoluble medications get absorbed in the blood stream?

Well, firstly the capsule itself is not entirely insoluble. You can see this for yourself - if you drop one into water, it will start to dissolve very slowly and the water will turn cloudy. This process usually goes faster in the stomach, which has a very acidic environment, or the small intestine which is more alkaline. The capsule material is designed to dissolve when it's eaten, hence releasing the drug - if it didn't dissolve, the whole capsule would just pass through your digestive system and come out... at the other end.

On to the issue about the drug itself. In general designing a drug is not very simple at all, because it has to survive the very same enzymes that break down our food - if the enzymes break down the drug then it becomes useless. For example, if you tried to eat insulin it wouldn't do you any good - insulin is a protein and will be immediately digested by protease enzymes. If the drug survives those enzymes, it has to be able to diffuse through the lining of the small intestine in order for it to enter the bloodstream. The cell membrane, being a non-polar lipid bilayer, doesn't easily let polar or charged molecules diffuse across it. So normally drugs have to be somewhat lipophilic, which is a fancy way of saying non-polar. But at the same time, as you rightly note, they also have to be water-soluble. Most drugs try to strike a balance between the two.

One way that drugs are absorbed is by reversible protonation. Many drugs have amine functional groups with a $\text{p}K_{\text{aH}}$ of around 6-8. At the physiological pH of 7.4, an equilibrium will exist in which both the unprotonated and protonated forms exist in significant amounts. The unprotonated form can cross cell membranes more easily, and the protonated form is more soluble in water.

There are other drugs that don't diffuse across cell membranes very well, due to their large polarity. They may be absorbed via different means, such as active transport (in which the cell uses energy in order to let the molecule enter), or facilitated diffusion (which is the same as diffusion, except that a protein helps out). An example is levodopa; it is structurally similar to the amino acid phenylalanine and the transport protein for phenylalanine can't quite tell the difference. Being polar molecules, these drugs are already water-soluble.

In fact, that's only half the story, because different drugs have different targets. Some need to go to the brain and for this to happen, they have to cross the blood-brain barrier. Some need to stay in the intestines to kill off bacteria - so you have to make them really polar so that they can't diffuse into the bloodstream. Absorption is only a small part of a drug's journey in the human body. If you are interested in finding out more, look for a textbook on medicinal chemistry and read the section about pharmacokinetics - it should have all the info you want.


Understanding Moseley’s law from the Rydberg-type equation?

The concept to understand here is that most theories in physics and chemistry are obtained by observing then theorizing, this results in the theories and models being made in such a way that they can explain most (if not all) observable phenomenon.

So, when Bohr made his model of the atom, it had to explain the the phenomenon of "X-ray emission" otherwise it wouldn't be accepted, because, X-ray emission is an observed phenomenon of atoms. Following on, it was discovered that the Bohr-model couldn't explain all observable phenomenon as equipment got better and more things could be seen (notably the splitting of spectral lines in magnetic fields). So any theory that followed on to explain this splitting of spectral lines had to explain X-ray emission as well, without disturbing the results of Bohr's theory.

A suitable analogy would be the discovery of special relativity by Einstein, whatever discovered had to explain why Newton's Law had stuck around for hundred's of years without people realizing it was wrong. (And it did, the correction at velocities we know are less than one part in a million and that's why we didn't think Newton was wrong)

So the deeper question is, even a one part in a million error is still an error and so we should know that the theory is wrong even early on... this is not the case as most instrumental errors on their own are much larger in this regard (at least in Newton's time), and the same applies here in that older instruments weren't usually sensitive enough to notice the change brought on by quantum mechanics on X-ray emission to that of Moseley's rough equation. (Though now it is usually possible to detect differences to a good degree.)

So to the final part of your question, there is a horrendous equation waiting if you want to plug in exact values to find the exact wavenumber, but most like the instrument barely notices a difference, so in a sense, it doesn't matter which theory we use. (It's very well possible a hundred years down the road, another genius comes up with an even smaller correction that can be observed only with futuristic devices)


When it rains, it puddles. Spilled salt cycles between wet and dry with humidity. Is this akin to a phase change?

What is this process called exactly? As far as I understand
"deliquescence" is a property and not a process. Does the exchange of
water between my pile of salt and the air have a specific terminology?
Something that would go along with terms like evaporation,
sublimation, condensation, etc.? Perhaps some mash-up of adsorption
and dissolution?

The property of salts to become wet as I mentioned in another post Scientific Reason for Salt Solution Gaining Volume is deliquescence. The corresponding verb is to deliquesce. It is a pretty old word. From OED (not open access):

To melt or become liquid by absorbing moisture from the air, as certain salts.
1756   C. Lucas Ess. Waters i. 14   They attract the humidity of the
air, and deliquesce, or run liquid.
1781   Philos. Trans. 1780 (Royal Soc.) 70 349   This
pot-ash..deliquesces a little in moist air.
1876   D. Page Adv. Text-bk. Geol. (ed. 6) xvi. 299   Pure chloride of
sodium is not liable to deliquesce.*

Regarding phase changes, there were many phases changes involved in that process. The water as gas became liquid. The solid NaCl phase become one phase with a liquid. Once it evaporated, you got the solid again.


How do/did chemists observer/infer the structure of organic molecules and reactions?

Various crystallographic techniques exist, the lab technique I have used most is x-ray crystallography

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1186895/pdf/mp53000008.pdf

Currently I specialize in computational chemistry, which can estimate (with very high accuracy) the geometry of a molecule which gives the lowest energy (aka the equilibrium geoemetry) through a process called geoemetry optimization

https://www.shodor.org/chemviz/optimization/students/background.html


ChemDraw: Automatic Caption with Number for Molecules

There is no fully automatic method for creating labels. However, you can manually generate property labels, which are automatically placed and centred without disturbing any existing alignment, as follows.

Select the entire molecule, right-click the selected structure, then click Analysis, and then click one of the available options (Name, Chemical Formula, Exact Mass, Molecular Weight, m/z, or Elemental Analysis):

Use the Text tool (on the tool bar) to modify the text of the label (after your edit, you will receive the warning “This change to the text has caused auto-updating to be disabled.”):

You can use the Text menu, the Style toolbar, or shortcut keys (e.g. CTRL+B) to change the text format (structure labels are usually written in bold). Repeat this procedure for all relevant structures:


How can we tell whether a phenomena was a chemical reaction or not?

For some chemical reactions, it would be very difficult to prove without chemical analysis of the materials before and after the reaction. However, with modern analysis techniques, you could show whether a reaction took place. For example, if you tempered steel or converted diamond to graphite, analysis would show that it was a physical change or change in the crystal structure that occurred.

For a low-tech example, consider the combustion of some condensed-phase hydrocarbon in oxygen. If you ignite the hydrocarbon under oxygen in a container sealed except for an oxygen source and an exit to a condenser, you would see the hydrocarbon vanish and water would appear in the condenser. For this we'll just ignore carbon dioxide production. You then easily confirm that the water is not just another form of the hydrocarbon by measuring it's freezing and boiling temperatures. You could then repeat the experiment without ignition of the hydrocarbon and see that you get nothing.
This would confirm beyond any reasonable doubt that you've had a chemical reaction that converted the hydrocarbon to water.


Esterification of racemic acid with optically active alcohol

It's a pretty basic concept. Two diastereoisomeric esters will be formed. Now diastereoisomers have different physical and chemical properties just like normal isomers. Hence, obviously, their rate of formation will be different. So, different amount of R and S acid will react

Thus, the third fraction will contain unequal amount of the acids which will show optical activity, and the other two fractions are the two diastereoisomers which are also optically active.


Would there be a difference if deuterium is embedded instead of protium (regular hydrogen) in acids?

Yes, there would be a difference. The acids would be weaker.  Why? Because the heavier isotope lowers the zero-point energy (the lowest possible energy) in the acid molecule, making the acid's bond to the D stronger than it would have been to the H. Acids in $\rm D_2O$ are weaker than they are in ordinary water, too.

Any reaction that involves dissociation of these acids will run more slowly, too. This is called "the kinetic isotope effect". You can read more about that here: http://en.wikipedia.org/wiki/Kinetic_isotope_effect

You might want to ask further questions: why is the zero point energy lower for a bond to D than a bond to H? How exactly does the lower zero point energy make the bond stronger? How exactly does it affect the rates of reactions that include acid dissociation?

Ask away!

If you'd like to see how substituting a D for an H in acids affects the acid dissociation constant for particular acids, here's a good reference:

R. A. Robinson, M. Paabo, R. G. Bates, "Deuterium Isotope Effect on the Dissociation of Weak Acids in Water and Deuterium Oxide", J. Res. Nat. Bureau. Standards, 73A(3), 1969, pp 299-308.


The Big Bang started with about 3/4 hydrogen atoms and 1/4 helium atoms.  Add a pinch of lithium and a whisper of boron.  Supernovae since then have released a lot of helium.  Heavier elements are still rare.

"The most common [molecular] material in the universe" must survive photolysis, radiolysis, and reaction.  Water is poor candidate.

http://www2.astro.psu.edu/users/alex/astro497_7.pdf
What is out there.  CO is the marker.
http://physics.nist.gov/cgi-bin/micro/table5/start.pl
Molecules in space, isotopically and by
http://science.gsfc.nasa.gov/691/cosmicice/interstellar.html
http://www.astro.uni-koeln.de/cdms/molecules
atom count


What is the rationale for calling something as an 'inorganic [organic compound]'?

I would not call graphite organic, but there is no clear-cut way of defining organic and inorganic compounds.
To your question, the pairs of compounds are isoelectronic. That means that if we assume that molecular orbitals (MOs) arise from atomic orbitals, the corresponding MOs in the two compounds have the same occupancy. In these cases it arises because the number of electrons in the dimer BN is the same as in the dimer CC. Another example of isoelectronic compounds would be N$_2$ and CO. Being isoelectronic, they do share some properties, e.g. both N$_2$ and CO have a triple bond. If only the valence electrons are identically occupied, we call them isovalent, which would include e.g. H$_2$O and H$_2$S.
The bonding of benzene and borazine differs in that not all sites are equivalent, but the form of the MOs is broadly similar and the same applies to boron nitride. Similarity in MOs is then reflected in similarity in reactivity. (Even though they are not that similar in reactivity.)


Pencil lead writes on rough surfaces by abrading into tiny particles and getting pushed into surface irregularities. If the surface isn't rough, you would get no visible residue of graphite. A pencil might sometimes even leave a scratch on dense wood that is so smooth that it has almost no blackness from graphite.

Wood is hydrophilic and even contains some water (unless bone dry). Graphite is not so hydrophilic or polar. Acetone wets the wood surface and either swells it by diffusing into watery regions or simply gets between a graphite particle and the wood surface, loosening any van Der Waals forces holding the graphite to the wood. While the surface is wet, the graphite can be wiped off (or smeared out so much that the original mark is not visible any more). Then the acetone evaporates quickly and the wood looks clean(er). Ethanol or isopropanol might do as well - even water (especially with a surfactant to coat and disperse the graphite) might work.

A grease pencil would be much more difficult to remove because so much material is pushed into crevices that a solvent can't easily loosen it without spreading it.


Essentialy yes, this will work for an unsaturated acid. It will work better if you dissolve your acid substrate in a volatile organic solvent (ether or dichloromethane would be good) and stir it efficiently, wash with sodium thiosulfate solution to remove any unreacted iodine, then allow the solvent to evaporate.


This is not really a standalone answer, rather a pointer or an exeption. There are few complexes where deprotonated thiourea $\ce{S=C(NH2)NH}^-$ acts as a bidentate ligand. One example is (aminomethanethioamido)-hydrido-tetrakis(trimethylphosphine)-technetium(III) $\ce{[Tc(P(CH3)3)4(SCN2H3)]+}$ complex[1]:

The fact that thiourea ligand is deprotonated and is coordinated asymmetrically within equatorial plane in $\eta^2-\ce{N,S}$ bonding mode has been proven by NMR and SCXRD studies. From the article [1, p. 4642] (emphasis mine; $\ce{tu}$ = thiourea $\ce{S=C(NH2)2}$):

The $\ce{Tc-N}$ distance, 2.190(13) Å, is long when compared to
known $\ce{Tc(III)}$ complexes. The range for $\ce{Tc-N}$ bonds was found
to be 2.041 Å ... to 2.124 Å ... . This reflects delocalization of negative charge within the bidentate deprotonated thiourea ligand. This elongation is also seen in the $\ce{Tc-S}$ bond, 2.543(4) Å, which is significantly longer than $\ce{Tc-S}$ bonds in the parent compound $\ce{[Tc(\textit{S}-tu)6][Cl]3}$ ... .

As Martin pointed out, the reasons for the observed denticity are not always transparent. Simply put, it seems like in this case in order to be a bidentate ligand, thiourea molecule needs to increase its nucleophilicity and remove steric hindrance caused by $\ce{H}$-atoms lying within the $\sigma_\mathrm{h}$ plane.   Both can be achieved by eliminating a single proton from the thiourea molecule.
Reference

Watson, P. L.; Albanese, J. A.; Calabrese, J. C.; Ovenall, D. W.; Smith, R. G. Inorg. Chem. 1991, 30 (24), 4638–4643. DOI: 10.1021/ic00024a035.


You need to know the oxygen content (partial pressure) of the steam-water mixture with which you are purging the system.

The water being purged will come to equilibrium with the purge gas based on Henry's law, just as in your example of atmospheric conditions, except that the partial pressure of oxygen in the steam should be much lower.


Why does the pH of a weak acid not increase by 1 when diluted by a factor of 10?

In a weak acid  $\ce{HB}$ solution, with a nominal concentration $c$, a tiny amount of its molecules are dissociated into $\ce{H^+}$ and $\ce{B^-}$. Let's call this concentration $[\ce{H^+}]$ = $\ce{[B^-]}$ = $x <<c$, so that the following approximation can be made : $c - x$= $c$. The dissociation equilibrium constant $K_a$ of this weak acid $\ce{HB}$ can be approximated by$$K_a\ce{= \frac {[H^+][B^-]}{$c - x$} = \frac{$x$^2}{$c-x$} = \frac{$x$^2}{$c$}}$$ $$\ce{[H^+] = x = \sqrt{K_ac}}$$ $$\ce{$p$H = \frac{1}{2} (log$K$_a - log $c$)}$$ Look at the coefficient $1/2$ before the logarithm. If the concentration $c$ is multiplied by $10$, the log c increases by $1$, but the $p$H changes by $1/2$.


1. A pragmatic view
As the comments suggested you can use different functions than slater determinants.
But this would be more or less as bad as an idea
as describing a movement in a circle without using polar coordinates.
If you include the constraints of your system implicitly in the coordinate system or basis of your description, nearly everything gets better.
More elegant equations, faster convergence... .
Determinants are a fast and elegant way to include constraints like the Pauli Principle in your description.
2. A bit more mathematical view
Let's go back to the definition of a Determinant in general.
From Wikipedia:

[...] Another way to define the determinant is expressed in terms of the columns of the matrix.  If we write an $n \times n$ matrix $A$ in terms of its column vectors
$$A = \begin{bmatrix} a_1 & a_2 & \cdots & a_n \end{bmatrix}$$
where the $a_j$ are vectors of size $n$, then the determinant of $A$
is defined so that
$$
\begin{align}
  \det \begin{bmatrix} a_1 & \cdots & b a_j + c v & \cdots & a_n \end{bmatrix}
          &= b\det(A) + c \det \begin{bmatrix} a_1 & \cdots & v & \cdots & a_n \end{bmatrix} \\
  \det \begin{bmatrix} a_1 & \cdots & a_j & a_{j+1} & \cdots & a_n \end{bmatrix}
          &= -\det \begin{bmatrix} a_1 & \cdots & a_{j+1} & a_j & \cdots & a_n \end{bmatrix} \\
  \det(I) &= 1
\end{align}
$$
[...]

The last constraint comes from the normalization of the atomic orbitals and the second constraint is just the Pauli-Principle.
To the best of my knowledge you don't have to force the linearity in the first constraint, but if you do, you end up with a determinant.


Comparing the Lewis acidity of carbon tetrafluoride and silicon tetrafluoride

Both molecules can theoretically act as a Lewis acid. This would involve the approach of a Lewis base (i.e. a lone pair-carrying group), its interaction with the central atom and the formation of a four-electron-three-centre bond. These bonds with a bond order of $0.5$ can be understood by invoking the following two resonance structures:

$$\ce{\overset{-}{X}\bond{...}AF3-F <-> X-AF3\bond{...}\overset{-}{F}}$$

Where $\ce{A}$ is carbon or silicon in the context of this question. The structure that this arrangement would require is a trigonal bipyramid; the central atom is $\mathrm{sp^2}$ hybridised with its unhybridised p orbital forming the 4e3c bond to the electronegative groups.

You may immediately recognise this as the transition state of an $\mathrm{S_N2}$ reaction which is possible both with carbon and silicon centres.

These four-electron-three-centre bonds are typically more stable if the central atom has a lower and the terminal atoms have a higher electronegativity. If you compare interhalogen compounds, only iodine is able to form such bonds with chlorine ($\ce{(ICl3)2}$) while all other halogens can only form these bonds with fluorine (e.g. $\ce{ClF3}$). This already should allow us to conclude that $\ce{SiF4}$ is the stronger Lewis acid. But the discussion is not finished here.

We also need to discuss steric effects. Carbon tetrahalides are very crowded molecules — even with fluorine even though it is the smallest halogen. Giving up the tetrahedral structure of $\ce{CF4}$ to arrive at trigonal bipyramidal $\ce{CF4X-}$ would require the three fluorines that do not take part in the 4e3c bond to move closer together, increasing steric crowing even more. Indeed, the required interaction would be the Lewis base’s lone pair with $\sigma_{\ce{C-F}}^*$ — an orbital which is already well hidden by the three other fluorines. Thus, steric effects further reduce $\ce{CF4}$’s ability to be a Lewis acid to the point where we shouldn’t even be saying it be a weak Lewis acid: we should just call it Lewis inactive.

Silicon is larger than carbon and better able to stabilise 4e3c bonds due to its lower electronegativity. Thus, 4e3c bonds are sufficiently accessable and stable for silicon in $\ce{SiF4}$ to act as a Lewis acid. Depending on the nature of $\ce{X-}$, the pentacoordinated $\ce{SiF4X-}$ species may either break down under liberation of either $\ce{F-}$ ($\mathrm{S_NSi}$ substitution) or $\ce{X-}$ (reverse reaction) or — especially if $\ce{X} = \ce{F}$ — it may add another to give $\ce{[SiF4X2]^2-}$ like the well-known hexafluoridosilicate anion $\ce{[SiF6]^2-}$.


When I checked the SDS of acetanilide on Sigma-Aldrich website, I saw that it is harmful to aquatic life.
So, there is no exception whether it is an alga or fish to kill. I suggest you not use acetanilide in the aquarium.


KCN is a ionic compound consisting of $\ce{K+}$ and $\ce{CN-}$. You can know that the cyanide ion will have a negative charge, because a potassium ion always has a positive. Also the elements in the cyanide ion are more electronegative - that means they will have the negative oxidation number.

So we know that since potassium has an oxidation state of +1, the cyanide ion has an overall oxidation state of -1. As stated above, ions have oxidation states corresponding to their charge. But you want to know what the oxidation state of the NITROGEN atom has.

There are rules you can use for polyatomic ions (ions with multiple atoms in it), such as $\ce{CN-}$. Nitrogen is more electronegative than carbon. Thus it will have the negative oxidation state (think: it's better at grabbing electrons, which are negatively charged). It is 3 columns away from being a noble gas, thus its oxidation state it -3. The overall oxidation state of the ion is -1, thus carbon must have an oxidation state of +2.

To sum up: K = +1, C = +2, N = -3. If you're trying to balance an equation though, you don't need to look at C and N individually. You just look at the $\ce{CN-}$ as a whole and the $\ce{K+}$.

How did I know $\ce{CN-}$ was an ion? Since potassium is a metal, and only forms ionic bonds. C and N do not form ionic bonds, they bond covalently, thus they act as a single ion in ionic compounds. Does that make sense? Other than that it's just something you learn to recognize with time!

Read rules for determining oxidation states here:
http://www.tmcleod.org/cgi-bin/chem1/wiki.cgi?action=browse&diff=3&id=OxidationNumber

There's a comment to this answer explaining what causes electronegativity. A general rule of thumb is that electronegativity increases as you go towards higher group number and low period number. You can study it for yourself in the picture below:

(source: wikispaces.com)


Why enol content of ethyl acetoacetate is more in toluene than water?

The extent of enolization is affected by the solvent, concentration and temperature. Ethyl acetoacetate has enol content of 0.4% in water and 19.8% in toluene. This is because water reduces the enol content by hydrogen bonding with the carbonyl group making this group less available for intramolecular hydrogen bonding. The effectiveness of intramolecular hydrogen bonding in stabilizing the enol, with respect to the keto form is thus seen on varying the solvent and particularly on transfer to a hydroxylic solvent.


Green Potassium Manganate and Sodium Hydroxide Equation

$$\ce{ 3K2MnO4 + 2NaOH + 2H2O -> MnO2 +2NaMnO4 + 6KOH }$$

As you said the brown solid A is $\ce{MnO2}$ while the purple solution is due to the permanganate ion .


Can someone walk me through this gas mixture question?

Solve for the amount of gas in A: $n_A=p_AV_A/RT=\pu{0.727 mol}$. You'll need to perform necessary unit conversions or use R in units that match the given T, p and V units.
Write the volume of chamber B in terms of knowns: $V_B=n_BRT/p_B=n_B \times\pu{4.13\times 10^3 m^3/mol}$.
Write the total volume when mixed: $V_{tot}=n_{tot}RT/p_{tot}=n_{tot} \times\pu{6.89\times 10^3 m^3/mol}$.
Solve this last expression for $n_B$ after inserting the equalities $V_{tot}=V_A+V_B$, $n_{tot}=n_A+n_B$  and the result of step 2:
$$\begin{align} V_A+V_B &=(n_A+n_B)\times \pu{6.89\times 10^{-3} m^3/mol}\\ &=V_A + n_B \times\pu{4.13\times 10^{-3} m^3/mol}\\\rightarrow n_B &= \frac{V_A - n_A\times \pu{6.89\times 10^{-3} m^3/mol}}{\pu{2.75\times 10^{-3} m^3/mol}} = \pu{1.45 mol}\end{align} $$
Compute the volume in chamber B: $V_B=\frac{n_{B}RT}{p_B}=\pu{6.00\times 10^{-3} m^3}$

It's usually a good idea to perform a check to see if we did things right. Compute back the final pressure: $p_{tot}=\frac{n_{tot}RT}{V_{tot}}=\pu{3.60 bar}$.

Answer: $V_B = \pu{6.00 L}$


What would be the physical interpretation of the equation dG = Vdp − SdT?

For a continuous flow system operating at steady state in contact with a constant temperature reservoir at the same temperature as that of the entering fluid, and with negligible change in kinetic energy and gravitational potential energy, the reversible shaft work per mole that the system does on the surrounding is equal to the integral of -vdP, where v is the fluid molar volume.  So $\Delta G$ per mole is equal to minus the shaft work per mole.


So basically this problem is bad and incomplete (they tend to be at this level) in that it doesn't give you the concentrations/volumes of the reagents/solutions, so we end up making several assumptions. First of all, precipitates cannot be aqueous, by definition they are solids. So the way you end up solving this question is by assuming one of the products precipitates out completely, and it is the least soluble one that precipitates out (barium sulfide in this case). Unfortunately, this isn't exactly accurate.

In a real world scenario, given enough water you could have no precipitate at all, but most likely you would have a mix (not all of the barium sulfide would precipitate out of solution). At higher levels of chemistry (AP/IB -> post-secondary) you'll learn that none of these reactions (including the solubility ones) ever go to completion (some just go so far that you can assume they go to completion), and what you end up doing is using solubility constants to find out how much of each product is dissolved and/or precipitated. If you're interested in more info google "solubility constants" and "common ion effect".


How to calculate the volume or mass of carbon dioxide gas absorbed by a calcium hydroxide solution?

(a) It consumes $\pu{4.4 L}$ of carbon dioxide gas

(wrong)

You've correctly calculated the volume of carbon dioxide $V(\ce{CO2})$ via the molar volume of ideal gas $V_\mathrm{m} = \pu{24.790 L mol-3}$ at SATP (Standard Ambient Temperature and Pressure):

$$V(\ce{CO2}) = nV_\mathrm{m} = \pu{0.1 mol}\cdot\pu{24.790 L mol-3} \approx \pu{2.5 L}$$

I personally wouldn't recommend to remember molar volumes at STP, NTP, SATP etc. as the numbers are close, but still different, and it brings up a lot of confusion.
Instead, only remember the ideal gas constant $R$, and remember it well as it's being used ubiquitously.
Pretending you forgot the molar volume for these conditions, you can use ideal gas law directly:

$$V(\ce{CO2}) = \frac{nRT}{p} = \frac{\pu{0.1 mol}\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}}{\pu{1e5 Pa}} \approx\pu{2.5 L}$$

Here, keep in mind that $\pu{1 Pa} = \pu{1 J m-3}$ and $\pu{1 m3} = \pu{1e3 L}$.

  (b) It produces $\pu{24.5 L}$ of liquid water

(wrong)

Liquid water and SATP both imply that the density of water $ρ(\ce{H2O})\approx\pu{1 g mL-1}$, and its volume can be found via molecular mass $M(\ce{H2O})$:

$$V(\ce{H2O}) = \frac{m(\ce{H2O})}{ρ(\ce{H2O})} = \frac{nM(\ce{H2O})}{ρ(\ce{H2O})} = \frac{\pu{0.1 mol}\cdot\pu{18.02 g mol-1}}{\pu{1 g mL-1}} \approx \pu{1.8 mL}$$

  (c) It consumes $\pu{0.5 mol}$ of carbon dioxide

(wrong)

The simplest out of four.
As you've already ruled out, there are equimolar amounts of both reactants and products according to the stoichiometry, so $n(\ce{CO2}) = n(\ce{Ca(OH)2} = \pu{0.1 mol})$:

$$\ce{Ca(OH)2 (aq) + CO2 (g) -> CaCO3 (s) + H2O (l)}$$

  (d) It produces $\pu{10.0 g}$ of solid calcium carbonate

(correct)

Mass of calcium carbonate $m(\ce{CaCO3})$ can be found trivially from its molar mass $M(\ce{CaCO3})$:

$$m(\ce{CaCO3}) = nM(\ce{CaCO3}) = \pu{0.1 mol}\cdot\pu{100.09 g mol-1} \approx \pu{10.0 g}$$


Does the benzene ring in styrene show a negative inductive effect on the ethenyl group?

This can be explained by looking at the molecular orbitals. The molecular LUMO of benzene (which accepts electrons) is closer to the HOMO of ethylene (which given electrons) as compared to the large gap between HOMO of benzene and LUMO of ethylene.

Surely, the HOMO, LUMO don't exist separately and only exist as a molecular orbital. But the separation makes for easy transfer of electrons from ethylene to benzene unlike the opposite.
Because I have involved pi orbitals into the solution, I believe mesomeric effect is more apt here. To talk about inductive effect we have to involve a more complex idea (group electronegativities). Resources for such data are not easily accessible. I would appreciate if someone can find the group electronegativities of benzene and ethylene and post them here for the analysis of inductive effect as well.
Surely, if present, a soft nucleophile will it will attack the 2-carbon on ethylene side-group.


How is enthalpy change equal to heat supplied at constant volume?

No the book is not completely wrong.If you read the next 6 lines then you will find that it wants to point that this approximately true only for solids and liquids, not for gases.

From the First Law,

$$\Delta U = q + w$$
and since work done is $0$ in an isochoric process (constant $V$),

$$\Delta U = q_V$$

Furthermore,

$$\begin{align}
\Delta H &= \Delta U + \Delta (pV) \\
&= q_V + \Delta (pV)
\end{align}$$

Since we are considering solids and liquids, the changes in the volume of these matter upon changing pressure are negligible (as they are incompressible matter) when compared to gases, so the $\Delta (pV)$ term can be considered negligible.

Therefore

$$\boxed{\Delta H \approx \Delta U = q_V}$$

for these incompressible solids and liquids. However this fails when we consider gases as the term $\Delta (pV)$ will no longer be negligible.

Comparison between the $\Delta (pV)$ values for different states of matter, being heated from $20\ ^\circ\mathrm{C}$ to $30\ ^\circ\mathrm{C}$:

$$\begin{array}{c|c}
\text{Substance} & \Delta (pV)\text{ / J} \\
\hline
\text{Air} & 2850 \\
\text{Water} & 0.1 \\
\text{Iron} & 0.0004
\end{array}$$


Amorphous to crystalline: Opposite trend?

What they say in your book is right. Take the case of polyethylene terephthalate (PET) which is a kind of polyester used in water bottles for example. It's an amorphous polymer (remember the bottles are transparent). When it is heated, the strained chains have the possibility to move and reorder themselves and the polymer becomes partially crystalline. This fact can readily be proved experimentally by heating a piece of PET (over its glass transition temperature, around $74 ^0\mathrm{C}$), then quenching the sample. It becomes white and not transparent which means it's partially crystalline.
We can monitor this transformation using a specific instrument called differential scanning calorimeter, where we can notice the formation of an exothermic peak corresponding to crystallization.


How to separate copper(II) chloride from 4-iodophthalic anhydride?

I just realized I can add sodium hydroxide, and filter off the resulting insoluble copper(II) hydroxide and sodium chloride, then boil down the result. If you can think of a better answer, feel free to suggest it to me.


What are the important implications to the field of chemistry, if any, of the recently confirmed hexavalent carbon species?

The carbon is not  hexavalent, it is hexacoordinated. A covalent bond does not equal to a total of two electrons between the bonding partners and the nature of  the chemical bond may lie somewhere between totally covalent and totally ionic.
Examples of this include boranes, with their three-centre-two-electron bonds. But we don't need to stop there; any π coordination has fewer electrons in the bonding orbitals than double the bonding partners.

We struggle understanding the bonding in simple molecules like carbon monoxide, while more complex molecules like many organic molecules have all similar bonding situations which are quite easy to grasp. The concept of aromaticity is still not fully understood and while we believe it is as simple as drawing a MO diagram, it most certainly is not. However, molecular orbitals certainly enhance our understanding.

Chemist in general are interested in unusual bonding situations, since they challenge our understanding of the bonding and the chemistry of such molecules itself.
One of the prominent examples is the 2-norbonyl cation, the structure of which remained a mystery for a long time since it didn't fit within the common constraint of organic chemistry. Such ions are nowadays usually referred to as non-classical ions. Their bonding is different from the more common two-electron-two-centre bonds we expect in organic (and inorganic) molecules. In a first approximation they can be described with resonance (see here: What is resonance, and are resonance structures real?), but the understanding comes with a few misconceptions. A MO description involves multi-centre bonds and typically bond orders less than one.
Another interesting example of unusual bonding are fluxional molecules like bullvalene. (See also here: What is the conformer distribution in monosubstituted fluoro bullvalene?) Similarly the bonding situation in these molecules is quite fluid, which allow them to change shape in such a way and at room temperature we obtain a single signal in the proton NMR (Addison Ault. J. Chem. Educ. 2001, 78 (7), 924-927.).

As such the hexamethylbenzene dication is another representative of unusual bonding situations. While it has been prepared more than forty years ago, it took until now to confirm the actual structure. It is special since it contains a pyramidal coordinated carbon exclusively bonded to other carbons, and it marks the highest observed coordination number for carbon so far. The authors actually state the general motivation behind such approaches in the first two sentences:

  The tetravalency of carbon and the hexagonal-planar ring structure of benzene are fundamental axioms of organic chemistry, and were developed 150 years ago by Kekulé. Chemists have long been fascinated by finding exceptions from these rules.

Let's go a little deeper, and look at the questions you are asking step by step:

  So, I understand how the octet rule is not actually violated. But then, can these molecules really contain carbons with 6 primarily covalently bound carbon neighbors?

The covalent-to-ionic character of a bonding orbital is completely independent of whether it is occupied or not. I believe there still is no unilaterally accepted criterion for what constitutes a covalent bond and what does an ionic bond.
As previously stated, a bond does quite often deviate from the traditional Lewis-concept. This is to be expected as the Lewis concept is quite crude and cannot account for many chemical phenomena.
We can analyse the electron density with the Quantum Theory of Atoms in Molecules, which defines us a set of parameters on which we can judge whether a bond is predominantly covalent of ionic. First and foremost we can obtain bond paths, i.e. these are paths that connect nuclei in a way that the value of the electron density is maximum. On this path we will find a bond critical point, where the electron density is minimal. The Laplacian at this point gives us a measure whether the bond is predominantly covalent (negative) or ionic (positive). In the picture below areas of positive Laplacian are marked with solid lines, i.e. charge depletion, and negative values with dashed lines, i.e. charge accumulation. The location of the plane is given in the 3D model. The hexa-coordinated carbon is C10.

        rho/a.u.    Lapl./a.u.
BCP44   0.259       -0.811
BCP47   0.294       -0.121
BCP53   0.157        0.226
BCP61   0.270       -0.882
BCP64   0.288       -0.117

From the pure visual inspection we can see that the area between C10 and C12 clearly is a predominantly covalent bond, which is obvious from the dashed lines. This is also supported by the numbers.
Visual inspection of C10 to C4 would suggest the same. However, the bond critical point lies outside of the charge accumulation area. This might well be an issue with the methodology. I have reproduced the calculation from the paper only on their geometry. Therefore the optimum MP2 density might be slightly different and completely describes the bond as covalent. Also note that there are no symmetry restrictions, which could also lead to slight variations. From what we can see, we can derive that there is a significant covalent contribution though.
The values of the electron density (rho) also show that the pyramidal bonds are only about half as strong as the bonds in the 1-ethylium-1-ylidene or the five-membered ring.
(Fun fact: The AIM analysis shows no cage critical point. From a technical point of view, the five-membered ring is no ring. The whole molecule is rather a goblet (of fire, lol).)

  In what orbitals do all these electrons lie?

Since it is a cation of benzene, the atomic orbitals involved in bonding are the same as in regular benzene. The total number of molecular orbitals that are formed are one fewer than in the neutral case. Obviously due to the different geometric arrangement they combine also differently.
The canonical orbitals are actually quite pretty and reproduce the (almost) symmetry quite nicely as some of the orbitals are nearly degenerate (for example HOMO and HOMO-1, or LUMO and LUMO+1), but have a look for yourself:

  Should we not be surprised as similar hypervalency occurs elsewhere on the periodic table?

Well, no. Hypercoordinate bonding is one of the most common patterns for anything outside of carbon chemistry.

A few words of advice: It has nothing to do with hypervalency. The term is an ancient relic from the times where people believed that d-orbitals are necessary to describe molecules that exceed the coordination number dictated by Lewis theory. The gold book writes on hypervalency:

  The ability of an atom in a molecular entity to expand its valence shell beyond the limits of the Lewis octet rule. Hypervalent compounds are common for the second and subsequent row elements in groups 15–18 of the periodic table. A description of the hypervalent bonding implies a transfer of the electrons from the central (hypervalent) atom to the nonbonding molecular orbitals which it forms with (usually more electronegative) ligands.

Note that this orbital expansion is not necessary when considering MO theory. As such, the term valency, and especially hypervalency, should be avoided at all cost.
It is important to know that according to the gold book, valence is an absolute property of an element, it does not change:

  The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.

Carbon is tetravalent. Always. (Seriously, from now on never use *valen{ce .. cy .. t} again, please.)

  Is this really a breakthrough of science or one of semantics and scientific media hype?

Of course it is. It is always something incredible when we refine our understanding of the universe, challenge our own theories. Is it as ground breaking as the discovery of electricity; probably not. However, with such advances in synthetic chemistry we can obtain more robust data to develop our theoretical models. We can in general expand our knowledge of bonding, think outside the box.
With advances like this we achieve the complete break down of our theories; without such we would probably still be using Lewis structures (for everything, and that would be ridiculous).


Why is the empirical formula of sodium chloride NaCl?

The empirical formula of, say, $\ce{P4O10}$ is $\ce{P2O5}$ but that doesn't mean that each phosphorus is bonded to only 2.5 oxygens. It's perfectly possible (and in fact it is pretty much a requirement) that if Na has 6 nearest-neighbour Cl and Cl has 6 nearest-neighbour Na, then the overall stoichiometry is NaCl.

Let's say, for example, that there are $n$ sodium atoms in a sample of NaCl. Then, if each $\ce{Na+}$ atom has 6 nearest-neighbour $\ce{Cl-}$ ions, then at first glance we might say there are $6n$ chloride ions.

However, since each $\ce{Cl-}$ ion has 6 nearest-neighbour $\ce{Na+}$ ions, this means that in the earlier calculation we counted each $\ce{Cl-}$ ion 6 times (once per $\ce{Na+}$ neighbour). So in fact there are only $6n/6 = n$ chloride ions, leading to a stoichiometry of $1:1$.

Maybe if this is too complicated for you, you can reduce it to one dimension. Consider the following (infinite) arrangement of red and green trees along a road:

Surely you'd agree that each red tree has two green neighbours, and each green tree has two red neighbours. Yet the "stoichiometry" is obviously $1:1$.

Now if you tile that in three dimensions, you get the $\ce{NaCl}$ structure.


Why does tomato soup "explode" when boiled?

I think it's probably a combination of things you mention. First, tomato soup is a suspension of solids (tomato tissue) in water. So it's the water that's boiling. That water is turning into steam, which bubbles out of the water. The solid suspension likely traps some of that steam until the pressure builds and you get a big explosive bubble.

I happen to cook, and make a fair bit of tomato sauce, which is very similar to tomato soup. I disagree with your statement that tomato soup explodes only once upon boiling. I see bubbles popping quite continuously, and you have to stir often to prevent this. Which happens to also be the way you prevent bumping in laboratory conditions. Cooking is probably the most comparable chemistry application most people are exposed to.


What is the name of cyclopropane with a double bond?

Ethylidenecyclopropane

 Vinylcyclopropane (or Ethenylcyclopropane)


Why is CCSD(T)/aug-cc-pVTZ not a suitable method for the geometry optimisation of omeprazole?

To transform the question into something simpler, let's assume CISD instead CCSD(T), which is cheaper and share the common expensive step.

The molecule has 182 electrons and 1545 basis functions in the basis you specified, that means approx. 1400 virtual orbitals.

Therefore, the number of double excitation $ij \rightarrow ab$ is approximately $180 \cdot 180 \cdot 1400 \cdot 1400 = 32400 \cdot 1960000 = 63504\cdot10^6$

Unfortunately, in each of this excitation you have to calculate integral of type $(ab|cd)$, and there are $1400^4$ of them.

Such a big problem is out of reach for traditional CCSD, but tractable with more approximate methods.


How good is aromaticity at stabilization?

The situation in S and P is very similar.
Essentially you have a negative charge $\alpha$ to two double bonds which will delocalize these charges.
In the case of cyclopentadiene, this will be stabilized because the anion is aromatic.
But still, there will be a charge which is supported by carbon atoms.

In the case of the 1,3-diketone, the charge will also be delocalized by two double bonds, but instead of an aromatic anion of five members, the charge will be supported by only three atoms.
However, two of them are oxygens which can hold a negative charge much better than a carbon.

The electronegativity of carbon is 2.55 while for oxygen it is 3.44.
Note that the electronegativity scale ranges from 0.7 for $\ce{Fr}$ to 4.0 for $\ce{F}$.
This ability of supporting negative charge is much bigger than the small contribution of aromaticity to the stabilization of a carbanion.