SELECT * FROM question_mgmt as q WHERE id=2076 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=62 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.
The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D
Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate of formation of D/mol L - 1 min - 1 |
I | 0.1 | 0.1 |
6.0 × 10 - 3 |
II | 0.3 | 0.2 |
7.2 × 10 - 2 |
III | 0.3 | 0.4 |
2.88 × 10 - 1 |
IV | 0.4 | 0.1 |
2.40 × 10 - 2 |
Determine the rate law and the rate constant for the reaction.
Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal' pchloroform' and pacetoneas a function of xacetone. The experimental data observed for different compositions of mixture is.
100 ×xacetone |
0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 |
pacetone /mm Hg |
0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
pchloroform/mm Hg |
632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 |
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