SELECT * FROM question_mgmt as q WHERE id=2033 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=69 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
D-glucose reacts with hydroxylamine (NH2OH) to form an oxime because of the presence of aldehydic (-CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH2OH to give an oxime.
But pentaacetate of D-glucose does not react with NH2OH. This is because pentaacetate does not form an open chain structure.
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 |
Comments
Taking Screenshots on your Samsung Galaxy M31s is very easy and quick.
Report a problem on Specifications:
Taking Screenshots on your Samsung Galaxy M31s is very easy and quick.
Report a problem on Specifications:
Taking Screenshots on your Samsung Galaxy M31s is very easy and quick.
Report a problem on Specifications: