Answer:
Reduction of Aldehydes and Ketones. Hydride reacts with the carbonyl group, C=O, in aldehydes or ketones to give alcohols. ... Reduction of ketones gives secondary alcohols. The acidic work-up converts an intermediate metal alkoxide salt into the desired alcohol via a simple acid base reaction.
The carbon atom of a carboxyl group is in a relatively high oxidation state. Diborane, B2H6, reduces the carboxyl group in a similar fashion. ... Sodium borohydride, NaBH4, does not reduce carboxylic acids; however, hydrogen gas is liberated and salts of the acid are formed.
Primary alcohols can be oxidized to form aldehydes and carboxylic acids; secondary alcohols can be oxidized to give ketones. Tertiary alcohols, in contrast, cannot be oxidized without breaking the molecule's C–C bonds.
A secondary alcohol can be oxidised into a ketone using acidified potassium dichromate and heating under reflux. The orange-red dichromate ion, Cr2O72−, is reduced to the green Cr3+ ion. This reaction was once used in an alcohol breath test.
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Answer: 52.17%
Explanation:
Number of matches in the box = 46
Number of matches that lit on the first strike = 22
Number of matches that did not light on the first strike = 46 - 22 = 24
Therefore, the percentage of the matches in the box did not light on the first strike will be:
= (Number of matches that did not light on the first strike / Number of matches in the box) × 100
= 24/46 × 100
= 52.17%
Therefore, the percentage of the matches in the box that did not light on the first strike is 52.17%.
When a change in PH = 10^-ΔPH
so the change = 10^-3.2
change depends on two factor 0.00063 (10^-3.2)and factor 1585 (10^3.2)depending on the way which the change goes.if PH change from PH=0 to PH= 3.2 so the change is decreasing from concentration from 1 mol to 0.00063 and if PH change from PH = 3.2 to PH=0 so the change is increasing by a factor of 1585.
1.78L x (3.00M/1L) = 5.34
