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2 years ago

Can anyone explain oxidation, and reduction reactions of:

Chemistry

1 answer:

____ [38]2 years ago

4 0

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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For the galvanic (voltaic) cell Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s) (E°= 0.77 V at 25°C), what is [Fe2+] if [Mn2+] = 0.040 M and

avanturin [10]

Answer:

0.01836 M

Explanation:

Again the reaction equation is;

Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)

E°cell= 0.77 V

Ecell= 0.78 V

[Mn2+] = 0.040 M

[Fe2+] = the unknown

n=2

From Nernst's equation;

Ecell= E°cell- 0.0592/n log Q

0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]

0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]

0.01/ -0.0296= log [Fe2+] /[0.040]

-0.3378= log [Fe2+] /[0.040]

Antilog(-0.3378) = [Fe2+] /[0.040]

0.459= [Fe2+] /[0.040]

[Fe2+] = 0.459 × 0.040

[Fe2+] = 0.01836 M

7 0

2 years ago

The experimental apparatus represented above is used to demonstrate the rates at which gases diffuse. When the cotton balls are

cestrela7 [59]

The correct answer is —-C

5 0

2 years ago

Compounds can be made by combining ________

Anni [7]

Answer:

hfew hv ihrv i

Explanation:

3 0

2 years ago

Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?

chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now we start with 150 grams of Thorium-234

$150 \times \frac{1}{2}=24.1$