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sdas [7]
3 years ago
9

The half reaction of an oxidation reduction reaction shows that iron gains electrons. what does this electron gain mean for iron

?
A. it is neutralized
B. it is oxidized
C. it is reduced
D. it has dissolved
E. it has precipitated

PLEASE HELP ME ASAP, ITS URGENT
Chemistry
2 answers:
Sidana [21]3 years ago
8 0
Hi,
The answer should be letter C.
chubhunter [2.5K]3 years ago
6 0

Answer: The correct option is C.

Explanation:

Oxidation reactions are a type of chemical reactions in which a substance looses electrons and forms positive charge. The substance undergoing this reaction is being oxidized.

A\rightarrow A^++e^-

Reduction reactions are a type of chemical reactions in which a substance gains electrons and forms negative charge. The substance undergoing this reaction is being reduced.

A^++e^-\rightarrow A

For the half reaction in which iron gains electrons means that iron is getting reduced.

Hence, the correct option is C.

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Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir
guajiro [1.7K]

Answer:

y_{O2} =4.3%

Explanation:

The ethanol combustion reaction is:

C_{2}H_{5} OH+3O_{2}→2CO_{2}+3H_{2}O

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}

Dividing the previous equation by x:

1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles

Calculate the number of moles of CO2 and water considering the same:

0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)

0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)

The total number of moles at the reactor output would be:

N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)

So, the oxygen mole fraction would be:

y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3%

6 0
3 years ago
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3.4752 moles of water

Explanation:

There are 13.84 mole in one cup of water so,

13.84 divided by 4= 3.4725 :)

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Answer:

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Explanation:

First of all, we determine the reaction:

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Initial        x

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