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Sergio039 [100]
3 years ago
6

______________ is a reaction that causes the growth of a polymer chain. In this reaction, a relatively small molecule is formed

as a result of the polymerization reaction and may involve different monomers as starting or precursor molecules.
a. Pseudo polymerization
b. Simple molecule polymerization
c. Byproduct polymerization
d. Condensation polymerization
Chemistry
1 answer:
WITCHER [35]3 years ago
5 0

Answer:

Condensation polymerization

Explanation:

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A mixture of 50.0 mL of ammonia gas and 60.0 mL of oxygen gas react. If all the gases are at the same temperature and pressure,
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Answer:

72.0 mL of steam is formed.

Explanation:

The reaction is :

4 NH_{3} + 5O_{2} \rightarrow 4NO+6H_{2} O

You can treat coefficient of compounds as amount of volume used.

Therefore for 4 mL of ammonia 5 mL of oxygen is used to form 4 mL of nitric oxide gas and 6 mL of steam.

For 1 mL of ammonia \frac{5}{4} (=1.25) mL of oxygen is used to form \frac{4}{4} (=1) mL of nitric oxide gas and \frac{6}{4} (=1.5) mL of steam.

OR

Just transform the chemical equation by dividing the whole equation by 4 so that the coefficient of NH_{3} become one like this

NH_{3} + \frac{5}{4} O_{2} \rightarrow \frac{4}{4}NO+\frac{6}{4}H_{2} O

We don't know which one will be completely exhausted and which one will be left so we have to consider two cases :

<em>1.  </em><em>Assume ammonia to be completely exhausted</em>

For 50 mL of ammonia 50 \times \frac{5}{4} (= 62.5) mL of oxygen is needed. But we have just 60 mL of oxygen so this assumption is false.

2.  <em>Assume oxygen to be completely exhausted</em>

For 60 mL of oxygen only 60 \times \frac{4}{5} (=48) mL of ammonia is needed. In this case we have sufficient amount of ammonia. So this case is true.

60\times\frac{4}{5}NH_{3} + 60\ O_{2} \rightarrow 60\times\frac{4}{5}NO+60\times\frac{6}{5}H_{2} O\\\\48NH_{3} + 60\ O_{2} \rightarrow 48NO+72H_{2} O

Now we know that during complete reaction 48 mL of ammonia and 60 mL of oxygen is used which will form 60 \times \frac{4}{5} (= 48) mL of nitic oxide gas and 60 \times \frac{6}{5} (= 72) mL of steam.

Therefore <em>72 mL of steam </em>is formed.

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