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Leno4ka [110]
3 years ago
14

The molar mass of oxygen is 16.00 g/mol. how many moles of oxygen are in copper oxide?

Chemistry
1 answer:
BabaBlast [244]3 years ago
5 0
 oxygen added to the copper is 3.300 grams. 

<span>89.920 - 83.620 = 3.300 g oxygen </span>

<span>which is 0.2063 moles of oxygen. </span>

<span>3.300 g O x (1 mol O / 16.00 g O) = 0.20625 mol </span>
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Answer:

See Explanation

Explanation:

It is a common observation that a strip of aluminium metal in aqueous copper(II)Sulfate does not show any visible reaction. Aluminium is normally expected to displace copper in solution since it is higher than copper in the electrochemical series.

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7 0
3 years ago
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.

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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.

5 0
3 years ago
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Answer:D

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Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

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