1mol N2 + 3mol H2 → 2mol NH3
<span>Because you are dealing with gases, at the same temperature and pressure you can say: </span>
<span>1volume N2 + 3 volumes H2 → 2 volumes NH3 </span>
<span>OR: 4 volumes reactants = 2 volumes products. </span>
<span>You have 1.9+5.7 = 7.6 volumes reactants, which must give 7.6/2 = 3.8 volumes of NH3 </span>
<span>Answer: 3.8 litres of NH3 will be formed</span>
Answer:
The will be increased by three times of the original volume
Explanation:
The answer is B. I just took the quiz and got it right. Hope it's the same thing for you.
F. <em>None of the above
</em>
<em>No O atoms are present</em> as reacting substances, only O_2 and H_2O molecules.
O_2 + 2H_2O + 2e^(-) → 4OH^(-)
We must use <em>oxidation numbers</em> to decide whether oxygen or water is the substance reduced.
The oxidation number of O changes from 0 in O_2 to -2 in OH^(-).
A decrease in oxidation number is <em>reduction</em>, so O_2 is the substance reduced.
The oxidation number of O is -2 in both H_2O and OH^(-), so water is <em>neither oxidized nor reduced</em>.
Answer:
water is an example of mixture
