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max2010maxim [7]
3 years ago
14

The change in entropy, Δ????∘rxn , is related to the the change in the number of moles of gas molecules, Δ????gas . Determine th

e change in the moles of gas for each of the reactions and decide if the entropy increases, decreases, or has little or no change. A. K(s)+O2(g) ⟶ KO2(s)
Chemistry
1 answer:
olga nikolaevna [1]3 years ago
5 0

Answer:

The entropy decreases.

Explanation:

The change in entropy in a chemical reaction is lower when the moles of the gas product decrease. In the same way, the change in entropy is higher when the moles of the gas product increase.

For the reaction:

K(s) + O₂(g) ⟶ KO₂(s)

The number of moles of gas reactant is one (1 mole of K and 1 mole of O₂) and there are not moles of gas in product. The change in moles of gas is 1.

As the number of moles of product decrease <em>the entropy decreses.</em>

<em></em>

I hope it helps!

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zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol

The given reaction equation is as follows.

C + 2H_{2} \rightarrow CH_{4}

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

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