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irina [24]
4 years ago
5

Which 1-mole sample has the least entropy?

Chemistry
1 answer:
Alenkinab [10]4 years ago
8 0
<span>The correct answer is option 1. Entropy is the degree of disorderliness in a system. Among the phases, solid has the most ordered structure. Also, at lower temperatures the molecules will have less kinetic energy therefore less motion. Thus, bromine in solid form at 266K has the least entropy.</span>
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The tube that stores sperm cells
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Answer:

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What are various ways in which elements are classified?
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If 588 grams of FeS2 is allowed to react with 352 grams of O2 according to the following equation, how many grams of Fe2O3 are p
Rainbow [258]

<u>Answer:</u> The mass of iron (III) oxide  produced is 782.5 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For </u>FeS_2<u> :</u>

Given mass of FeS_2 = 588 g

Molar mass of FeS_2 = 120 g/mol

Putting values in equation 1, we get:

\text{Moles of }FeS_2=\frac{588g}{120g/mol}=4.9mol

  • <u>For </u>O_2<u> :</u>

Given mass of O_2 = 352 g

Molar mass of O_2 = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of }O_2=\frac{352g}{32g/mol}=11mol

The chemical equation for the reaction of FeS_2 and oxygen gas follows:

FeS_2+O_2\rightarrow Fe_2O_3+SO_2

By Stoichiometry of the reaction:

1 mole of FeS_2 reacts with 1 mole of oxygen gas

So, 4.9 moles of FeS_2 will react with = \frac{1}{1}\times 4.9=4.9mol of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, FeS_2 is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of FeS_2 produces 1 mole of iron (III) oxide

So, 4.9 moles of FeS_2 will produce = \frac{1}{1}\times 4.9=4.9moles of iron (III) oxide

Now, calculating the mass of iron (III) oxide  from equation 1, we get:

Molar mass of iron (III) oxide  = 159.7 g/mol

Moles of iron (III) oxide  = 4.9 moles

Putting values in equation 1, we get:

4.9mol=\frac{\text{Mass of iron (III) oxide}}{159.7g/mol}\\\\\text{Mass of iron (III) oxide}=(4.9mol\times 159.7g/mol)=782.5g

Hence, the mass of iron (III) oxide  produced is 782.5 grams

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