Which of the following gases would be most likely to experience ideal behavior at high pressures?

1 answer:

monitta4 years ago

5 0

<span>Which of the following gases would be most likely to experience ideal behavior at high pressures?
a. F2
b. Ne
c. C2H6

Yes. The answer is (b) Ne (Neon). This is because monoatomic gases such as neon do not experience </span><span>intermolecular attractions and thus most likely to be close to ideal gases behavior. Not only that, Neon is a noble gas and is unreactive. </span>

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Choose the number of significant figures indicated. 0.00666

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0.006660 has four significant figures because there is a zero after nonzeroes.

0.00666 only has three significant figures because the zeros before a significant figure are not significant.

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When 50.5 g iron(III) oxide reacts with carbon monoxide, 32.2 g iron is produced. What is the percent yield of the reaction?

xz_007 [3.2K]

Taking into account definition of percent yield, the percent yield for the reaction is 91.17%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Fe₂O₃: 1 mole
CO: 3 moles
Fe: 2 moles
CO₂: 3 moles

The molar mass of the compounds is:

Fe₂O₃: 159.7 g/mole
CO: 28 g/mole
Fe: 55.85 g/mole
CO₂: 44 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Fe₂O₃: 1 mole ×159.7 g/mole= 159.7 grams
CO: 3 moles ×28 g/mole= 84 grams
Fe: 2 moles ×55.85 g/mole= 111.7 grams
CO₂: 3 moles ×44 g/mole= 132 grams

<h3>Mass of Fe formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 159.7 grams of Fe₂O₃ form 111.7 grams of Fe, 50.5 grams of Fe₂O₃ form how much mass of Fe?

$mass of Fe=\frac{50.5 grams of Fe_{2} O_{3} x111.7 grams of Fe}{159.7 grams of Fe_{2} O_{3}}$

<u><em>mass of Fe= 35.32 grams</em></u>

Then, 35.32 grams of Fe can be produced from 50.5 g iron(III) oxide.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

$percent yield=\frac{actual yield}{theorical yield}x100$

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, you know: