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AlexFokin [52]
2 years ago
13

Using the balanced equation for the next few questions: 4 Fe(s) + 30 (9) - 2Fe2O3(s)

Chemistry
1 answer:
stira [4]2 years ago
5 0

Answer:

0.075 moles of iron oxide would be produced by complete reaction of 0.15  moles of iron.

Explanation:

The balanced reaction is:

4 Fe + 3 O₂ → 2 Fe₂O₃

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:

  • Fe: 4 moles
  • O₂: 3 moles
  • Fe₂O₃: 2 moles

You can apply the following rule of three: if by stoichiometry 4 moles of Fe produce 2 moles of Fe₂O₃, 0.15 moles of Fe produce how many moles of Fe₂O₃?

moles of Fe_{2} O_{3} =\frac{0.15 moles of Fe*2 moles of Fe_{2} O_{3}  }{4 moles of Fe}

moles of Fe₂O₃= 0.075

<u><em>0.075 moles of iron oxide would be produced by complete reaction of 0.15  moles of iron.</em></u>

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Answer:

8.68 L is the new volume

Explanation:

You use Boyle's law for this.

P_{1}V_{1}=P_{2}V_{2}

P_{1} = first pressure

P_{2} = second pressure

V_{1} = first volume

V_{2} = second volume

Convert pressure from atm to mmHg (use same units):

5.97 x 760 = 4537.2 -> 4.54 x 10³

...maintain 3 significant figures in calculation, and round as needed...

(4.54 x 10³ mmHg)(2.79 L) = (1460 mmHg)(V_{2})

(4.54 x 10³ mmHg)(2.79 L) / (1460 mmHg) = V_{2} = 8.68 L

Hope this helps :)

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What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances
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The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope                    mass amu        Relative abundance

1                                77.9                     14.4

2                               81.9                     14.3

3                               85.9                      71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9  

% abundance of isotope 1 = 14.4% = \frac{14.4}{100}=0.144

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = \frac{14.3}{100}=0.143

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = \frac{71.3}{100}=0.713

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]

A=84.2amu

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

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