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lubasha [3.4K]
3 years ago
5

Find the formula for the compound that contains 72.40% iron and 27.60% oxygen.

Chemistry
1 answer:
Vikki [24]3 years ago
8 0
Oxygen = 16
Iron = 55.8
16 x 27.6% = 4.4 /4 = 1.1
55.8 x 72.4% = 40.4 /4 = 10.1
1 oxygen and 10 iron, so Fe10 O

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The answer for the following problem is mentioned below.

  • <u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

Explanation:

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Initial volume (V_{1}) = 230 ml

Final volume (V_{2}) = 860 ml

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To find:

Final moles (n_{2})

We know;

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where;

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R represents the universal gas constant

T represents the temperature of the gas

So;

    V ∝ n

\frac{V_{1} }{V_{2} } = \frac{n_{1} }{n_{2} }

where,

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(n_{1}) represents the initial  moles of the gas

(n_{2}) represents the final moles of the gas

Substituting the above values;

   \frac{230}{860} = \frac{3.8 * 10^-4}{n_{2} }

  n_{2} = 14.2 × 10^{-4} moles

<u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

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Answer:

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