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Anestetic [448]
3 years ago
13

Calculate the mass percentage of oxygen in dry air

Chemistry
2 answers:
grigory [225]3 years ago
5 0
              <span> (0.20948 x 31.998) / (
(0.78084 x 28.013) +
(0.20948 x 31.998) +
(0.00934 x 39.948) +
(0.000375 x 44.0099) +
(0.00001818 x 20.183) +
(0.00000524 x 4.003) +
(0.000002 x 16.043) +
(0.00000114 x 83.80) +
(0.0000005 x 2.0159) +
(0.0000005 x 44.0128) ) = 0.23140 = 23.140% O by mass </span>
Oxana [17]3 years ago
4 0

Answer:

0.23 = 23% O by mass

Explanation:

Hi! Let's solve this!

We have the percentage data of each component in the air and its masses.

Nitrogen 0.78084 28.013

Oxygen 0.20948 31.998

Argon 0.00934 39.948

Carbon dioxide 0.000375 44.0099

Neon 0.00001818 20.183

Helium 0.00000524 4.003

Methane 0.000002 16.043

Krypton 0.00000114 83.80

Hydrogen 0.0000005 2.0159

Nitrous oxide 0.0000005 44.0128

Now we have to calculate a molar fraction.

(0.20948 * 31.998) = (0.78084 * 28.013)

+ (0.20948 * 31.998) + (0.00934 * 39.948) + (0.000375 * 44.0099) + (0.00001818 * 20.183) + (0.00000524 * 4.003) + (0.000002 * 16.043) + (0.00000114 * 83.80) + (0.0000005 * 2.0159) + ( 0.0000005 * 44.0128) = 0.23 = 23% O by mass

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Explanation:

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Answer:

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-\frac{d[A]}{dt} =kt

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