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il63 [147K]
3 years ago
10

If there are 3 moles of hydrogen, 6 moles of oxygen, and 9 moles of helium in a 10 L vessel at 27°C, the total pressure of the g

as
mixture will be:
A. 55 atm
B.
105 atm
C.
18 atm
D. 44 atm
Reset
Submit
Chemistry
1 answer:
harkovskaia [24]3 years ago
4 0

Answer:

The correct answer is D) 44 atm

Explanation:

We apply Dalton's law, where for a gas mixture the total pressure is the sum of the partial pressures of each gas that makes up that mixture. You can add the amount of moles and then calculate the pressure, using the ideal gas formula. We convert the temperature in Celsiud into Kelvin : 0°C= 273K---->

27°C=27 + 273= 300K

PV= (nH+ n0+ nHe)x RT

P=((nH+ n0+ nHe)x RT)/V=

P=((3 mol + 6 mol +9 mol)x 0,082 l atm/ Kmol x300K)/10 L

<em>P=44, 28 atm</em>

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Explanation:

Data given:

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atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

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mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

3 0
3 years ago
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Answer:

Explanation:

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Answer:

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