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Lelu [443]
3 years ago
9

what pressure will be needed to reduce the volume of 77.4 l of helium at 98.0 kpa to a volume of 60.0 l? 76 kpa 126 kpa

Chemistry
1 answer:
nekit [7.7K]3 years ago
6 0
The system follow the Boyle's Law:

P_1\cdot V_1 = P_2\cdot V_2\ \to\ P_2 = \frac{P_1\cdot V_1}{V_2} = \frac{98\ kPa\cdot 77.4\ L}{60.0\ L} = \bf 126.4\ kPa
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Answer:

The number of hydrogen atoms is 4.96x10²⁴.

 

Explanation:

The number of atoms can be found with the following equation:

n = N*\eta_{H}

Where:

N: is the Avogadro's number = 6.022x10²³ atoms/mol

η: is the number of moles of hydrogen

n: is the number of hydrogen atoms

First, we need to find the number of hydrogen moles. The number of moles of CH₄ is:

\eta_{CH_{4}} = \frac{m}{M}

Where:

m: is the mass of methane = 33 g

M: is the molar mass of methane = 16.04 g/mol                      

\eta_{CH_{4}} = \frac{33 g}{16.04 g/mol} = 2.06 mol

Now, since we have 4 hydrogen atoms in 1 mol of methane, the number of moles of hydrogen is:

\eta_{H} = 2.06\: mol\: CH_{4}*4 \frac{mol\: H}{mol \: CH_{4}} = 8.24 mol

Hence, the number of hydrogen atoms is:

n = N*\eta_{H} = 6.022 \cdot 10^{23} \: atoms/mol*8.24 mol = 4.96 \cdot 10^{24} atoms

Therefore, the number of hydrogen atoms is 4.96x10²⁴.

I hope it helps you!      

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Answer:

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