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Archy [21]
3 years ago
12

How many molecules of he are contained in a 10.0 l tank at 7.53 atm and 485 k?

Chemistry
1 answer:
ankoles [38]3 years ago
4 0

Answer:

1.126 x 10^22

Explanation:

pV = nRT

7.53 x 10 = n x 8.31 x 485

n = (7.53 x 10) / (8.31 x 485) = 0.0187 moles

M = n x Avogadros number

0.0187 x 6.02 x 10^23 = 1.126 x 10^22

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Sodium Sulfate = Na2(SO4) meaning there are two ions of Na+ in one mole of Sodium Sulfate the M stands for Molarity, defined as Molarity = (moles of solute)/(Liters of solution), So if the Na2SO4 solution is 3.65M that means one Liter of has 3.65 moles of Na2SO4, the stoichiometry of Na2SO4 shows that there would be two Na+ ions in solution for every one Na2SO4.

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Finally we find [Na+] = 2*3.65 = 7.3M

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