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3 years ago

How many moles of hypomanganous acid. H3 MnO4, are contained in 22.912 g?

Chemistry

1 answer:

gregori [183]3 years ago

6 0

Answer:

0.188mol

Explanation:

Using the formula;

mole = mass/molar mass

Molar mass of hypomanganous acid. (H3MnO4) = 1(3) + 55 + 16(4)

= 3 + 55 + 64

= 122g/mol

According to this question, there are 22.912g of H3MnO4

mole = 22.912g ÷ 122g/mol

mole = 0.188mol

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3 years ago

A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling

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The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

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Hence, the concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

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