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Anton [14]
3 years ago
15

What amount of pure acid must be added to 600 mL of a 40% acid solution to produce a 60% acid solution?

Chemistry
1 answer:
julia-pushkina [17]3 years ago
5 0

Answer:

You need to add 300mL of a pure acid to produce a 60% acid solution.

Explanation:

Pure acid is a 100% acid. If you suppose X is the volume addition of the pure acid it is possible to write:

100%X + 600×40% = (600+X)60% (M₁V₁=M₂V₂)

100X + 24000 = 36000 + 60x

40X = 12000

X = 300

That means <em>you need to add 300mL of a pure acid to produce a 60% acid solution.</em>

<em></em>

I hope it helps!

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A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m
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0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

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\frac{8.3333}{0.01225 mol}=\frac{2}{x}

x = 2.9 ≈ 3

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MCl_3\rightarrow M^{3+}+Cl^-

Formulas for the oxide and sulfate of M will be:

Oxide of M is M_2O_3 and sulfate of M_2(SO_4)_3.

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