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IrinaK [193]
3 years ago
14

1) What mass of Na2CO3 is required to make 50cc of its seminormal solution?

Chemistry
1 answer:
love history [14]3 years ago
7 0

Answer:

m=1.325gNa_2CO_3

Explanation:

Hello,

In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:

eq-g=0.5eq-g/L*0.050L=0.025eq-g

Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:

mol=0.025eq-gNa_2CO_3*\frac{1molNa_2CO_3}{2eq-gNa_2CO_3}\\ \\mol=0.0125molNa_2CO_3

Finally, the mass is computed by using its molar mass (106 g/mol)

m=0.0125molNa_2CO_3*\frac{106gNa_2CO_3}{1molNa_2CO_3} \\\\m=1.325gNa_2CO_3

Regards.

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Explanation:
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An aqueous solution contains the following ions Cl⁻, Ag⁺, Pb²⁺, NO₃⁻ & SO₄²⁻ and more than one precipitate will form are AgCl, PbCl₂, PbSO₄ & Ag₂SO₄.

<h3>What is precipitate?</h3>

Precipitate is the insoluble compound which is present at the bottom of any chemical reaction in the solid state.

If in an aqueous solution Cl⁻, Ag⁺, Pb²⁺, NO₃⁻ & SO₄²⁻ ions are present then:

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Hence precipitates are AgCl, PbCl₂, PbSO₄ & Ag₂SO₄.

To know more about precipitates, visit the below link:

brainly.com/question/2437408

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