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abruzzese [7]
3 years ago
9

Please help with all 3 parts!

Chemistry
2 answers:
Murljashka [212]3 years ago
8 0

Answer:

1:Part A.

\bold{42.2 g C_{12}H_{22}O_{11} \:in \:528 g H₂O}

Mass Percent=\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}

=\frac{42.2}{528}*100\%=\bold{\underline{7.99\: or \:8\%}}

Part B.

\bold{198\:m g\: C_{6}H_{12}O_{6} \:in\:4.71 g\: H₂O}

mass of solute: 198mg

mass of solvent :4.71g=4710g

Mass Percent=\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}

=\frac{198}{4710}*100\%=\bold{\underline{4.20\%}}

Part C.

\bold{8.85 g NaCl \:in \:190 g\: H₂O}

Mass Percent=\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}

=\frac{8.85}{190}*100\%=\bold{\underline{4.66\%}}

Alecsey [184]3 years ago
6 0

Answer:

It will help you !!!!!!!!!!

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[Ag⁺] = 5.0x10⁻¹⁴M

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The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

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Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ <em>just begin to precipitate when the product  [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>

<em />

As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

<h3>[I⁻] = 1.67x10⁻³</h3><h3 />

So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:

[Ag⁺] [I⁻] = 8.3x10⁻¹⁷

[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷

<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>

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3 years ago
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