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sergij07 [2.7K]
3 years ago
12

The molar solubility of zns is 1.6 Ã 10-12 m in pure water. Calculate the ksp for zns.

Chemistry
1 answer:
Oksi-84 [34.3K]3 years ago
5 0

Answer:

The solubility product of zinc sulfide is 2.56\times 10^{-24}.

Explanation:

Solubility of the zinc sulfide = S=1.6\times 10^{-12}

ZnS\rightleftharpoons Zn^{2+}+S^{2-}

                S        S

The expression of solubility product is given by :

K_{sp}=[Zn^{2+}]\times [S^{2-}]

K_{sp}=S\times S=S^2

=1.6\times 10^{-12}\times 1.6\times 10^{-12}=2.56\times 10^{-24}

The solubility product of zinc sulfide is 2.56\times 10^{-24}.

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Answer:

The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams

Explanation:

The question is with regards to density calculations

The density of the given sodium carbonate solution, ρ = 0.4 g/dm³

The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³

Density \ of \ an \ object, \rho  = \dfrac{The \ mass \ of \ the \ object, \ m }{\ The \ volume \ of \ the \ object, \ V }

\rho = \dfrac{m}{V}

Therefore, we have;

The \ Density \ of \ the \ sodium \ carbonate, \ \rho  = 0.4 \ g/dm^3 =  \dfrac{m }{ 0.01 \ dm^3 }

The mass, "m", of the sodium carbonate in  = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g

The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.

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To titrate 45.00 milliliters of an unknown HCl sample, use 25.00 milliliters of a standard 0.2000 M NaOH titrant. What is the mo
gtnhenbr [62]
V ( HCl ) = 45.00 mL in liters : 45.00 / 1000 => 0.045 L

M ( HCl ) = ?

V ( NaOH ) = 25.00 / 1000 => 0.025 L 

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number of moles NaOH :

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1 mole HCl ---------- 1 mole NaOH
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hope this helps!




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