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MrRa [10]
3 years ago
7

Question 1 of 10

Chemistry
2 answers:
aivan3 [116]3 years ago
8 0

Answer:

Convert the 200 mol of water to kilograms of water.

Explanation:

patriot [66]3 years ago
7 0

A. Convert the 200 mol of water to kilograms of water.

<h3>Further explanation</h3>

Molality (m) shows the number of moles dissolved in every 1000 grams(1 kg) of solvent.

\tt m=n\times \dfrac{1000}{p}

m = Molality

n = number of moles of solute

p = Solvent mass (gram)

or

\tt m=\dfrac{mol~solute}{kg~solvent}

solute = 10 mol NaCl

solvent = 200 mol water⇒Convert to kilograms of water.

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The ksp of manganese(ii) carbonate, mnco3, is 2.42 × 10-11. calculate the solubility of this compound in g/l.
Vikki [24]

Answer :  The solubility of this compound in g/L is 565.414\times 10^{-6}g/L.

Solution : Given,

K_{sp}=2.42\times 10^{-11}

Molar mass of MnCO_3 = 114.945g/mole

The balanced equilibrium reaction is,

                      MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3

At equilibrium                         s       s

The expression for solubility constant is,

K_{sp}=[Mn^{2+}][CO^{2-}_3]

Now put the given values in this expression, we get

2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L

The value of 's' is the molar concentration of manganese ion and carbonate ion.

Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.

s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L

Therefore, the solubility of this compound in g/L is 565.414\times 10^{-6}g/L.


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What property is used to calculate the ph of a solution
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What is the total number of hydrogen atoms contained in one molecule of (NH4)2C204?
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=8 atoms

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