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oee [108]
3 years ago
13

Determine the molar solubility of agi in pure water. ksp (agi) = 8.51 × 10-17.

Chemistry
1 answer:
Alex787 [66]3 years ago
3 0
When the salt AgI dissolves in water it dissociates as follows;
AgI ---> Ag⁺ + I⁻
molar solubility is the number of moles that are dissolved in 1 L of solution.
If molar solubility of AgI is x, then molar solubility of Ag⁺ is x and I⁻ is x.
the formula for solubility product constant - ksp of AgI is given below 
ksp = [Ag⁺][I⁻]
ksp = (x)(x)
ksp = 8.51 x 10⁻¹⁷ 
therefore,
x² = 8.51 x 10⁻¹⁷
x = 9.22 x 10⁻⁹
since molar solubility of AgI is x, then molar solubility of AgI is 9.22 x 10⁻⁹ M
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2Li + H2SO4=Li2SO4 + H2 How many liters of hydrogen gas, H2 at STP can be produced from 3.0 moles of Li? The molar volume of a g
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volume of H_2=33.6 litre

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2Li + H_2SO_4=Li_2SO_4 + H_2

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3 mole Li⇔1.5 mole H_2

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therefore volume of gas produced from 3 mole Li at STP = 1.5\times22.4 \frac{L}{mol}

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Tungsten (W) and chlorine (Cl) form a series of compounds with the following compositions:_______.
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Answer:

WCl₂, WCl₄, WCl₅, WCl₆

Explanation:

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Mass of Chlorine = 35.5 g/mol

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Percentage of tungsten = 72.17 %

Upon solving;

72.17 % = 184

100 % = Total mass

Total mass of compound = 254.95g

Mass of chlorine = 254.95 - 184 = 70.95 (Dividing by 35.35; This is approximately 2 Chlorine atoms.

The Formular is WCl₂

In the second compound;

Percentage of tungsten = 56.45 %

Upon solving;

56.45 % = 184

100 % = Total mass

Total mass of compound = 325.95 g

Mass of chlorine = 325.95 - 184 = 141.95g (Dividing by 35.35; This is approximately 4 Chlorine atoms.

The Formular is WCl₄

In the third compound;

Percentage of tungsten = 50.91 %

Upon solving;

50.91 % = 184

100 % = Total mass

Total mass of compound = 361.42 g

Mass of chlorine =  361.42 - 184 = 177.42 (Dividing by 35.35; This is approximately 5 Chlorine atoms.

The Formular is WCl₅

In the fourth compound;

Percentage of tungsten = 46.39 %

Upon solving;

46.39 % = 184

100 % = Total mass

Total mass of compound = 396.64 g

Mass of chlorine = 396.64 - 184 = 212.64 (Dividing by 35.35; This is approximately 6 Chlorine atoms.

The Formular is WCl₆

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3 years ago
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