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marshall27 [118]
4 years ago
7

1) AgI has a Ksp of 8.3 × 10-17, ZnS has a Ksp of 1.6 × 10-24, PbS has a Ksp of 8.3 × 10-28, and CaSO4 has a Ksp of 9.6 × 10-6.

Chemistry
1 answer:
vodka [1.7K]4 years ago
4 0
For the first question, I think the correct answer would be D. Calcium sulfate would be the most soluble salt in the given choices. The solubility product constant or Ksp expresses the equilibrium of the a solid and the ions it dissociates when in solution. Such that a higher value of Ksp would mean the compound is more soluble in solution. For the second question, the answer would be the first option. Given the amounts of the reactants and the Ksp of AgBr, the product silver bromide would precipitate in the product solution. This is because the resulting concentration of this product is way higher than the solubility of the compound.<span />
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What are the boiling points and freezing points (in oC) of a solution of 50.3 g of I2 in 350 g of chloroform? The kb = 3.63 oC/m
patriot [66]

Answer:

Boiling point: 63.3°C

Freezing point: -66.2°C.

Explanation:

The boiling point of a solution increases regard to boiling point of the pure solvent. In the same way, freezing point decreases regard to pure solvent. The equations are:

<em>Boiling point increasing:</em>

ΔT = kb*m*i

<em>Freezing point depression:</em>

ΔT = kf*m*i

ΔT are the °C that change boiling or freezing point.

m is molality of the solution (moles / kg)

And i is Van't Hoff factor (1 for I₂ in chloroform)

Molality of 50.3g of I₂ in 350g of chloroform is:

50.3g * (1mol / 253.8g) = 0.198 moles in 350g = 0.350kg:

0.198 moles / 0.350kg = 0.566m

Replacing:

<em>Boiling point:</em>

ΔT = kb*m*i

ΔT = 3.63°C/m*0.566m*1

ΔT = 2.1°C

As boiling point of pure substance is 61.2°C, boiling point of the solution is:

61.2°C + 2.1°C = 63.3°C

<em>Freezing point:</em>

ΔT = kf*m*i

ΔT = 4.70°C/m*0.566m*1

ΔT = 2.7°C

As freezing point is -63.5°C, the freezing point of the solution is:

-63.5°C - 2.7°C = -66.2°C

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Caffeine, a stimulant in coffee and tea, has a molar mass of 194.19 g/mol and a mass percentage composition of 49.48% C, 5.19% H
lozanna [386]

Answer : The molecular formula of a caffeine is, C_8H_{10}N_4O_2

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.48g}{12g/mole}=4.12moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.19g}{1g/mole}=5.19moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28.85g}{14g/mole}=2.06moles

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{16.48g}{16g/mole}=1.03moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{4.12}{1.03}=4

For H = \frac{5.19}{1.03}=5.03\approx 5

For N = \frac{2.06}{1.03}=2

For O = \frac{1.03}{1.03}=1

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_4H_5N_2O_1=C_4H_5N_2O

The empirical formula weight = 4(12) + 5(1) + 2(14) + 16 = 97 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{194.19}{97}=2

Molecular formula = (C_4H_5N_2O)_n=(C_4H_5N_2O)_2=C_8H_{10}N_4O_2

Therefore, the molecular of the caffeine is, C_8H_{10}N_4O_2

5 0
3 years ago
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