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3 years ago

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What is the percent composition by mass of sulfur in the compound mgso4

1 answer:

Margaret [11]3 years ago

6 0

Answer:

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

Explanation:

Molar mass of compound = 120 g/mol

Number of sulfur atom = 1

Atomic mass of sulfur = 32 g/mol

Percentage of element in compound :

$=\frac{\text{Number of atoms}\times \text{Atomic mass}}{\text{molar mas of compound}}\times 100$

Sulfur :

$=\frac{1\times 32 g/mol}{120 g/mol}\times 100=26.7\%$

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

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Calculate the pH of each of the following solutions. a. 0.100 M propanoic acid (HC3H5O2, Ka 1.3 105 ) b. 0.100 M sodium propanoa

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a) pH = 2.95

b) pH = 8.94

c) pH = 7.00

d) pH = 4.89

Explanation:

a) The reaction is:

CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)(1)

At equlibrium: 0.1 - x x x

<u>The dissociation constant of equation (1), Ka, is:</u>

$K_{a} = \frac{[C_{3}H_{5}O_{2}^{-}][H_{3}O^{+}]}{[C_{3}H_{6}O_{2}]}$

$1.3\cdot 10^{-5} = \frac{x^{2}}{(0.1 - x)}$ (2)

<u>Solving equation (2) for x, we have:</u>

x = 0.00113 = [H₃O⁺] = [CH₃CH₂COO⁻]

Hence, the pH is:

$pH = -log [H_{3}O^{+}] = -log (0.00113) = 2.95$

b) The reaction is:

CH₃CH₂COO⁻(aq) + H₂O(l) ⇄ CH₃CH₂COOH(aq) + OH⁻(aq)

At equlibrium: 0.1 - x x x

<u>The dissociation constant of the above equation, Kb, is:</u>

$K_{b} = \frac{[C_{3}H_{6}O_{2}][OH^{-}]}{[C_{3}H_{5}O_{2}^{-}]}$

<u>Also, we have that:</u>

$K_{w} = K_{a}*K_{b} \rightarrow K_{b} = \frac{K_{w}}{K_{a}} = \frac{1 \cdot 10^{-14}}{1.3 \cdot 10^{-5}} = 7.69 \cdot 10^{-10}$

$7.69\cdot 10^{-10} = \frac{x^{2}}{0.1 - x}$ (3)

<u>Solving equation (3) for x, we have:</u>

x = 8.77x10⁻⁶ = [OH⁻] = [CH₃CH₂COOH]

Hence, the pH is:

$pOH = -log[OH^{-}] = -log(8.77\cdot 10^{-6}) = 5.06 \rightarrow pH = 14 - pOH = 8.94$

c) Pure water:

H₂O ⇄ H⁺ + OH⁻

$K_{w} = [H^{+}][OH^{-}] \rightarrow 1\cdot 10^{-14} = [H^{+}][OH^{-}]$

<em>Since is pure water: [H⁺] = [OH⁻]</em>

$1\cdot 10^{-14} = [H^{+}]^{2} \rightarrow [H^{+}] = \sqrt{1\cdot 10^{-14}} = 1 \cdot 10^{-7}$

Therefore, the pH is:

$pH = -log [H^{+}] = -log (1 \cdot 10^{-7}) = 7.00$

d) The reaction is:

CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)

At equlibrium: 0.1 0.1 x

The pH is:

$pH = pKa + log(\frac{[C_{3}H_{5}O_{2}^{-}]}{[C_{3}H_{6}O_{2}]}) = -log(1.3 \cdot 10^{-5}) + log(\frac{0.1}{0.1}) = 4.89$

I hope it helps you!

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