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3 years ago

A hydrocarbon contains 85.7% carbon and the remainder

Chemistry

1 answer:

Viefleur [7K]3 years ago

6 0

Answer:

CH₂ ; 67.1 %

Explanation:

To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation

Assume 100 grams of the compound.

# mol C = 85.7 g / 12.01 g/mol = 7.14 mol

# mol H = 14.3 g / 1.008 g/mol = 14.19 mol

The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C

So the empirical formula is CH₂

For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.

We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )

1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol

Theoretical yield from balanced chemical equation:

0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆

Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g

% yield = 0.295 g/ 0.440 g x 100 = 67.1 %

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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

Allisa [31]

Answer:

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

The dissociation constant of equation (1), Ka, is:

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

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

Solving equation (2) for x, we have:

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

The dissociation constant of the above equation, Kb, is:

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

Also, we have that:

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

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

Solving equation (3) for x, we have:

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⁻

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

Since is pure water: [H⁺] = [OH⁻]

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

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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