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Serggg [28]
3 years ago
7

A buffer solution contains 0.11 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L. What is the pH of this buffer?

Chemistry
1 answer:
aleksley [76]3 years ago
5 0

Answer:

Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]

You need to know the pKa for acetic acid. Looking it up one finds it to be 4.76

(a). pH = 4.76 + log [0.13]/[0.10]

= 4.76 + 0.11

= 4.87

(b) KOH + CH3COOH =>H2O + CH3COOK so (acid)goes down and (salt)goes up. Assuming no change in volume, you have 0.10 mol acid - 0.02 mol = 0.08 mol acid and 0.13 mol salt + 0.02 mol = 0.15 mol salt

pH = 4.76 + log [0.15]/[0.08]

= 4.76 + 0.27

= 5.03

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11.45

Explanation:

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14-2.55=ph=11.45

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The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd
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Answer:

The answer is "= 0.078 \ kg \ H_2".

Explanation:

calculating the moles in CH_4 =\frac{PV}{RT}

                                                =\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol

Eqution:

CH_4 +H_2O \to  3H_2+ CO \ (g)

Calculating the amount of H_2 produced:

= 12.9 \ mol CH_4 \times  \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2

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5 0
3 years ago
What is significant about 42?​
maks197457 [2]

Answer:

It’s twice as much as 9+10

Explanation:

5 0
2 years ago
The reactant concentration in a first-order reaction was 8.10×10−2 M M after 15.0 s s and 1.80×10−3 M M after 90.0 s s . What is
kipiarov [429]

Answer:

The answer to the question is

The rate constant for the reaction is 1.056×10⁻³ M/s

Explanation:

To solve the question, e note that

For a zero order reaction, the rate law is given by

[A] = -k×t + [A]₀

This can be represented by the linear equation y = mx + c

Such that y = [A], m which is the gradient is = -k, and the intercept c = [A]₀

Therefore the rate constant k which is the gradient is given by

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= \frac{1.80*10^{-3} M- 8.10*10^{-2} M}{90 s - 15 s} = -0.001056 M/s = -1.056×10⁻³ M/s

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3 years ago
Why is frying an egg an example of a chemical change
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3 years ago
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