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adell [148]
3 years ago
10

sarah measures out 151 grams of SO2. how many moles is this? Express your answer to three significant figures. ​

Chemistry
1 answer:
svp [43]3 years ago
4 0

150 g SO₂ = 2.36 mol SO₂


M_r = 32.06 + 2×16.00 = 32.07 + 32.00 = 64.06


Moles of SO₂ = 151 g SO₂ × [1 mol SO₂/64.06 g SO₂] = 2.36 mol SO₂



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ook at sample problem 18.12 in the 8th ed Silberberg book. Write a balanced chemical equation (salt hydrolysis). So acetate ion
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Here's what I get  

Explanation:

1. Write the chemical equation

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻; Kₐ = 2 × 10⁻⁵

Let's rewrite the equation as

A⁻ + H₂O ⇌ HA + OH⁻

2. Calculate Kb

K_{\text{b}} = \dfrac{K_{\text{w}}}{K_{\text{a}}} = \dfrac{1.00 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10}

3. Set up an ICE table

                      A⁻ + H₂O ⇌ HA + OH⁻

I/mol·L⁻¹:      0.35                 0       0

C/mol·L⁻¹:       -x                  +x      +x

E/mol·L⁻¹:    0.35-x               x        x

4. Solve for x

\dfrac{\text{[HA ][OH$^{-}$]}}{\text{[A$^{-}$]}} = \dfrac{x^{2}}{0.35-x} = 5 \times 10^{-10}

Check for negligibility,

\dfrac{\text{[HA]}}{K_{\text{b}}} = \dfrac{0.35}{5 \times 10^{-10}} = 7 \times 10^{8}> 400\\\\\therefore x \ll 0.35\\\\\dfrac{x^{2}}{0.35} = 5 \times 10^{-10}\\\\x^{2} = 0.35 \times 5 \times 10^{-10} = 1.8\times 10^{-10}\\\\x = \sqrt{1.8\times 10^{-10}} = \mathbf{1 \times 10^{-5}}

5. Calculate the pOH

[OH⁻] = 1 × 10⁻⁵ mol·L⁻¹

pOH = -log[OH⁻] = -log(1 × 10⁻⁵) = 4.88

6. Calculate the pH.

pH + pOH = 14.00

pH + 4.88 = 14.00

pH = 9.12

Note: The answer differs from that given by Silberberg because you used only one significant figure for the Kₐ of acetic acid.

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