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3 years ago

Calculate the density of a block of water measuring 2 cm x 4 cm x 1 cm, with a mass of 5.4

Chemistry

1 answer:

Nastasia [14]3 years ago

3 0

the formula for density is p=m/v

m aka the mass = 5.4; the v aka the volume = 8

so, p=5.4/8

p= 0.675 g/cm^3

*let's just say the mass' unit is gram so the unit for the density is g/cm^3

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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

vfiekz [6]

Answer:

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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