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Phoenix [80]
3 years ago
5

What is the volume, in milliliters, of 6.56 g of acetone?

Chemistry
1 answer:
PIT_PIT [208]3 years ago
4 0

Answer:

8.29 mL

Explanation:

Density of acetone in g/mL = 0.791 g/ml

Dimensional analysis: 6.56 g x (1 mL/0.791 g) = 8.29 mL

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Calculate the solubility of CaF2 in g/L (Ksp = 4.0 x 10-8). 2. What is the pH of a solution containing a hydrogen ion concentrat
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Answer:

\large \boxed{1. \text{ 0.17 g/L; 2. 3.52; 3. Cl; 4. (a) +3; (b) +4; (c) +6}}

Explanation:

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(a) Molar solubility

CaF₂ ⇌ Ca²⁺ + 2F⁻

K_{\text{sp }} = \text{[Ca$^{2+}$]}\text{[F$^{-}$]}^{2}= 4.0 \times 10^{-8}\\s(2s)^{2}=4.0 \times 10^{-8}\\4s^{3} = 4.0 \times 10^{-8}\\s^{3} = 1.0 \times 10^{-8}\\s =2.2 \times 10^{-3}\text{ mol/L}

(b) Mass solubility

\text{Solubility} = 2.2 \times 10^{-3} \text{ mol/L} \times \dfrac{\text{78.07 g}}{\text{1 L }} = \text{0.17 g/L}\\\\\text{The solubility of CaF$_{2}$ is $\large \boxed{\textbf{0.17 g/L}}$}

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pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52

3. Oxidizing and reducing agents

Zn + Cl₂ ⟶ ZnCl₂

\rm \stackrel{\hbox{0}}{\hbox{Zn}} + \stackrel{\hbox{0}}{\hbox{ Cl}_{2} }\longrightarrow \stackrel{\hbox{+2}}{\hbox{Zn}}\stackrel{\hbox{-1}}{\hbox{Cl}_{2}}

The oxidation number of Cl has decreased from 0 to -1.

Cl has been reduced, so Cl is the oxidizing agent.

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(a) Al₂O₃

\stackrel{\hbox{$\mathbf{+3}$}}{\hbox{Al}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{3}}

1O = -2; 3O = -6; 2Al  = +6; 1Al = +3

(b) XeF₄

\stackrel{\hbox{$\mathbf{+4}$}}{\hbox{Xe}}\stackrel{\hbox{-1}}{\hbox{F}_{4}}

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(c) K₂Cr₂O₇

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1K = +1; 2K = +2; 1O = -2; 7O = -14

+2 - 14 = -12

2Cr = + 12; 1 Cr = +6

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