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Shtirlitz [24]
3 years ago
11

Acetic acid only partially ionizes in water. A solution contains a large quantity of acetic acid dissolved in water. How can thi

s acid solution best be described?
A. strong and concentrated
B. weak and concentrated
C. strong and dilute
D. weak and dilute
Chemistry
2 answers:
vazorg [7]3 years ago
8 0

Answer: Option (B) is the correct answer.

Explanation:

It is known that acetic acid is a weak acid. Hence, it only ionizes partially into a solution.

Since, a solution consists of solute and solvent. A substance present in smaller quantity is known as solute and a substance present in larger quantity is known as solvent.

So, when we add more and more of a solute into a solvent then there will occur an increase in the concentration of solute. As a result, the solution becomes concentrated in nature.

Therefore, we can conclude that the given acid solution best be described as  weak and concentrated.

dusya [7]3 years ago
4 0
B because acetic acid is a weak acid and large quantity means you make it become concentrated
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Pepsi [2]

Answer:

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

Explanation:

1. Solubility of CaF_2

(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 $\large \boxed{\textbf{0.17 g/L}}$}

2. pH

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.

4. Oxidation numbers

(a) Al₂O₃

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

1O = -2; 3O = -6; 2Al  = +6; 1Al = +3

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

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\stackrel{\hbox{${+1}$}}{\hbox{K}_{2}}\stackrel{\hbox{$\mathbf{+6}$}}{\hbox{Cr}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{7}}

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+2 - 14 = -12

2Cr = + 12; 1 Cr = +6

8 0
3 years ago
Analysis of a compound of sulfur, oxygen and fluorine showed that it is 31.42% S and 31.35% O, with F accounting for the remaind
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Answer:

The molecular formula is  SO2F2

Explanation:

Step 1: Data given

Suppose the mass of compound = 100 grams

The compound contains:

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31.35 % O = 31.35 grams O

100 - 31.42 - 31.35 = 37.23 F

Molar mass of S = 32.065 g/mol

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Step 2: Calculate moles

Moles = mass / molar mass

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Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.9799 / 0.9799 = 1

F: 1.959/ 0.9799 = 2

O : 1.959 / 0.9799 = 2

The empirical formula is SO2F2

This formula has a molecular mass of 102.06 g/mol

This means the empirical formula is also the molecular formula : SO2F2

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Answer:

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