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jarptica [38.1K]
3 years ago
11

Which flask has the solution with the lowest freezing point

Chemistry
1 answer:
scZoUnD [109]3 years ago
8 0

Answer:

Here's what I get  

Explanation:

Assume you have 0. 1 mol·kg⁻¹ solutions of A) KCl, B) CH₃OH, C) Ba(OH)₂, and D) CH₃COOH.

The formula for freezing point depression is

\Delta T_{f} = iK_{f}b

Where

  b = the molal concentration

K_f = the freezing point depression constant

    i = the van't Hoff i factor

The i-factor is the number of solute particles produced by one formula unit of the substance.

The only difference in the solutions is the i-factor of the solutes.

A) KCl

KCl(aq) ⟶ K⁺(aq) +Cl⁻(aq); i = 2

B) CH₃OH

Methanol is a nonelectrolyte.

CH₃OH(aq) ⟶ CH₃OH(aq); i = 1

C) Ba(OH)₂

Ba(OH)₂(aq) ⟶ Ba²⁺(aq) + 2OH⁻(aq); i = 3

D) CH₃COOH

CH₃COOH(aq) ⇌ CH₃COO⁻(aq)+ H⁺(aq); i ⪆ 1

The Ba(OH)₂ solution has the greatest i-value. It therefore has the greatest freezing point depression and the lowest freezing point.

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Explanation:

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Show your work with good use of units, rounding, and significant figures. [Hint: it is good practice to show the value of your a
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Heat required : 4.8 kJ

<h3>Further explanation </h3>

The heat to change the phase can be formulated :

Q = mLf (melting/freezing)

Q = mLv (vaporization/condensation)

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Lv=latent heat of vaporization

The heat needed to raise the temperature

Q = m . c . Δt

1. heat to raise temperature from -20 °C to 0 °C

\tt Q=10\times 2.09\times (0-(-20)=418~J

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\tt Q=10\times 333=3330~J

3. heat to raise temperature from 0 °C to 25 °C

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3 0
2 years ago
A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h
frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]

Where;

\frac{f_2}{f_1}  is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1}  = 10^{0.7289}\\\\\frac{f_2}{f_1}  = 5.356

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

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What is the function of these organelles?
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Answer:

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Explanation:

took test :D

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3 years ago
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