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3 years ago

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Phase changes!a. Write out the Clapeyron equation and explain what it describes.b. Write out the Clausius-Clapeyron equation and

explain what it describes.c. Calculate the boiling temperature of water at an elevation of 1700 m (approximately the height of Boulder, CO) where the atmospheric pressure is ~0.83 atm. You may use the fact that water boils at 373 K, 1 atm, and that the enthalpy of vaporization is 6.01 kJ/mol.

1 answer:

djverab [1.8K]3 years ago

4 0

Answer:

Solutions in the attachment.

Explanation:

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jek_recluse [69]

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Biological Oxygen demand is the amount of dissolved oxygen needed by aerobic biological organisms to break down organic material present in a given water sample at certain temperature over a specific time period.

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3 years ago

Read 2 more answers

Addition of 50. J to a 10.0-g sample of a metal will cause the temperature of a metal to rise from 25ºC to 35ºC. The specific he

Snowcat [4.5K]

Answer:

b) C = 0.50 J/(g°C)

Explanation:

Q = mCΔT

∴ Q = 50 J

∴ m = 10.0 g

∴ ΔT = 35 - 25 = 10 °C

specific heat (C) :

⇒ C = Q / mΔT

⇒ C = 50 J / (10.0 g)(10 °C)

⇒ C = 0.50 J/(g°C)

7 0

3 years ago

Need help fast please 15 points!!!

In-s [12.5K]

Answer:

C. disposition and condensation

Explanation:

8 0

3 years ago

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

6 0

3 years ago

For each of the following choose the variable that will be more favorable in solution formation. A) Weak or Strong Solvent inter

Deffense [45]

Based on factors affecting solution formation;

Strong solvent-solute intermolecular forces favors solution formation.
Increase in entropy favours solution formation
Decrease in enthalpy favours solution formation

<h3>What is a solution?</h3>

A solution is a substance formed when a substance known as solute dissolves in another substance know as solvent.

Factors that affect solution formation include:

strength of intermolecular forces between solute and solvent
entropy
enthalpy

Strong solvent-solute intermolecular forces favors solution formation.

Increase in entropy favours solution formation

Decrease in enthalpy favours solution formation.

Learn more about solutions at: brainly.com/question/6675586

6 0

2 years ago