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mestny [16]
3 years ago
12

Colligative properties depend upon the ____. Select one: a. nature of the solute b. nature of the solvent c. number of solute pa

rticles in a solution d. freezing point of a solute e. Choices a and b only. f. Choices a and c only. g. Choices b and c only.
Chemistry
2 answers:
Olin [163]3 years ago
5 0

Answer:

c. number of solute particles in a solution

Explanation:

Colligative properties depend upon the number of solute particles in a solution. Option C is correct.  

Colligative properties does not depend nature of the solute or solvent particles. Hence options A, B and E are incorrect.  

Freezing point of a solvent is a colligative property which itself depends on the number of solute particles in a solution. Hence option D is incorrect.  

Other options F and G are also incorrect.  

Flauer [41]3 years ago
3 0
The correct answer is C. Colligative properties only depend upon the number of solute particles in a solution but not on the identity or nature of the solute and solvent particles. I hope this anwers your question. 
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In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
How many miles of O2 can be produced by letting 12.00 miles of KC103 react?
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    12                               6    (moles) 
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il63 [147K]
The answer would be NaOH
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What can you calculate with the Henderson-Hasselbalch equation
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Answer:

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A

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