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Viefleur [7K]
3 years ago
5

Select all that apply.

Chemistry
2 answers:
Marina86 [1]3 years ago
3 0
Consider this system: CH3COOH(aq) H+(aq) + CH3COO- Ka = 1.8 x 10-5 
<span>The equilibrium as OH- ions are added;</span>

<span>A) H+ will combine with OH- to form water. </span>

<span>C) The acid will dissociate to supply more H+. </span>
<span>D) The reaction will move to the right. </span>

Scrat [10]3 years ago
3 0

Answer : The correct options are, (A), (C) and (D)

Explanation :

Le-Chatelier's principle : It states that if any change in the variables of the reaction, then the equilibrium will shift in that direction where the effect will be minimum.

The balanced equilibrium reaction will be,

CH_3COOH(aq)\rightleftharpoons H^+(aq)+CH_3COO^-

As the concentration of H^+ are decreased, the equilibrium will shift in a direction where concentration of H^+ is increasing and thus the dissociation of acid will increase to supply more of H^ ions. During this process the reaction will move to the right direction.

In the given equilibrium reaction when the OH^- are added then the H^+ ion will combine with OH^- ion to form water.

Hence, the correct options are, (A), (C) and (D)

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DCPIP is a redox dye that starts off blue but will turn reddish-pink in acidic solution.

This chemical is often used in general chemistry experiments with vitamin C (ascorbic acid), which is a good reducing agent. The DCPIP become colorless when it is reduced (remember, it's blue when oxidized).
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Describe how spead, velocity and<br> brcoloration are related.
Flura [38]

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3 years ago
What is 694,563,239 rounded to the nearest thousand
lina2011 [118]
694,563,239 rounded to the nearest thousand is 694,563.
 
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6 0
3 years ago
Read 2 more answers
Mosquitoes lays eggs that must hatch in water. The young, called larvae live their first few days in water
nevsk [136]

Answer:

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I believe he is C

3 0
3 years ago
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
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