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Bond [772]
2 years ago
12

Be sure to answer all parts. liquid ammonia autoionizes like water: 2nh3(l) → nh4+(am) + nh2−(am) where (am) represents solvatio

n by ammonia. at the boiling point (−33°c), kam = 5.1 × 10−27. calculate [nh4+] at this temperature. [nh4+] = × 10 m(enter your answer in scientific notation.)
Chemistry
1 answer:
Genrish500 [490]2 years ago
5 0
Answer is: concentration of ammonium ions are 7,14·10⁻¹⁴ M.
Chemical reaction: 2NH₃(l) → NH₄⁺(am) + NH₂⁻(am).
Kam = 5,1·10⁻²⁷.
[NH₄⁺] · [NH₂⁻] = x; equilibrium concentration of cations and anions.
Kam = [NH₄⁺] · [NH₂⁻].
Kam = x².
x = [NH₄⁺] = √5,1·10⁻²⁷.
[NH₄⁺] = 7,14·10⁻¹⁴ M.

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You wish to extract an organic compound from an aqueous phase into an organic layer (three to six extractions on a marco scale).
AVprozaik [17]

Answer:

Explanation:

It will be better to use solvents that are lighter than water, because their density has an influence on the miscibility . This will give you a better separation during extraction.

7 0
2 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

The value of x = C\alpha = 0.28 M

Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

Concentration of N_2O_4 = 1.0 - x  = 1.0 - 0.28 = 0.72 M

Concentration of NO_2 = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

Now put all the values in this expression, we get :

K_c=\frac{(0.56)^2}{0.72}=0.44

Therefore, the equilibrium constant K_c for the reaction is, 0.44

8 0
2 years ago
Why couldn't you substitute 3m H2SO4 for concentrated HNO3 when oxidizing copper?
schepotkina [342]

This question seems to be an essay question from experiment. Different solution of oxidizing agent will have different strength.  Sulfuric acid or H2SO4 is weaker oxidizing agent when compared to nitric acid (HNO3). In this case, if you subtitute the H2SO4 you wouldn't be able to get the same result for the experiment.



5 0
2 years ago
Read 2 more answers
What mitigation measures can communities do to reduce the damage and impact of sudden geologic hazards?
jasenka [17]

Explanation:

require an emergency support immediately

4 0
2 years ago
A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m
ioda
In this kind of exercises, you should  use the "ideal gas" rules: PV = nRT
P should be in Pascal: 
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = \frac{PV}{nT} = constant
\frac{P1 V1}{n T} = \frac{P2 V2}{n T}
==> P1 * V1 = P2 * V2
V2 = \frac{P1 V1}{P2} = \frac{445 0.016}{1225}
V2 = 0.00581 m3 = 5.81 L


7 0
2 years ago
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