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Vesnalui [34]
3 years ago
5

Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of th

e following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. (a) AgCl(s) in 0.025 M NaCl (b) CaF2(s) in 0.00133 M KF (c) Ag2SO4(s) in 0.500 L of a solution containing 19.50 g of K2SO4 (d) Zn(OH)2(s) in a solution buffered at a pH of 11.45\
Chemistry
1 answer:
kirill [66]3 years ago
8 0

Answer:

Explanation:

<u>a) AgCl(s) in 0.025 M NaCl</u>

Equation:  AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

Initial conc :    S            O               O

equili conc :    O            S                S

                  NaCl(s) ⇒ Na⁺ (aq) + Cl⁻ (aq)

Initial conc :  0.025      0           0

equili conc :     0          0.025    0.025

Therefore the concentration:  Ag⁺ = 6.4 * 10^-9 M,  Cl⁻  = 0.025 M

attached below is the detailed solution of the

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True or false when matter changes it’s state it is a physical change because it is the same substance the whole time.
Blababa [14]

Answer:

true

Explanation:

5 0
3 years ago
What is the ph of an aqueous solution with the hydronium ion concentration : [h3o+] = 3 x 10-5 m ?
BaLLatris [955]

The concentration of hydrogen can be shown as:

[H+ ] = 3 * 10-5 M

pH can be determined as:

pH = - log [H+ ]

= - log (3 * 10-5)

= 4.53

Thus the pH of solution is 4.53


8 0
3 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration 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

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

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

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
3 years ago
An electrolytic cell is set up to plate Zr(s) from a solution containing Zr4 (aq). A current of 7.92 amps is run through this so
iren [92.7K]

The mass of Zr deposited in the process is 41.4 g.

<h3>What is electrolytic cell?</h3>

An electrolytic cell is a chemical cell which produces electrical energy by non-spontaneous chemical processes.

From the question;

Zr^4+(aq) + 4e ------> Zr(s)

We know that;

91 g of Zr is deposited by 4(96500) C

xg of Zr is deposited by (7.92 × 6.16 × 60 × 60) C

xg = 91 g ×  (7.92 × 6.16 × 60 × 60) C/4(96500) C

x g = 41.4 g

Learn more about electrolysis: brainly.com/question/12054569

8 0
2 years ago
If 12.4 mol of Ne gas occupies 122.8 L, how many mol of Ne would occupy 339.2 L under the same temperature and pressure? Record
8090 [49]

Answer:

3.43×10¹ mol

Explanation:

Given data:

Initial number of  moles = 12.4 mol

Initial volume = 122.8 L

Final number of moles = ?

Final volume = 339.2 L

Solution:

The number of moles and volume are directly proportional to each other at same temperature and pressure.

V₁/n₁  =  V₂/n₂

122.8 L/ 12.4 mol  =  339.2 L / n₂

n₂ = 339.2 L× 12.4 mol  / 122.8 L

n₂ = 4206.08 L.mol /122.8 L

n₂ = 34.3mol

In scientific notation:

3.43×10¹ mol

7 0
3 years ago
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