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Pavel [41]
2 years ago
12

(PLZ ANSWER SOON)

Chemistry
1 answer:
almond37 [142]2 years ago
7 0

Answer:

10.5 k.J.

Explanation:

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

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I was checking my answers for a chem test and came across a solution for a problem where the ∆S was pos, ∆H was neg and ∆G was n
nevsk [136]

Answer:

there is only 15 points

Explanation:

The second law of thermodynamics says that the entropy of the universe always increases for a spontaneous process: \Delta \text {S}_{\text{universe}}=\Delta \text {S}_{\text{system}} + \Delta \text {S}_{\text{surroundings}} > 0ΔS  

universe

​

=ΔS  

system

​

+ΔS  

surroundings

​

>0delta, start text, S, end text, start subscript, start text, u, n, i, v, e, r, s, e, end text, end subscript, equals, delta, start text, S, end text, start subscript, start text, s, y, s, t, e, m, end text, end subscript, plus, delta, start text, S, end text, start subscript, start text, s, u, r, r, o, u, n, d, i, n, g, s, end text, end subscript, is greater than, 0

At constant temperature and pressure, the change in Gibbs free energy is defined as \Delta \text G = \Delta \text H - \text{T}\Delta \text SΔG=ΔH−TΔSdelta, start text, G, end text, equals, delta, start text, H, end text, minus, start text, T, end text, delta, start text, S, end text.

When \Delta \text GΔGdelta, start text, G, end text is negative, a process will proceed spontaneously and is referred to as exergonic.

The spontaneity of a process can depend on the temperature.

Spontaneous processes

In chemistry, a spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly, because spontaneity is not related to kinetics or reaction rate. A classic example is the process of carbon in the form of a diamond turning into graphite, which can be written as the following reaction:

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