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lilavasa [31]
4 years ago
13

Which of the following metals will react with water to produce a metal hydroxide and hydrogen gas? (2 points)

Chemistry
2 answers:
vazorg [7]4 years ago
8 0
The answer would be the first option - k
serg [7]4 years ago
8 0

<u>Answer:</u> The correct answer is Potassium (K).

<u>Explanation:</u>

Alkali (Group 1) and Alkaline earth (Group 2) elements react with water to produce metal hydroxide and also releases hydrogen gas.

Reaction of group 1 elements with water:

2M+2H_2O\rightarrow 2MOH+H_2

Reaction of group 2 elements with water:

M+2H_2O\rightarrow M(OH)_2+H_2

From the given options:

Potassium (K): This metal belongs to Group 1 and readily reacts with water to form potassium hydroxide and hydrogen gas.

2K+2H_2O\rightarrow 2KOH+H_2

Tin (Sn): This element belongs to Group 14 and does not readily reacts with water.

Nickel (Ni): This element belongs to Group 10 and does not readily reacts with water.

Manganese (Mn): This element belongs to Group 7 and does not readily reacts with water.

Hence, the correct answer is potassium (K).

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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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