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Alenkinab [10]
3 years ago
5

Is it possible for a molecular with polar bonds to be nonpolar?

Chemistry
1 answer:
marin [14]3 years ago
5 0
A molecule will either be polar or nonpolar. if the molecule has unevenly distributed charge, it has a pole and is polar. if charge is evenly distributed it is a nonpolar molecule because the net charge is 0.
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Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)5(OH)+(aq) + H3O+(aq) for wh
Elanso [62]

Answer: The pH of 0.10 M Cu(NO_{3})_{2}(aq) is 4.49.

Explanation:

Given: Initial concentration of Cu(H_{2}O)^{2+}_{6} = 0.10 M

K_{a} = 1.0 \times 10^{-8}

Let us assume that amount of Cu(H_{2}O)^{2+}_{6} dissociates is x. So, ICE table for dissociation of  Cu(H_{2}O)^{2+}_{6}  is as follows.

                               Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}

Initial:                       0.10 M                       0                       0

Change:                    -x                              +x                      +x

Equilibrium:           (0.10 - x) M                    x                        x

As the value of K_{a} is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.

And, (0.10 - x) will be approximately equal to 0.10 M.

The expression for K_{a} value is as follows.

K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}

Hence, [H_{3}O^{+}] = 3.2 \times 10^{-5}

Formula to calculate pH is as follows.

pH = -log [H^{+}]

Substitute the values into above formula as follows.

pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49

Thus, we can conclude that the pH of 0.10 M Cu(NO_{3})_{2}(aq) is 4.49.

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3 years ago
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6 [ I cannot answer that]

7)6,1,2

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