All categories

2 years ago

solubility of cu(oh)2 is 2.42 x 10-3 mol/l. calculate ksp and in which direction the reaction is spontaneous if δgº

Chemistry

1 answer:

Simora [160]2 years ago

7 0

Ksp of copper(II) hydroxide Cu(OH)2 is 4.9 x 10-8.

Chemical reaction (dissociation) of copper(II) hydroxide in water:

Cu(OH)2(s) → Cu²⁺(aq) + 2OH⁻(aq).

Ksp(Cu(OH)2) = [Cu²⁺]·[OH⁻]².

[Cu²⁺] = 2.42 x 10-3 mol/l; solubility od copper ions

[OH⁻] = 2[Cu²⁺] = 2 x 2.42 x 10-3 mol/l

[OH⁻] = 4.48 x 10-3 mol/l; solubility of hydroxide ions

Ksp = 2.42 x 10-3 mol/l x (4.48 x 10-3 mol/l)²

Ksp = 4.9 x 10-8

Ksp is the solubility product constant for a solid substance dissolving in an aqueous solution.

Solubility of the compound depends on the temperature of the solution and the structure of that compound.

More info about: brainly.com/question/23946616

#SPJ4

You might be interested in

The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution

Alex_Xolod [135]

The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is 2.64 x 10⁻⁵ M.

4 0

3 years ago

Read 2 more answers

Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The

qaws [65]

Answer:

$E_{ionization}=5.45\times 10^{-19}\ J$

Explanation:

$E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

So, $n_i=2$ and $n_f=\infty$ (As the hydrogen has to ionize)

Thus,

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J$

$E_{ionization}=5.45\times 10^{-19}\ J$

4 0

3 years ago

Label the liquids from least to most dense, Vegetable oil, water, Dish soap, Corn syrup, Honey

masya89 [10]

Answer:

1. Vegetable oil

2. Water

3. Dish soap

4. Corn syrup

5. Honey

8 0

3 years ago

If 316 mL nitrogen is combined with 178 mL oxygen, what volume of N2O is produced at constant temperature and pressure if the re

lord [1]

Answer;

=259 ml

Explanation;

-According to Gay Lussac's Law of Combining Volumes when gases react, they do so in volumes which have a simple ratio to one another, and to the volume of the product formed if gaseous, provided the temperature and pressure remain constant.

-Thus; from the volume of nitrogen and oxygen gases; we have; 316 / 178 = 1.775 moles of nitrogen gas per mole of oxygen gas.

-Therefore, nitrogen gas is the limiting reactant, and for each mole of nitrogen gas used, we will get 1 mole of N2O. This means the resulting volume of N2O with 100% yield will be the same as the volume of nitrogen gas used, thus, 100% yield will produce 316 mL.

However, with 82% yield the volume would be; 316 × 82/100 =259 ml

Therefore; the volume of N2O at 82% yield will be 259 ml

7 0

3 years ago

Read 2 more answers

a swimming-pool worker wants the pools pH to be 7.2. She comes by to test the water and finds that the pH is 6.2. What should sh

nikdorinn [45]

Add an alkaline compound to raise the pH to 7.2.

3 0

3 years ago