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Gwar [14]
1 year ago
9

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]1 year 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

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A sample contains 16 mg of polonium-218. After 12 minutes, the sample will contain 1.0 mg of polonium-218. What is the half life
mariarad [96]

Answer:

Half-life = 3 minutes

Explanation:

Using the radioactive decay equation we can solve for reaction constant, k. And by using:

K = ln2 / Half-life

We can find half-life of polonium-218

Radioactive decay:

Ln[A] = -kt + ln [A]₀

Where:

[A] could be taken as mass of polonium after t time: 1.0mg

k is Reaction constant, our incognite

t are 12 min

[A]₀ initial amount of polonium-218: 16mg

Ln[A] = -kt + ln [A]₀

Ln[1.0mg] = -k*12min + ln [16mg]

-2.7726 = - k*12min

k = 0.231min⁻¹

Half-life = ln 2 / 0.231min⁻¹

<h3>Half-life = 3 minutes</h3>

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50 POINTS*** Which of the following is not a correct chemical equation for a double displacement reaction?
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5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
3 years ago
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