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Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L

Oduvanchick [21]

Answer:

Ksp = [ Cu+² ] [ OH-] ²

molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol

Ksp = [ Cu+² ] [ OH-] ²

Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰

|__________|___Cu+² __|_2OH-____|

|Initial concentration(M)|___0__|_0______|

|Change in concentration(M)|_+S |__+2S__|

|Equilibrium concentration(M)|_S _|2S___|

Ksp = [ Cu+² ] [ OH-] ²

2.2 ×10-²⁰ = (S)(2S)²= 4S³

$s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7}$

S = 1.8 × 10-⁷ M

The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

$Cu (OH)2 = \frac{1.8 \times {10}^{ - 7} mol \:Cu (OH)2 }{1L} \times \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2} \\ = 1.75428 \times 10 ^{ - 5}$

<h3>Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>

I hope I helped you^_^

8 0

3 years ago

An artifact originally had 16 grams of carbon-14 present. the decay model upper a equals 16 e superscript negative 0.000121 ta=

xeze [42]

Given:

Initial amount of carbon, A₀ = 16 g

Decay model = 16exp(-0.000121t)

t = 90769076 years

To determine:

the amount of C-14 after 90769076 years

Explanation:

The radioactive decay model can be expressed as:

A = A₀exp(-kt)

where A = concentration of the radioactive species after time t

A₀ = initial concentration

k = decay constant

Based on the given data :

A = 16 * exp(-0.000121*90769076) = 16(0) = 0

Ans: Based on the decay model there will be no C-14 left after 90769076 years

3 0

3 years ago

Read 2 more answers

What are the reactants and products when butane combusts with excess oxygen?

larisa86 [58]

C3H8 + O2 --> CO2(g) + H2O(g) + energy(heat)

butane + oxygen --> carbon dioxide + water + heat

7 0

3 years ago

the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29

Vera_Pavlovna [14]

Answer:

ΔG = 16.218 KJ/mol

Explanation:

dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate

ΔG = ΔG° - RT Ln Q

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴ R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

5 0

3 years ago

The metal tantalum becomes superconducting at temperatures below 4.483 K. Calculate the temperature at which tantalum becomes su

masha68 [24]

Answer:

The correct answer is "-268.667°C".

Explanation:

Given:

Temperature,

= 4.483 K (below)

Now,

The formula of temperature conversion will be:

⇒ $T(^{\circ} C)=T(K)-273.15$

By putting the values, we get

⇒ $=4.483-273.15$

⇒ $=-268.667^{\circ} C$

Thus the above is the correct answer.

3 0

3 years ago