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3 years ago

The Ksp for Cu(OH)2 is 4.8 × 10-20. Determine the molar solubility of Cu(OH)2 in a buffer solution with a pH of 10.1.

Chemistry

1 answer:

Papessa [141]3 years ago

3 0

Answer:

E) 3.0x10⁻¹²

Explanation:

Ksp of Cu(OH)₂ is defined as:

Cu(OH)₂(s) ⇄ Cu²⁺(aq) + 2 OH⁻(aq)

Ksp = [Cu²⁺] [OH⁻]²

When pH is 10.1, [OH⁻] is:

pOH = 14 - pH

pOH = 3.9

pOH = -log [OH⁻]

[OH⁻] = 1.26x10⁻⁴M

Replacing in ksp formula:

4.8x10⁻²⁰= [Cu²⁺] [1.26x10⁻⁴]²

3.0x10⁻¹² = [Cu²⁺]

That means the maximum amount of Cu²⁺ that can be in solution is 3.0x10⁻¹²M, thus, molar solubility of Cu(OH)₂ is

E) 3.0x10⁻¹²

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Gasoline contains only hydrogen and carbon atoms, yet nitrogen oxide and nitrogen dioxide are produced when gasoline burns. What

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Answer:

The answer to the question is

The source of the nitro- gen and oxygen atoms is from the atmosphere

Explanation:

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3 years ago

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Which glassware would be used to make a 1.0 m kcl solution?

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Molarity of solution is expressed as,
Molarity = $\frac{\text{weight of solute(g)}}{\text{Molecular Weight X Vol. of solution (l)}}$

Molecular weight of KCl = 74.55 g/mol

For preparing 1 mol of KCl solution (1 liter), 74.44 g KCl must be dissolved in 1 liter of water.

Following glasswares are required for preparing the solution.
1) Watch Glass (for weight the KCl)
2) Beaker (for preparing the solution)
3) Stirrer (For dissolving KCl in water)
4) Standard measuring flask (To ensure final volume of solution is 1 liter)

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3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

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Lithium diisopropylamide (LDA) is used as a strong base in organic synthesis. LDA is itself prepared by an acid-base reaction be

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Explanation:

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