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Flauer [41]
3 years ago
11

Calculate the solubility of CuX (Ksp=[Cu2+][X2−]=1.27×10−36) in a solution that is 0.200 M in NaCN.

Chemistry
2 answers:
Advocard [28]3 years ago
5 0

Answer:

The solubility of CuX is 1.425x10⁻⁷M

Explanation:

Given:

initial concentration of NaCN=0.2M

Ksp=1.27x10⁻³⁶

The reaction are:

CuX → Cu²⁺ + X²⁺, Ksp=1.27x10⁻³⁶

Cu²⁺ + 4CN⁻ → (Cu(CN)₄)²⁻, Kf=1x10²⁵

The overall reaction is:

CuX + 4CN⁻ → (Cu(CN)₄)²⁻ + X²⁺

The equilibrium constant is:

K=Ksp*Kf=1.27x10⁻³⁶*1x10²⁵=1.27x10⁻¹¹

              CuX      +        4CN⁻      →       (Cu(CN)₄)²⁻ +        X²⁺

I              -                      0.2                    0                         0

C            -                       -4                     +x                        +x

E             -                      0.2-4                x                          x

The equation for equilibrium is:

K=\frac{[Cu(CN)4]^{2} [X]}{[CN]^{4} } \\1.27x10^{-11} =\frac{x^{2} }{(0.2-x)^{4} }

Here, solving for x:

x=1.425x10⁻⁷M=CuX

Tomtit [17]3 years ago
3 0

Answer:

Solubility= 1.08×10-12

Explanation:

Take the cube root of 1.27×10-36

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Molarity of Ag+ is less than the molar solubility thus ppt will not occur.

Balanced reaction-:

<h3>2AgNO3(aq)+K2CrO4(aq)→Ag2CrO4(s)+2KNO3(aq)</h3>

Moles of AgNO3=mass(g)molar mass (g/mol) =2.7×10−5g / 169.86 gmol

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<h3>What is the molarity calculation formula?</h3>

The volume of solvent required to dissolve the provided solute is multiplied by the ratio of the moles of the solute whose molarity has to be computed. (M=frac{n}{V}) The molality of the solution that needs to be computed in this case is M. n is the solute's molecular weight in moles.

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Explanation:

THIS IS THE COMPLETE QUESTION BELOW;

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In the first step, calcium carbide and water react to form acetylene and calcium hydroxide: CaC (s) + 2 H2O(g) - CH (9) + Ca(OH),(s) AH -414. kJ In the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6 C H (9) + 3 CO2(9) + 4H2O(g) - SCH,CHCO,H) AH-132. kJ Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ?

The two equations from the reaction can be written as;

a)CaC₂(s) + 2H₂O(l) ------->C₂H₂(g) + CaOH₂(s)

Δ H= -414Kj ........................ equation (a)

b)6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj ...................... equation (b)

In equation (b)acrylic acid was produced by the reaction between Acetylene carbon dioxide and water

Then we can multiply equation(a) by factor of 6 and the ΔH Then we have (6× -414Kj)= ΔH= -2484Kj.

6CaC₂(s) + 12H₂O(l) ------->6C₂H₂(g) + 6CaOH₂(s)

Δ H= -2484Kj.................. equation (c)

6C₂H₂(g) +3CO₂(g)+4H₂O(g) -------> 5CH₂CHCO₂H(g) Δ H= 132Kj

Then add equation (c) and equation(b) then we have

6CaC₂(s) + 16H₂O(l)+3CO₂(g)------> 5CH₂CHCO₂H(g) + 6CaOH₂(s) ΔH= -2352Kj

ΔH(net)= -2352Kj/5moles

=-471Kj/mole

therefore, net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. Round your answer to the nearest kJ. x 5 ? is -471Kj/mole acrylic acid

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