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2 years ago

Which one of the following salts has the highest molar solubility in water?

Chemistry

1 answer:

svetlana [45]2 years ago

4 0

Answer:

Correct option is

; K

=6.2×10

−12

(A) For CaCO

,S=

Ksp

8.7×10

−9

=9.3×10

−5

(B) For CuS,S=

Ksp

8.5×10

−45

=9.2×10

−23

,S=(Ksp/4)

1/3

=(6.2×10

−12

/4)

1/3

=1.15×10

−4

(D) For Pb(IO

)

,S=(Ksp/4)

1/3

=(2.6×10

−13

/4)

1/3

=4×10

−5

Hence, Ag

has the greatest molar solubility in pure water.

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3 years ago

What is the main difference between ionic<br> compounds and molecular compounds?

Sophie [7]

Answer:

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2 years ago

For a particular isomer of C 8 H 18 , C8H18, the combustion reaction produces 5104.1 kJ 5104.1 kJ of heat per mole of C 8 H 18 (

vlada-n [284]

Answer:

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

Explanation:

Step 1: Data given

The combustion reaction of octane produces 5104.1 kJ per mol octane

Step 2: The balanced equation

C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol

Step 3:

∆H°rxn = ∆H°f of products minus the ∆H° of reactants

∆H°rxn = ∆H°f products - [∆H°f reactants]

-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)

∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol

∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ /mol

∆H°f C8H18 = -220.1 kJ/mol

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

7 0

2 years ago

Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl

tensa zangetsu [6.8K]

Answer:

$[SO_2Cl_2] = 0.09983 M$

Explanation:

Write the balance chemical equation ,

$SO_2Cl_2((g) = SO_2(g) + Cl_2(g)$

initial concenration of $SO_2Cl_2((g) =0.1M$

lets assume that degree of dissociation= $\alpha$

concenration of each component at equilibrium:

$[SO_2Cl_2] = 0.1-0.1\alpha$

$[SO_2] = 0.1\alpha$

$[Cl_2] = 0.1\alpha$

$Kc =\frac{0.1\alpha \times 0.1\alpha}{0.1-0.1\alpha}$

$Kc =\frac{0.1\alpha \times \alpha}{1-\alpha}$

as $\alpha$ is very small then we can neglect $1-\alpha$

therefore ,

$Kc ={0.1\alpha \times \alpha}$

$\alpha =\sqrt{\frac{Kc}{0.1}}$

$\alpha = 1.73 \times 10^{-3}$

Eqilibrium concenration of $[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173$

$[SO_2Cl_2] = 0.09983 M$

4 0

2 years ago