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3 years ago

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Which of the following is true regarding Keq?

1 answer:

grandymaker [24]3 years ago

6 0

Answer:

option b is true

Explanation:

Since the equilibrium constant Kp is defined as the ratio of [products]/[reactants] ( option c is false) , then Kp units depends on the order of reaction of the reactants and the number of reactants and products involved ( option a is false).

for example if the concentration is expressed as M (molarity)

Keq= [A]/[B]*[C] then Kep has as units 1/M

Also t a specific temperature , the equilibrium constant represents the relationship between concentrations of reactants and products , and its fixed at constant temperature (option b is true).

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Is Kl covalent or ionic?

kotegsom [21]

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3 years ago

A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan

Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = $\frac{18.0 g}{60 g/mol}=0.3 mol$

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

$Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}$

Molarity of the urea solution ;

$M=\frac{0.3 mol}{0.200 L}=1.50 M$

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

$m=d\times V=0.95 g/mL\times 200.0 mL=190 g$

$m = 190 g = 190 \times 0.001 kg = 0.19 kg$

(1 g = 0.001 kg)

Molality of the urea solution ;

$Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}$

$m=\frac{0.3 mol}{0.19 kg}=1.58 m$

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

7 0

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Explanation:

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At 25°C, the equilibrium constant Kc for the reaction in thesolvent CCl4 2BrCl <----> Br2 + Cl2 is 0.141. If the initial c

ivolga24 [154]

<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M

<u>Explanation:</u>

We are given:

Initial concentration of chlorine gas = 0.0300 M

Initial concentration of bromine monochlorine = 0.0200 M

For the given chemical equation:

$2BrCl\rightleftharpoons Br_2+Cl_2$

<u>Initial:</u> 0.02 0.03

<u>At eqllm:</u> 0.02-2x x 0.03+x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}$

We are given:

$K_c=0.141$

Putting values in above equation, we get:

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Neglecting the value of x = -0.96 because, concentration cannot be negative

So, equilibrium concentration of bromine gas = x = 0.00135 M

Hence, the equilibrium concentration of bromine gas is 0.00135 M

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