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3 years ago

For the reaction H2O) =H% + OH", if x moles of H and y moles of OH are formed K. is y A. *?/(1 – x) C. 1/xy D. xy

Chemistry

1 answer:

hjlf3 years ago

5 0

Answer: Correct option is B)

The standard enthalpy change of formation of a compound is the enthalpy change which occurs when one mole of the compound is formed from its elements under standard conditions. The equation showing the standard enthalpy change of formation of water is

H
2
(g)+

2
1

 O

 (g)⟶H

2 ↑

 O(l) 1 mole of water formed.

∴ Enthalpy of formation is −X2kJ/mol.

________________________________________________________

Explanation:I hope this helped

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Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

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

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

3 years ago

If a piece of metal were to cut in half would the density change

nadezda [96]

Density depends on both the mass and the volume of an object. If you cut a bar of gold in half, you would have two bars with half the mass of the original bar. However, each bar would also have half the volume of the original bar. The density of gold does not change.

8 0

3 years ago

Read 2 more answers

This green solution of chromium(III) can further be reduced by zinc metal to a blue solution of chromium(II) ions. Write the bal

Contact [7]

Answer:

Half-reactions:

Cr³⁺ + 1e⁻ → Cr²⁺; Zn → Zn²⁺ + 2e⁻

Net ionic equation:

2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺

Explanation:

The Cr³⁺ is reduced to Cr²⁺:

<h3>Cr³⁺ + 1e⁻ → Cr²⁺ -Half-reaction 1-</h3>

Zn is oxidized to Zn²⁺:

<h3>Zn → Zn²⁺ + 2e⁻ -Half-reaction 2-</h3>

Twice the reduction of Cr:

2Cr³⁺ + 2e⁻ → 2Cr²⁺

Now this reaction + Oxidation of Zn:

2Cr³⁺ + 2e⁻ + Zn → 2Cr²⁺ + Zn²⁺ + 2e⁻

<h3>2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ - Net ionic equation</h3>

6 0

3 years ago

When baking soda and vinegar react, the product includes bubbles. what is most likely occurred

Varvara68 [4.7K]

The gases that get released form bubbles in the solution

4 0

3 years ago

Read 2 more answers

A chemist prepares a solution of mercury(I) chloride Hg2Cl2 by measuring out

Genrish500 [490]

Answer:

1.26 × 10^-8 M

Explanation:

We are given;

Number of moles of mercury (i) chloride as 0.000126 μmol

Volume is 100 mL

We are required to calculate the concentration of the solution.

We need to know that;

Concentration is also known as molarity is given by;

Molarity = Number of moles ÷ Volume

Number of moles = 1.26 × 10^-10 Moles

Volume = 0.01 L

Therefore;

Concentration = 1.26 × 10^-10 Moles ÷ 0.01 L

= 1.26 × 10^-8 M

Thus, the molarity of the solution is 1.26 × 10^-8 M

6 0

4 years ago

For the reaction H2O) =H% + OH", if x moles of H and y moles of OH are formed K. is y A. *?/(1 – x) C. 1/xy D. xy ​

For the reaction H2O) =H% + OH", if x moles of H and y moles of OH are formed K. is y A. *?/(1 – x) C. 1/xy D. xy