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3 years ago

Consider the following system at equilibrium where \Delta H° = 18.8 kJ, and Kc = 9.52×10-2, at 350 K.

Chemistry

1 answer:

slega [8]3 years ago

7 0

Answer:

1. C. remains the same. 2. C. is less than Kc. 3. B. run in the reverse direction to reestablish equilibrium. 4. A. increase.

Explanation:

At constant temperature, the equilibrium concentration has not effect on the equilibrium constant because the rate constants do not change with change in the concentrations or amounts of the reactants or products. Change in the concentration of one reactant or product causes the concentration of the others to change so as to maintain a constant value for the equilibrium constant. On the other hand, the reaction quotient is used to measure the relative amounts of reactants and products during a chemical reaction at any point in time. The value of the reaction quotient shows the direction of the chemical reaction.

Therefore, when 0.31 moles of CCl4(g) are removed from the equilibrium system at constant temperature:

1. the value of Kc remains the same

2. the value of Qc is less than Kc

3. the reaction must run in the reverse direction to reestablish equilibrium

4. the concentration of $CH_{4}$ will increase because product will be converted to reactants to reestablish equilibrium.

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A cup of coffee has 71 mL of coffee and 127 mL of water. What is the percent volume of the coffee solution?

Anna [14]

Answer:

35.9%

Explanation:

The percent volume of the coffee solution can be calculated as follows:

% volume of coffee solution = volume of coffee/total volume of coffee solution × 100

According to this question, a cup of coffee has 71 mL of coffee and 127 mL of water. This means that, the total volume of coffee solution is;

71mL + 127mL = 198mL

% volume = 71/198 × 100

= 0.359 × 100

Percent volume of coffee solution = 35.9%

3 0

3 years ago

When two gases of a chemical reaction are both at the same temperature and pressure and they have the same molar volume, what is

Rudik [331]

When two gases of a chemical reaction are at the same temperature, pressure and molar volume, then the stoichiometric ratio of the gases would be 1 is to 1. Molar volume is the volume of the gas per mole of the gas. Having the same conditions for both gases would mean that they are present with the same number of moles.

4 0

3 years ago

Read 2 more answers

An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To

antoniya [11.8K]

Answer: The new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}$ ........(1)

We are given:

$P_{PCl_5}=217.0torr$

$P_{PCl_3}=13.2torr$

$P_{Cl_2}=13.2torr$

Putting values in above equation, we get:

$K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24$

Now we have to calculate the new partial pressure of $Cl_2$ .

$P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}$

$217.0torr+13.2torr+P_{Cl_2}=263.0torr$

$P_{Cl_2}=32.8torr$

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of $Cl_2$ .

Now, the equilibrium is shifting to the reactant side. The equation follows:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initial: 13.2 32.8 217.0

At eqm: 13.2-x 32.8-x 217.0+x

Putting values in expression 1, we get:

$1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38$

Neglecting the 40.4 value of 'x' because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of $PCl_5$ = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of $PCl_3$ = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of $Cl_2$ = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0

3 years ago

In a chemical reaction, a certain amount of hydrochloric acid reacts with 10 grams of sodium hydroxide to produce 19 grams of a

11Alexandr11 [23.1K]

Sodium chloride is the answer I believe

8 0

3 years ago

Newton’s second law of motion is F=ma A net force of 60 N north acts on an object with a of 30 kg. Use Newton’s second law of mo

Luda [366]

<u>Answer:</u> The acceleration of the object is $2m/s^2$ . If net force increases, acceleration will also increase and if mass increases, the acceleration will decrease.

<u>Explanation:</u>

Force is defined as the product of object's mass and acceleration.

Mathematically,

$F=ma$ ......(1)

or,

$a=\frac{F}{m}$ .....(2)

where,

F = Force exerted on an object = 60N

m = mass of an object = 30kg

a = acceleration of the object = ?

Putting values in above equation, we get:

$a=\frac{60kg.m/s^2}{30kg}=2m/s^2$

The acceleration of the car is $2m/s^2$ .

From equation 2, it is visible that acceleration is directly proportional to force. This means that \if force increases, acceleration also increases.

And acceleration is inversely proportional to mass of the object. This means that if mass increases, the acceleration decreases.

Hence, if net force increases, acceleration will also increase and if mass increases, the acceleration will decrease.

8 0

3 years ago