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statuscvo [17]
2 years ago
11

What is the value of for this aqueous reaction at 298 k? a b↽−−⇀c dδ°=20. 46 kj/mol

Chemistry
1 answer:
sineoko [7]2 years ago
5 0

The value for this aqueous reaction at 298 k? a b↽−−⇀c dδ°=20. 46 KJ/mol is 9.91 mol. in equilibrium

<h3>What is an aqueous reaction in equilibrium?</h3>

When a chemical reaction happens at the liquid state and the formation of reactant and product is the same then the reaction is known as an aqueous reaction in equilibrium denoted by K.

δG = − R T ln

          R = universal gas constant 8.313

          δG= 20. 46 kj/mol

           T =  298 k or 24.4 in celcius.

substituting the value in the equation.

20. 46 kj/mol = 8.313 × 24.4 in celcius × K

K =  8.313 × 24.4 in celcius / 20. 46 kj/mo

k = 9.91 mol .

Therefore, The value of this aqueous reaction at 298 k? a b↽−−⇀c dδ°=20. 46 KJ/mol is 9.91 mol. in equilibrium

Learn more about the aqueous reaction in  equilibrium, here:

brainly.com/question/8983893

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For the following elements in the choices, these are their values of Z*:

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Gaseous ammonia chemically reacts with oxygen O2 gas to produce nitrogen monoxide gas and water vapor. Calculate the moles of ox
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Now we have to calculate the moles of oxygen.

From the balanced chemical reaction we conclude that,

As, 6 moles of water vapor produces from 5 moles of oxygen

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the pressure in a sealed plastic container is 108 kPa at 41 degrees Celsius. What is the pressure when the temperature drops to
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<u>Answer:</u> The new pressure will be 101.46 kPa.

<u>Explanation:</u>

To calculate the new pressure, we use the equation given by Gay-Lussac Law. This law states that pressure is directly proportional to the temperature of the gas at constant volume.

The equation given by this law is:

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are initial pressure and temperature.

P_2\text{ and }T_2 are final pressure and temperature.

We are given:

By using conversion factor:   T(K)=T(^oC)+273

P_1=108kPa\\T_1=41^oC=314K\\P_2=?kPa\\T_2=22^oC=295K

Putting values in above equation, we get:

\frac{108kPa}{314K}=\frac{P_2}{295K}\\\\P_2=101.46kPa

Hence, the new pressure will be 101.46 kPa.

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