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s2008m [1.1K]
3 years ago
5

Calculate the standard potential, e°, for this reaction from its equilibrium constant at 298 k. 6.47 x 10^5

Chemistry
1 answer:
Elina [12.6K]3 years ago
4 0
The working equation for this is written below:

E° = 0.0592logK/n

So, we have to know the value of n first which represents the number of moles electron in the reaction. We cannot answer this because we are not given with the reaction. However, just suppose the reaction is:

Cu⁺ + 2e⁻ --> Cu

Then, n=2. Continuing,

E° = 0.0592log(6.47×10⁵)/2 =<em> 0.172 V</em>
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Answer:

B. summer

Explanation:

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Defines chemicals and their<br> threats
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A flask is filled with 6.0 atm of N2 and 6.0 atm of H2. The gases react and NH3 is formed. What is the pressure in the flask aft
777dan777 [17]

<u>Answer:</u> The total pressure in the flask is 8.0 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of nitrogen gas = 6.0 atm

Initial partial pressure of hydrogen gas = 6.0 atm

The chemical equation for the reaction of nitrogen and hydrogen gas follows:

N_2+3H_2\rightarrow 2NH_3

By Stoichiometry of the reaction:

3 atm of hydrogen gas reacts with 1 atm of nitrogen gas

So, 6.0 atm of hydrogen gas will react with = \frac{1}[3}\times 6=2atm of nitrogen gas

As, given amount of nitrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

Pressure of nitrogen gas left = [6 - 2] = 4.0 atm

By Stoichiometry of the reaction:

3 atm of hydrogen gas produces 2 atm of ammonia gas

So, 6.0 atm of hydrogen gas will produce = \frac{2}{3}\times 6=4atm of ammonia

Total pressure in the flask = [4 + 4] = 8.0 atm

Hence, the total pressure in the flask is 8.0 atm

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3 years ago
Fill in the blank for a stoichiometric iso-octane air mixture. C8H18 + ____ (O2+3.76N2)
kati45 [8]

Answer: Option 12.5 is the correct answer.

Explanation:

The given reaction is as follows.

           C_{8}H_{18} + x(O_{2} + 3.76N_{2}) \rightarrow Products

As it is a combustion reaction so, nitrogen will not take part in it and hence, it will remain the same on both sides of the reaction.

Also, it is known that in a combustion reaction oxygen reacts with a hydrocarbon and results in the formation of carbon dioxide and water. Therefore, for the above reaction we write the complete reaction equation as follows.

             C_{8}H_{18} + x(O_{2} + 3.76N_{2}) \rightarrow CO_{2} + H_{2}O + 3.76 N_{2}

or,         C_{8}H_{18} + xO_{2} \rightarrow CO_{2} + H_{2}O as nitrogen is not taking part in the reaction.

Number of atoms on reactant side are as follows.

C = 8

H = 18

O = 2

Number of atoms on product side are as follows.

C = 1

O = 3

H = 2

Therefore, to balance this equation we multiply oxygen on reactant side by 12.5. Also, we multiply carbondioxide by 8 and water by 9 on product side. Hence, the complete balanced chemical equation is as follows.

          C_{8}H_{18} + 12.5O_{2} \rightarrow 8CO_{2} + 9H_{2}O

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