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enot [183]
3 years ago
12

Each step in the following process has a yield of 70.0%. CH4+4Cl2⟶CCl4+4HCl CCl4+2HF⟶CCl2F2+2HCl The CCl4 formed in the first st

ep is used as a reactant in the second step. If 2.00 mol CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess. moles HCl : mol
Chemistry
1 answer:
Nadusha1986 [10]3 years ago
8 0

Answer:

n(HCl)=1.96 mol

Explanation:

CH4+4Cl2⟶CCl4+4HCl

CCl4+2HF⟶CCl2F2+2HCl

With ideal yields we will end up with 4 moles of HCl.

With 70% yields on every stage

n(HCl)=0.7*0.7*4=1.96 mol

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Using the first volume and temperature reading on the table as V1 and T1, solve for the unknown values in the table below. Remem
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<u>Answer:</u> Th value of A is 3.8\times 10^2K, value of B is 0.84 L, value of C is 1.1 L and the value of D is 2.2\times 10^2K

<u>Explanation:</u>

To calculate the missing values of volume and temperature, we will use Charles' Law.

This law states that volume is directly proportional to the temperature of the gas at constant number of moles and pressure. Equation given by this law is:

\frac{V_1}{T_1}=\frac{V_2}{T_2}      ....(1)

where,

V_1\text{ and }T_1 are initial volume and temperature of the gas

V_2\text{ and }T_2 are final volume and temperature of the gas

  • <u>For A:</u>

V_1=1.0L\\T_1=295K\\V_2=1.3L\\T_2=AK

Putting values in equation 1, we get:

\frac{1.0L}{295K}=\frac{1.3L}{A}\\\\A=383.5K\approx 3.8\times 10^2K

Whenever multiplication and division are involved, the answer must not contain, more number of significant figures as there are in the least precise term.

Hence, the value of A is 3.8\times 10^2K

  • <u>For B:</u>

V_1=1.0L\\T_1=295K\\V_2=BL\\T_2=250K

Putting values in equation 1, we get:

\frac{1.0L}{295K}=\frac{B}{250K}\\\\B=0.84L

Hence, the value of B is 0.84L

  • <u>For C:</u>

V_1=1.0L\\T_1=295K\\V_2=CL\\T_2=325K

Putting values in equation 1, we get:

\frac{1.0L}{295K}=\frac{C}{325K}\\\\C=1.1L

Hence, the value of C is 1.1L

  • <u>For D:</u>

V_1=1.0L\\T_1=295K\\V_2=0.75L\\T_2=DK

Putting values in equation 1, we get:

\frac{1.0L}{295K}=\frac{0.75L}{D}\\\\D=221.25\approx 2.2\times 10^2K

Hence, the value of D is 2.2\times 10^2K

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The chromosphere is the third and the lowermost layer of the sun and appears bright red. This chromosphere region is visible only during the total solar eclipse. Thus, option B is correct.

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During this eclipse, the chromosphere region of the sun is visible as it was hidden by the photosphere before the eclipse. It shines bright red and yellow as the spectral line of the hydrogen alpha is emitted.

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