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Nady [450]
3 years ago
6

Calculate the pressures of NO, Cl2, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 8.2 a

tm NO and 4.1 atm Cl2. (Hint: Kp is relatively large; assume the reaction goes to completion then comes back to equilibrium.)
Chemistry
1 answer:
IRISSAK [1]3 years ago
7 0

The given question is incomplete. The complete question is as follows.

Calculate the pressures of NO, Cl_{2}, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 8.2 atm NO and 4.1 atm Cl_{2}. (Hint: KP is small; assume the reverse reaction goes to completion then comes back to equilibrium.)

      2NO(g) + Cl_{2} (g) \rightleftharpoons 2NOCl(g)

 K_{P} = 2.5 \times 10^{3}

Explanation:

According to the ICE table,

               2NO(g) + Cl_{2} (g) \rightleftharpoons 2NOCl(g)

Initial:      8.2              4.1              0

Change:    -4.1x           -x               +4.1x

Equilbm: (8.2 - 4.1x)   (4.1 - x)        +4.1x

Now, expression for K_{p} of the reaction is as follows.

           K_{P} = \frac{[NOCl]^{2}}{[NO]^{2}[Cl_{2}]}

    2.5 \times 10^{3} = \frac{(2x)^{2}}{(8.2 - 4.1x)(4.1 - x)}

                   x = 1.9

Therefore, at equilibrium

    [NOCl] = 2 \times 1.9 = 3.8

    [NO] = (8.2 - 7.79) = 0.41

    Cl_{2} = 2 - 1.9 = 0.1        

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メタン分子は、1つの炭素原子と4つの水素原子から作られています。炭素の質量は12.011uで、水素の質量は1.008uです。これは、1つのメタン分子の質量が12.011 u +(4×1.008u)、つまり16.043uであることを意味します。これは、1モルのメタンの質量が16.043グラムであることを意味します。^>^

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The following chemical equation is a balanced equation true or false
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The sum of the first 10 terms of an arithmetic progression is 120 and the sum of first twenty is 840. find sum of first 30 terms
STatiana [176]

Answer:

The sum of first 30 terms of the arithmetic progression is <u>2160.</u>

Explanation:

For an arithmetic progression, the sum of first n terms with first term as a and common difference d is given as:

S_n=\frac{n}{2}(2a+(n-1)d)

Now, it is given that:

For\ n=10,S_n=120\\For\ n=20,S_n=840

Now, plug in these values and frame two equations in a\ and\ d

S_{10}=\frac{10}{2}(2a+(10-1)d)\\120=5(2a+9d)\\2a+9d=\frac{120}{5}\\2a+9d=24------------1

S_{20}=\frac{20}{2}(2a+(20-1)d)\\840=10(2a+19d)\\2a+19d=\frac{840}{10}\\2a+19d=84-----------2

Now, we solve equations (1) and (2) for a\ and\ d. Subtract equation (1) from equation (2). This gives,

2a+19d-2a-9d=84-24\\19d-9d=60\\10d=60\\d=\frac{60}{10}=6

Now, plug in the value of d=6 in equation (1) and solve for a.

2a+9(6)=24\\2a+54=24\\2a=24-54\\2a=-30\\a=\frac{-30}{2}=-15

Plug in the values of a=-15,\ n=30\ and\ d=6 in the sum formula to find the sum of first 30 terms.

Now, the sum of first 30 terms is given as:

S_{30}=\frac{30}{2}(2(-15)+(30-1)(6))\\S_{30}=15(-30+29(6))\\S_{30}=15(-30+174)\\S_{30}=15(144)=2160

Therefore, the sum of first 30 terms of the arithmetic progression is 2160.

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