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Vaselesa [24]
3 years ago
9

Which of the following will allow measurement of a liquids volume with the greatest precision ?​

Chemistry
2 answers:
Evgesh-ka [11]3 years ago
6 0
I personally think that it’s C.
(i’m sorry if it’s wrong)
Oliga [24]3 years ago
3 0

Answer:

the answer is B)........

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Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial concentration of NOBr = 0.878 M, what is the equilibrium
pav-90 [236]

Answer:

The equilibrium concentration of NO is 0.02124 M.

Explanation:

Given that,

Initial concentration of NOBr = 0.878 M

k_{c}=3.07\times10^{-4}

Temperature = 24°C

We know that,

The balance equation is

2NOBr\Rightarrow 2NO+Br_{2}

Initial concentration is,

0.878\Rightarrow 0+0

Concentration is,

-2x\Rightarrow 2x+x

Equilibrium concentration

0.878-2x\Rightarrow 2x+x

We need to calculate the value of x

Using formula of concentration

k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}

Put the value into the formula

3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}

2x^2=3.07\times10^{-4}\times(0.878)^2+3.07\times10^{-4}\times4x^2-2\times2x\times0.878\times3\times10^{-4}

2x^2=0.0002367+0.001228x^2-0.0010536x

2x^2-0.001228x^2+0.0010536x-0.0002367=0

1.998772x^2+0.0010536x-0.0002367=0

x=0, 0.01062

We need to calculate the equilibrium concentration of NO

Using formula of concentration of NO

concentration\ of\ NO=2x

Put the value of x

concentration\ of\ NO=2\times0.01062

concentration\ of\ NO=0.02124

Hence, The equilibrium concentration of NO is 0.02124 M.

7 0
4 years ago
An unknown gas effuses 1.66 times more rapidly than CO2. What is the molar mass of the unknown gas, given that the molar mass of
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The molar mass is 16.0 grams 
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What is the name of the process shown in the above picture?
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Answer:

C is the answer to the question

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A stable atom that has a large nucleus most likely contains
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The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati
a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80M^{-1}s^{-1}

t = time taken  = 95 s

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0
3 years ago
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