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Charra [1.4K]
3 years ago
5

Plzzz help will do anything ;) How many molecules are there in 2.25 moles of H2CO3

Chemistry
1 answer:
Sonbull [250]3 years ago
3 0
It’s 8 molecules !!!.8/8/8/
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Find the number of neutron in an atom represented by 45X21
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C-12 and C-13 are naturally-occurring isotopes of the element carbon. C-12 occurs 98.88% of the time and C-13 occurs 1.108% of t
Veseljchak [2.6K]

Answer:

c) (12×0.9889) + (13×0.01108)

Explanation:

Given data:

Percentage of C-12 = 98.89%

Percentage of C-13 = 1.108%

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Solution:

98.89/100 = 0.9889

1.108/ 100 = 0.01108

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3 years ago
Hello, <br><br> So I was wondering if this is correct... Is it?
Sonja [21]
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4 0
2 years ago
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What is a good essential question for the subject molecular compounds
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Why do molecules combined into chains?
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3 years ago
When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
Amanda [17]

<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
3 years ago
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