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3 years ago

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

Helga [31]

Answer:

My school has no stationary and my teacher has no stationary

Explanation:

cuz i am smart duh

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2 years ago

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%

Atomic mass = ?

Solution:

98.89/100 = 0.9889

1.108/ 100 = 0.01108

Atomic mass = (12×0.9889) + (13×0.01108)

Atomic mass = (11.8668 + 0.144034)

Atomic mass = 12.01084

6 0

3 years ago

Hello, So I was wondering if this is correct... Is it?

Sonja [21]

Looks correct but the second to last I would of put abiotic and biotic factors but I don’t know what’s right for you

4 0

2 years ago

Read 2 more answers

What is a good essential question for the subject molecular compounds

Mashutka [201]

Why do molecules combined into chains?

6 0

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]

Answer: The value of $K_c$ for the given reaction is 1.435

Explanation:

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:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 0.20

At eqllm: 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