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Dafna11 [192]
3 years ago
10

What molarity is prepared when a solution contains 0.277 moles of calcium bromide in 500 mL of solution (must change mL to Liter

s)? *
Chemistry
1 answer:
Vladimir79 [104]3 years ago
5 0

Answer:

0.554M of Calcium Bromide

Explanation:

Molarity by defintion is #of moles of something/litres of solution.

Therefore, here, we have 0.277 moles of calcium bromide and 500mL (divide 500ml by 1000 to go from mL to L because for every 1L there's 1000mL) or 0.5L.

Molarity= 0.277/0.5 = 0.554M of Calcium Bromide

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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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