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Zielflug [23.3K]
3 years ago
8

Calculate the morality of each of the following solutions: a. 15.4 g KCl in 289.2 mL solution b. 14.4 g of CaCl2 in 0.614 L solu

tion c. 28.0 mL of 3.00 M H2SO4 diluted to 0.250 L
Chemistry
1 answer:
Nesterboy [21]3 years ago
4 0
The answer is:
a. 0.712 M
b. 0.210 M
c. 0.336 M

Molarity is a measure of the concentration of solute in a solution.
It can be expressed as moles of solute ÷ volume of solution:
c = n ÷V
where:
c - concentration of solute,
n - moles of solute
V - volume of solution

n can be expressed as:
<span>n = m ÷ Mr
</span>where:
<span>n - moles of solute
</span>m - mass of solute
Mr - relative molecular mass

a. We know volume:
V = 289.2 mL = 0.2892 L
We need n and c.

n = m ÷ Mr
m = 15.4 g
Mr (<span>KCl) = 74.55 g/mol
n = </span>15.4 g ÷ <span>74.55 g/mol
n = 0.206 mol</span>

Thus, 
c = 0.206 mol ÷ <span>0.2892 L
c = 0.712 mol/L = 0.712 M

</span>b. We know volume:
V = 0.614 L
We need n and c.

n = m ÷ Mr
m = 14.4 g
Mr (CaCl₂<span>) = 110.98 g/mol
n = </span>14.4 g ÷ <span>110.98 g/mol
n = 0.129 mol</span>

Thus, 
c = 0.129 mol ÷ <span>0.614 L
c = 0.210 mol/L = 0.210 M
</span>
c. We can use formula:
m₁V₁ = m₂V₂
<span>m₁ = 3 M
</span><span>V₁ = 28 mL= 0.028 L
</span><span>m₂ = ?
</span><span>V₂ = 0.250 L
</span>Thus:
3 M × 0.028 L = m₂× <span>0.250 L
</span> m₂ = 0.336 M
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How many bromine atoms are present in 37.9 g of CH2Br2?
VARVARA [1.3K]

Answer:

The answer to your question is: 6.55 x 10 ²³ atoms of Br

Explanation:

CH2Br2 = 37.9 g

MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g

                   174 g of CH2Br2 ------------------  160 g of Br2

                   37.9 g of CH2Br2   ---------------     x

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