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Alborosie
3 years ago
14

What is the percentage concentration by mass of a solution that contains 0.098 kg of H2SO4 in 500.0 g of H2O?

Chemistry
2 answers:
Setler [38]3 years ago
6 0

Answer: The percentage concentration by mass of a solution that contains 0.098 kg of H_2SO_4 in 500.0 g of H_2O is 16.4%

Explanation:

To calculate the mass percentage ,we use the formula:

\text{Mass percent}=\frac{\text{Mass of} H_2SO_4}{\text{Molar mass of} H_2SO_4+mass of water}\times 100

Mass of H_2SO_4 = 0.098 kg= 98 g

Mass of water = 500 g

Putting values in above equation, we get:

\text{Mass percent}=\frac{98g}{98+500}\times 100=16.4\%

Hence, percentage concentration by mass of a solution that contains 0.098 kg of H_2SO_4 in 500.0 g of H_2O is 16.4%

Arada [10]3 years ago
4 0
Mass of solute  in g :

1 kg ---------- 1000 g
0.098 kg ------ ??

0.098 x 1000 / 1 => 98.0 g

mass of solvent = 500.0 g 

Therefore:

% ( w/w ) = 98.0 / 98.0 + 500.0

% (w/w ) = 98.0 / 598.0

0.163 x 100 =>  16.3 %


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Which of the following pairs of elements is most likely to form an ionic compound? a oxygen and fluorine b sodium and aluminum c
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<em>Option B:</em>

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4 0
3 years ago
If the temperature on 244 mL of a gas is changed to 488 mL and 6 atm, at constant
Fittoniya [83]

Answer:

<h2>12 atm</h2>

Explanation:

To find the initial pressure we use the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the initial pressure

P_1 =  \frac{P_2V_2}{V_1}  \\

From the question we have

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We have the final answer as

<h3>12 atm</h3>

Hope this helps you

6 0
3 years ago
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