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3 years ago

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

2 answers:

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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3 years ago

Consider this equilibrium reaction between carbon monoxide and hydrogen gas, occurring in a sealed flexible container. CO(g) + 3

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More H2(g) is added to the container : Towards products.

CO is removed from the container : Towards reactants.

More CH4(g) is added to the container : Towards reactants

H2O(g) is removed from the container : Towards products.

The contents of the container are heated up. : Towards the reactants.

The contents of the container are cooled down : Towards the products.

The pressure inside the container is increased. :Towards the products

The container is stretched to increase the volume: Towards the reactants.

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CO(g) + 3 H2g) → CH4(g) + H2O(g)+ heat

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If the H2O is removed, the system will try to change this situation by shifting the balance to the right, and thus the concentration of products will increase, the concentration of products will decrease. Towards products.

If the temperature is increased, the system will reduce the amount of heat released. So the balance will shift to the left. Towards the reactants.

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