Mass percentage of a solution is the amount of solute present in 100 g of the solution.
Given data:
Mass of solute H2SO4 = 571.3 g
Volume of the solution = 1 lit = 1000 ml
Density of solution = 1.329 g/cm3 = 1.329 g/ml
Calculations:
Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g
Therefore we have:
571.3 g of H2SO4 in 1329 g of the solution
Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987
Mass percentage of H2SO4 (%w/w) is 42.99 %
Answer:
- <em>The partial pressure of oxygen in the mixture is</em><u> 320.0 mm Hg</u>
Explanation:
<u>1) Take a base of 100 liters of mixture</u>:
- N: 60% × 100 liter = 60 liter
- O: 40 % × 100 liter = 40 liter.
<u>2) Volume fraction:</u>
At constant pressure and temperature, the volume of a gas is proportional to the number of molecules.
Then, the mole ratio is equal to the volume ratio. Callin n₁ and n₂, the number of moles of nitrogen and oxygen, respectively, and V₁, V₂ the volume of the respective gases you can set the proportion:
That means that the mole ratio is equal to the volume ratio, and the mole fraction is equal to the volume fraction.
Then, since the law of partial pressures of gases states that the partial pressure of each gas is equal to the mole fraction of the gas multiplied by the total pressure, you can draw the conclusion that the partial pressure of each gas is equal to the volume fraction of the gas in the mixture multiplied by the total pressure.
Then calculate the volume fractions:
- Volume fraction of a gas = volume of the gas / volume of the mixture
- N: 60 liter / 100 liter = 0.6 liter
- V: 40 liter / 100 liter = 0.4 liter
<u>3) Partial pressures:</u>
These are the final calculations and results:
- Partial pressure = volume fraction × total pressure
- Partial pressure of N = 0.6 × 800.0 mm Hg = 480.0 mm Hg
- Partial pressure of O = 0.4 × 800.0 mm Hg = 320.0 mm Hg
Protons, neutron, and elecrons
Answer:
It is a beta decay equation unknown
Explanation:
none
