The process of chemical weathering by oxygen and water on iron is known as rusting........rust is a loose reddish brown layer of hydrated ferric oxide(Fe2O3.H2O)
- The molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol.
- If this sample was placed under extreme pressure, the volume of the sample will decrease.
<h3>How to calculate molar mass?</h3>
The molar mass of a substance can be calculated by first calculating the number of moles using ideal gas law equation:
PV = nRT
Where;
- P = pressure
- V = volume
- T = temperature
- R = gas law constant
- n = no of moles
0.98 × 1.2 = n × 0.0821 × 287
1.18 = 23.56n
n = 1.18/23.56
n = 0.05moles
mole = mass/molar mass
0.05 = 0.458/mm
molar mass = 0.458/0.05
molar mass = 9.15g/mol
- Therefore, the molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol
- If this sample was placed under extreme pressure, the volume of the sample will decrease.
Learn more about gas law at: brainly.com/question/12667831
Answer:
V₂ → 106.6 mL
Explanation:
We apply the Ideal Gases Law to solve the problem. For the two situations:
P . V = n . R . T
Moles are still the same so → P. V / R. T = n
As R is a constant, the formula to solve this is: P . V / T
P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:
(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C
((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂
58.66 mL.atm = 0.55 atm . V₂
58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL
Answer:
437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.
Explanation:
Let the mass of the first solution be x and second solution be y.
Amount solution required = 1250 kg
x + y = 1250 kg....[1]
Percentage of ethanol in required solution = 12% of 1250 kg
Percentage of ethanol in solution-1 = 5% of x
Percentage of ethanol in required solution = 25% of y
5% of x + 25% of y =12% of 1250 kg
x + 5y = 3000 kg...[2]
Solving [1] and [2] we :
x = 437.5 kg , y = 812.5 kg
437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.
