1.59 moles
112211111111111112
- 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
Development of carbonation
You HAVE TO know that molarity (M) tells you the number of moles of a solute per Liters of solution.
M=mol/L
0.3M = 2.7 mol/ Volume
Volume = 2.7 mol/3.0 L
Volume = 0.9 L
Answer:
No.of moles of C is , n = mass/molar mass = 75.46 g / 12 (g/mol) = 6.3 moles No.of moles of H is , n' = mass/molar mass = 4.43 g / 1.0(g/mol) = 4.43 moles No.of moles of O is , n'' = mass/molar mass = 20.10 g / 16(g/mol) =1.25 moles Ratio to the no.of moles of C,H& O is 6.3 : 4.43 : 1.25 In the simple integer ratio is ( 6.3/1.25) : ( 4.43/1.25) : (1.25/1.25) 5.04 :3.5 : 1
Explanation:
