Answer:
- Sedimentation: Sedimentation is the process of allowing particles in suspension in water to settle out of the suspension under the effect of gravity.
- Decantation: Decantation is a process for the separation of mixtures of immiscible liquids or of a liquid and a solid mixture
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Answer is: the absolute pressure of the air in the balloon is 1.015 atm (102.84 kPa).
n = 0.250 mol; amount of substance.
V = 6.23 L; volume of the balloon.
T = 35°C = 308.15 K; temperature.
R = 0.08206 L·atm/mol·K, universal gas constant.
Ideal gas law: p·V = n·R·T.
p = n·R·T / V.
p = 0.250 mol · 0.08206 L·atm/mol·K · 308.15 K / 6.23 L.
p = 1.015 atm; presure of the air.
Answer:
25.42 atm
Explanation:
Data Given:
Volume of a gas ( V )= 2.00 L
temperature of a gas ( T ) = 310 K
number of moles (n) = 2 mol
Pressure of a gas ( P ) = to be find
Solution:
Formula to be used
PV= nRT
Rearrange the above formula
P = nRT / V . . . . . . . . . . (1)
Where R is ideal gas constant
R = 0.08205 L atm mol⁻¹ K⁻¹
Put values in equation 1
P = nRT / V
P = 2 mol x 0.08205 L atm mol⁻¹ K⁻¹ x 310 k / 2 L
P = 50.84 L atm / 2 L
P = 25.42 atm
P ressure of gas (P) will be = 25.42 atm