Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
Answer:
P = 13.5 atm
Explanation:
Given that
No. of moles, n = 20 moles
Volume of nitrogen gas = 36.2 L
Temperature = 25°C = 298 K
We need to find the pressure of the gas. Using the ideal gas equation
PV = nRT
Where
R is gas constant,
So,
so, the pressure of the gas is equal to 13.5 atm.
Answer:
1367.7 g of ethylene glycol was added to the solution
Explanation:
In order to find out the mass of glycol we added, we apply the colligative property of lowering vapor pressure: ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution(P')
525.8 mmHg - 451 mmHg = 451 mmHg . Xm
74.8 mmHg / 451 mmHg = Xm → 0.166 (mole fraction of solute)
Xm = Mole fraction of solute / Moles of solute + Moles of solvent
We can determine the moles of solvent → 2000 g . 1 mol/18 g = 111.1 mol
(Notice we converted the 2kg of water to g)
0.166 = Moles of solute / Moles of solute + 111.1 moles of solvent
0.166 (Moles of solute + 111.1 moles of solvent) = Moles of solute
18.4 moles = Moles of solute - 0.166 moles of solute
18.4 = 0.834 moles of solute → Moles of solute = 18.4/0.834 = 22.06 moles
Let's convert the moles to mass → 62 g/mol . 22.06 mol = 1367.7 g
Answer:
d. A water particle and an air particle
Explanation:
The force of gravity (F) between two objects of masses m1 and m2 and separated by a distance r is given as:
where G is the gravitational constant
This force is therefore, directly proportional to the masses and inversely related to the distance between them.
Based on the given options, since the masses of the water and air particles are very small (masses of earth, moon and sun is relatively huge), the gravitational force between them would be negligible and difficult to measure.