Answer:
Mass of KNO3= 10g
Molar mass of KNO3 = 101.1032g/mol
Volume = 250ml = 0.25L
No of mole on of KNO3 = mass of KNO3/Molar mass of KNO3
no of mole of KNO3 = 10/101.1032
No of mole of KNO3 = 0.09891
molarity of KNO3 = no of mole of KNO3/Vol (L)
Molarity = 0.09891/0.25 = 0.3956M
Molarity of KNO3 = 0.3956M
Answer:
Explanation:
1)
Given data:
Initial volume of balloon = 0.8 L
Initial temperature = 12°C ( 12+273= 285 K)
Final temperature = 300°C (300+273 = 573 K)
Final volume = ?
Solution:
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 0.8 L .573 K / 285 K
V₂ = 458.4 L / 285
V₂ = 1.61 L
2)
Initial pressure = 204 kpa
Initial temperature = 29°C ( 29 + 273 = 302 K)
Final temperature = ?
Final pressure = 300 kpa
Solution:
P₁/T₁ = P₂/T₂
T₂ = T₁P₂/P₁
T₂ = 302 K . 300 kpa / 204 kpa
T₂ = 90600 K/ 204
T₂ = 444.12 K
3)
Given data:
Initial volume = 14 L
Initial pressure = 2.1 atm
Initial temperature = 100 K
Final temperature = 450 K
Final volume = ?
Final pressure = 1.2 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2.1 atm × 14 L × 450 K / 100 K × 1.2 atm
V₂ = 13230 L / 120
V₂ = 110.25 L
This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
Answer:
238,485 Joules
Explanation:
The amount of energy required is a summation of heat of fusion, capacity and vaporization.
Q = mLf + mC∆T + mLv = m(Lf + C∆T + Lv)
m (mass of water) = 75 g
Lf (specific latent heat of fusion of water) = 336 J/g
C (specific heat capacity of water) = 4.2 J/g°C
∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C
Lv (specific latent heat of vaporization of water) = 2,260 J/g
Q = 75(336 + 4.2×139 + 2260) = 75(336 + 583.8 + 2260) = 75(3179.8) = 238,485 J
Answer:
D
Explanation:I alredy know this i am in 7th gread
