Answer: CoBr3 < K2SO4 < NH4 Cl
Justification:
1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.
2) The formula for the depression of freezing point is:
ΔTf = i * Kf * m
Where i is the van't Hoof factor which accounts for the dissociation of the solute.
Kf is the freezing molal constant and only depends on the solvent
m is the molality (molal concentration).
3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result in the solution with higher depression of the freezing point (lower freezing point).
4) These are the dissociations of the given solutes:
a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles
b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles
c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles
5) So, the rank of solutions by their freezing points is:
CoBr3 < K2SO4 < NH4 Cl
The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m
<h3>What is Stack Height ?</h3>
Stack height means the distance from the ground-level elevation at the base of the stack to the crown of the stack.
If a stack arises from a building or other structure, the ground-level elevation of that building or structure will be used as the base elevation of the stack.
Given is a steel stack that exhausts 1,200 cu.m/min of gases
P= 1 atm and
T= 400 K
maximum expected wind speed at stack height of 12 m/s
The formula for the diameter of chimney
Q =1200 cu.m/min
= 1200 * 0.0166 = 19.92 cu.m/sec
Velocity = 12m/s
d= 1.45 m
Therefore The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m.
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Answer:
The answers are in the explanation.
Explanation:
The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:
Increasing temperature of ice from -10°C - 0°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g
Q = 2.06J/g°C*10°C*10g
Q = 206J
Change from solid to liquid:
The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:
Q = 333.55J/g*10g
Q = 3335.5J
Increasing temperature of liquid water from 0°C - 100°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g
Q = 4.18J/g°C*100°C*10g
Q = 4180J
Change from liquid to gas:
The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:
Q = 2260J/g*10g
Q = 22600J
Increasing temperature of gas water from 100°C - 120°C:
Q = S*ΔT*m
Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g
Q = 1.87J/g°C*20°C*10g
Q = 374J
Total Energy:
206J + 3335.5 J + 4180J + 22600J + 374J =
30695.5J =
30.7kJ
D, Neon and argon are both noble gases and contain almost the same elements.