Answer:
Explanation:
Given:
Depth = h
Width = w
Length = 1
Now,
Let ρ be the density of water
Thus,
Pressure force at the bottom will be
= Pressure × Area
= ρgh × (w × 1)
also,
Pressure force at the sides will be = 
ρgh × (h × w)
Therefore,
The ratio of the pressure force at the bottom to the pressure force on the side of the tank will be
= 
=
Answer:
The molecules in a solid are close together and they vibrate in place. The molecules in a liquid move quickly and farther apart from eachother. The particles in a gas move freely at high speeds and slide past eachother.
Hot water rises cold water sinks. So the warm water will stay at the bottom. Transfer of heat through the molecules will make all the water boil
Answer:
+523 kJ.
Explanation:
The following data will be used to calculate the average C-S bond energy in CS2(l).
S(s) ---> S(g)
ΔH = 223 kJ/mol
C(s) ---> C(g)
ΔH = 715 kJ/mol
Enthalpy of formation of CS2(l)
ΔH = 88 kJ/mol
CS2(l) ---> CS2(g)
ΔH = 27 kJ/mol
CS2(g) --> C(g) + 2S(g)
So we must construct it stepwise.
1: C(s) ---> C(g) ΔH = 715 kJ
2: 2S(s) ---> 2S(g) ΔH = 446 kJ
adding 1 + 2 = 3
ΔH = 715 + 446
= 1161 kJ
3: C(s) + 2S(s) --> C(g) + 2S(g) ΔH = 1161 kJ
4: C(s) + 2S(s) --> CS2(l) ΔH = 88 kJ
adding (reversed 3) from 4 = 5
ΔH = -1161 + 88
= -1073 kJ
5: C(g) + 2S(g) --> CS2(l) ΔH = -1073 kJ
6: CS2(l) ---> CS2(g) ΔH = 27 kJ
adding 5 + 6 = 7
ΔH = -1073 + 27
= -1046 kJ
7. C(g) + 2S(g) --> CS2(g) ΔH = -1046 kJ
Reverse and divide by 2 for C-S bond enthalpy
= -(-1046)/2
= +523 kJ.
I think the answer is A, ozone depletion
