Answer:
Volume of water at this temperature is 27.2 mL
Explanation:
We know that 
Here density of water is 0.992 g/mL
Here mass of water is 27.0 g
So 
= 
= 27.2 mL
Answer:
Q = 2640.96 J
Explanation:
Given data:
Mass of He gas = 10.7 g
Initial temperature = 22.1°C
Final temperature = 39.4°C
Heat absorbed = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree. Specific heat capacity of He is 14.267 J/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 39.4°C - 22.1°C
ΔT = 17.3°C
Q = 10.7 g× 14.267 J/g.°C × 17.3°C
Q = 2640.96 J
D. 1,008 liters because you are looking for liters from a calculation of moles. Recognizing that you can do STP (22.4L) you multiply this number by 45 moles and it is 1,008 liters
Answer:
calcium oxide is injected into the final stage of the scubber, wich then reacts with the sulfur dioxide to form calcium sulfite.
Explanation:
B. Distillation
Distilling is when you boil a substance and collect the vapors.
