Answer:
Original temperature (T1) = - 37.16°C
Explanation:
Given:
Gas pressure (P1) = 2.75 bar
Temperature (T2) = - 20°C
Gas pressure (P2) = 1.48 bar
Find:
Original temperature (T1)
Computation:
Using Gay-Lussac's Law
⇒ P1 / T1 = P2 / T2
⇒ 2.75 / T1 = 1.48 / (-20)
⇒ T1 = (2.75)(-20) / 1.48
⇒ T1 = -55 / 1.48
⇒ T1 = - 37.16°C
Original temperature (T1) = - 37.16°C
Answer:
Explanation:
You simply have to multiply it by the molar mass =
m = 50mol SnSO4 x 214.773 g/mol
m = 10.738 g SnSO4
Answer:
I agree
Explanation:
For scientists working in classified areas, collaboration with university programs and researchers provides opportunities to expand and strengthen their science through the conduct of peer-reviewed, open literature research.
Answer:
a. 50KCal
b. 400KCal
c. Same as (a) above
Explanation:
Given
To raise the temperature of 1kg of liquid water at 1°C requires 1KCal
To raise the temperature of 1kg of ice or water vapour by 1°C requires 0.5KCal
To melt 1kg of ice at 0°C requires 80KCal
To evaporate 1kg of liquid water sitting at 100°C requires 540KCal
a. How much heat is required to raise the temperature of 5 kg of liquid water by 20 C?
To raise the temperature of 5 kg of ice by 20°C requires:
5 kg * (0.5 kcal / kgC) * 20C
= 50 KCal
b. How much heat is required to melt 5 kg of ice at 0 C?
To melt an ice of 5 kg of ice at 0 C requires:
5 kg * (80 kcal / kg)
= 400 KCal
c. Same as (a) above
