Answer:
T2 = 260 K
Explanation:
<em>Given data:</em>
P1 = 150.0 k Pa
T1 = (-23+ 273.15) K = 250.15 K
V1 = 1.75 L
P2 = 210.0 kPa
V2 = 1.30 L
<em>To find:</em>
T2 = ?
<em>Formula:</em>
<em>Calculation:</em>
T2 = (210.0 kPa) x (1.30 L) x (250.15 K) / (150.0 kPa) x (1.75 L)
T2 = 260 K
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
6 miles of H2O is equal to 12 Hydrogen molecules and 6 oxygen molecules. Equaling 18 in total.
They exist in the outer orbitals
