Answer:
209.3 Joules require to raise the temperature from 10 °C to 15 °C.
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m × c × ΔT
Given data:
mass of water = 10 g
initial temperature T1= 10 °C
final temperature T2= 15 °C
temperature change =ΔT= T2-T1 = 15°C - 10°C = 5 °C
Energy or joules added to increase the temperature Q = ?
Solution:
We know that specific heat of water is 4.186 J/g .°C
Q = m × c × ΔT
Q = 10 g × 4.186 J/g .°C × 5 °C
Q = 209.3 J
The reaction is
2H₂(g) + O₂(g) ---> 2H₂O
Thus as per balanced equation two moles of hydrogen will react with one moles of oxygen.
There is a directly relation between moles and volume. [One mole of each gas occupies 22.4 L of volume at STP]
Thus we can say that two unit volume of hydrogen will react with one unit volume of oxygen
Now as we have started with equal units of volume of both oxygen and hydrogen, half of oxygen will be consumed against complete volume of hydrogen
so the gas which will remain in excess is oxygen
Answer:
Cold Front. A side view of a cold front (A, top) and how it is represented on a weather map (B, bottom). ...
Warm Front. ...
Stationary Front. ...
Occluded Front.