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
Answer:
The number of energy levels increases as you move down a group as the number of electrons increases. Each subsequent energy level is further from the nucleus than the last. Therefore, the atomic radius increases as the group and energy levels increase.
Explanation:
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Answer:

Explanation:
From the question we are told that:
Pressure 
Temperature 
Volume 
Heat Produced 
Generally the equation for ideal gas is mathematically given by



Therefore


Since
Heat of combustion of Methane=889 kJ/mol
Heat of combustion of Propane=2220 kJ/mol
Therefore

Comparing Equation 1 and 2 and solving simultaneously




Therefore
Mole fraction 0f Methane is mathematically given as


