Answer:
Hydrogen: -141 kJ/g
Methane: -55kJ/g
The energy released per gram of hydrogen in its combustion is higher than the energy released per gram of methane in its combustion.
Explanation:
According to the law of conservation of the energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qc + Qb = 0
Qc = -Qb [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Q = C . ΔT
where,
C is the heat capacity
ΔT is the change in the temperature
<h3>Hydrogen</h3>
Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (14.3°C) = -162 kJ
The heat released per gram of hydrogen is:
<h3>Methane</h3>
Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (7.3°C) = -82 kJ
The heat released per gram of methane is:
Probably ion with a positive 1 charge.
Explained briefly
C would be the right answer
Answer:44.04mL
Explanation:Parameters given
V1 = 30.0mL
P1 = 36.7psi
P2 = 25.0psi
V2 = ??
From Boyle's gas law, which states that "the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature"
This means that,
the pressure of a gas tends to increase as the volume of the container decreases, and also the pressure of a gas tends to decrease as the volume of the container increases.
Mathematically, Boyle's can be represented as shown below
P= k/V
Where P = Pressure, V = Volume and k is constant
Therefore,
PV = k
P1V1 = P2V2 =PnVn
Using the formula
P1V1 = P2V2
V2 = P1V1/P2
V2 = (36.7psi × 30.0mL) / 25.0psi
V2 = 1101.0/25.0
V2 = 44.04mL
