Answer:
45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.
Explanation:
Using Ideal gas equation for same mole of gas as
Given ,
V₁ = 25.0 L
V₂ = ?
P₁ = 2575 mm Hg
Also, P (atm) = P (mm Hg) / 760
P₁ = 2575 / 760 atm = 3.39 atm
P₂ = 1.35 atm
T₁ = 353 K
T₂ = 253 K
Using above equation as:

Solving for V₂ , we get:
<u>V₂ = 45.0 L</u>
45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.
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:

The answer would be A) sand, it is not soluble in water
First, we need to get n1 (no.of moles of water ): when
mass of water = 0.0203 g and the volume = 1.39 L
∴ n1 = mass / molar mass of water
= 0.0203g / 18 g/mol
= 0.00113 moles
then we need to get n2 (no of moles of water) after the mass has changed:
when the mass of water = 0.146 g
n2 = mass / molar mass
= 0.146g / 18 g/ mol
= 0.008 moles
so by using the ideal gas formula and when the volume is not changed:
So, P1/n1 = P2/n2
when we have P1 = 1.02 atm
and n1= 0.00113 moles
and n2 = 0.008 moles
so we solve for P2 and get the pressure
∴P2 = P1*n2 / n1
=1.02 atm *0.008 moles / 0.00113 moles
= 7.22 atm
∴the new pressure will be 7.22 atm