Answer:
44.2 L
Explanation:
Use Charles Law:
We have all the values except for V₂; this is what we're solving for. Input the values:
- make sure that your temperature is in Kelvin
From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:
Therefore, V₂ = 44.2 L
It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.
The answer is letter (B).
Answer:
fH = - 3,255.7 kJ/mol
Explanation:
Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
Answer:
The new volume of the gas is 276.45 mL.
Explanation:
