Answer: The molecular weight of the gas 28. The gas is probably N2
Explanation:Please see attachment for explanation
Answer:
9.91 mL
Explanation:
Using the combined gas law equation as follows;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (torr)
P2 = final pressure (torr)
V1 = initial volume (mL)
V2 = final volume (mL)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
V1 = 15.0mL
V2 = ?
P1 = 760 torr
P2 = 1252 torr
T1 = 10°C = 10 + 273 = 283K
T2 = 35°C = 35 + 273 = 308K
Using P1V1/T1 = P2V2/T2
760 × 15/283 = 1252 × V2/308
11400/283 = 1252V2/308
Cross multiply
11400 × 308 = 283 × 1252V2
3511200 = 354316V2
V2 = 3511200 ÷ 354316
V2 = 9.91 mL
Answer:
The heat at constant pressure is -3,275.7413 kJ
Explanation:
The combustion equation is 2C₆H₆ (l) + 15O₂ (g) → 12CO₂ (g) + 6H₂O (l)
= (12 - 15)/2 = -3/2
We have;
Where R and T are constant, and ΔU is given we can write the relationship as follows;
Where;
H = The heat at constant pressure
U = The heat at constant volume = -3,272 kJ
= The change in the number of gas molecules per mole
R = The universal gas constant = 8.314 J/(mol·K)
T = The temperature = 300 K
Therefore, we get;
H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ
The heat at constant pressure, H = -3,275.7413 kJ.
It Is Called The Parent Nuclide