Answer:
Explanation:
mass % of C = 0.27/0.45*100 = 60%
mass % of H = 0.02/0.45*100 = 4.4%
mass % of O = 0.16/0.45*100 = 35.6%
Total = 60%+4.4%+ 35.6% = 100%
To estimate the molar mass of the gas, we use Graham's law of effusion. This relates the rates of effusion of gases with their molar mass. We calculate as follows:
r1/r2 = √(m2/m1)
where r1 would be the effusion rate of the gas and r2 is for CO2, M1 is the molar mass of the gas and M2 would be the molar mass of CO2 (44.01 g/mol)
r1 = 1.6r2
1.6 = √(44.01 / m1)
m1 = 17.19 g/mol
The boiling point of oxygen is higher than nitrogen's boiling
The reason the boiling point of O2 is higher is not because of increased van der Waals interactions, but simple physics. The mass of a molecule of O2 is greater than that of a molecule of N2, so the molecule of O2 traveling at a speed sufficient to break out of the liquid phase has a greater kinetic energy than an analogous N2 molecule.
The net effect is that more energy must be distributed throughout a sample of O2 to achieve a given vapor pressure (in this case equal to atmospheric pressure) than for a sample of N2. More energy means greater temperature.
Answer:
119.5 J
Explanation:
First we <u>calculate the temperature difference</u>:
- ΔT = 100 °C - 50 °C = 50 °C
Then we can <u>calculate the heat released</u> by using the following formula:
Where q is the heat, Cp is the specific heat, ΔT is the temperature difference and m is the mass.
We <u>input the data</u>:
- q = 0.239 J/g°C * 50 °C * 10.0 g
