The question is incomplete. Here is the full version
Select the substance that has the higher boiling point. Select the substance that has the higher boiling point. a)C3H8 b)CH3OH c)CH3OCH3
Answer:
b)CH3OH
Explanation:
a)C3H8 The dominant intemolecular force here are London dispersion forces.
Dipole-dipole forces cancel out because the molecule is symmetric
b)CH3OH - The dominant intermolecular force is hydrogen bonding
c)CH3OCH3 - The dominant intemolecular force here are London
dispersion forces. Dipole-dipole forces cancel out because
the molecule is symmetric
<u>Answer:</u> The number of moles of gas present are 0.390 moles
<u>Explanation:</u>
To calculate the number of moles of gas, we use the equation given by ideal gas which follows:
where,
P = pressure of the gas = 0.868 atm
V = Volume of the gas = 11 L
T = Temperature of the gas =
R = Gas constant =
n = number of moles of gas = ?
Putting values in above equation, we get:
Hence, the number of moles of gas present are 0.390 moles
The ionization energy or ionization potential is - the energy required to remove an electron from an element in its gaseous state.
Answer:
Explanation:
The density of a gas can be obtained using the gas ideal equation and the molar mass of the gas.
This is the decution of the final formula:
Now, you just need to substitute values:
- R = 0.08206 atm-liter / k-mol
- d = 32.0 g/mol × 0.9869 atm / [0.08206 atm-liter/k-mol × 273.15K]
- d = 1.4 g/liter (using two significant figures)
As you see, I have not used the 4.8 grams datum. That is because the density of the gases may be calculated from the temperature, pressure and molar mass of the gas, using the ideal gas equation.
Since, you have the mass of gas, you might use this other procedure:
- Volume of 1 mol of gas at STP: about 22.4 liter/mol
- Mass of 1 mol of oxygen gas: 32.0 g/mol (the molar mass)
- number of moles in 4.8 g of oxygen = 4.8 g / 32.0 g/mol = 0.15 mol
- Volume of 0.15 mol of oxygen: 0.15 mol × 22.4 liter/mol = 3.36 liter
- Density = mass / volume = 4.8 g / 3.36 liter = 1.4 g/liter (same result)