Answer:
See explanation
Explanation:
The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.
2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.
As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than 2-methylpropane.
The balanced chemical reaction is written as:
<span>3NO2 + H2O = 2HNO3 + NO
Assuming that the gases in this reaction are ideal gas, then we can use the conversion from L to moles which is 1 mol of ideal gas is equal to 22.4 L. We calculate as follows:
538 L NO2 ( 1 mol / 22.4L ) ( 1 mol NO / 3 mol NO2 ) ( 22.4 L / 1 mol ) = 179.33 L NO is produced</span>
Answer:
The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L
Explanation:
Given data:
Number of moles of HF = 6.62×10⁻³ mol
Volume of HF in litter at STP = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Standard temperature = 273 K
Standard pressure = 1 atm
Now we will put the values in formula.
1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K
V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm
V = 148.38×10⁻³ L
Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.
