Answer:
- nitrogen
Explanation:
SANA PO MAKATULONG PO YUNG SAGOT NA BINIGAY KO
Answer:
E. CH₄ < CH₃Cl < CH₃OH < RbCl
Explanation:
The molecule with the stronger intermolecular forces will have the higher boiling point.
The order of strength of intermolecular forces (strongest first) is
- Ion-Ion
- Hydrogen bonding
- Dipole-dipole
- London dispersion
RbCl is a compound of a metal and a nonmetal. It is an ionic compound, so it has the highest boiling point.
CH₃Cl has a C-Cl polar covalent bond. It has dipole-dipole forces, so it has the second lowest boiling point.
CH₃OH has an O-H bond. It has hydrogen bonding, so it has the second highest boiling point.
CH₄ has nonpolar covalent C-H bonds. It has only nonpolar bonds, so the only attractive forces are London dispersion forces. It has the lowest boiling point.
Thus, the order of increasing boiling points is
CH₄ < CH₃Cl < CH₃OH < RbCl
Answer:
P(O₂) = 287.41 mmHg
P(O₂) = 413.59 mmHg
Explanation:
Given data:
Total pressure = 701 mmHg
Mass of methane = 2.75 g
Mass of oxygen = 3.45 g
Partial pressure of each gas = ?
Solution:
Number of moles of methane:
Number of moles = mass/molar mass
Number of moles = 2.75 g/ 16 g/mol
Number of moles = 0.17 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 3.45 g/ 32 g/mol
Number of moles = 0.12 mol
Total number of moles = 0.12 mol + 0.17 mol = 0.29 mol
Partial pressure of oxygen:
P(O₂) = [ moles of oxygen / total moles ] × total pressure
P(O₂) = [0.12 / 0.29 ] × 701 mmHg
P(O₂) = 0.41 × 701 mmHg
P(O₂) = 287.41 mmHg
Partial pressure of methane:
P(O₂) = [ moles of oxygen / total moles ] × total pressure
P(O₂) = [0.17 / 0.29 ] × 701 mmHg
P(O₂) = 0.59 × 701 mmHg
P(O₂) = 413.59 mmHg
What change did the iv cause