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Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
Answer : Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.
Explanation :
The given molecule is, 
Three types of inter-molecular forces are present in this molecule which are Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.
- Hydrogen-bonding : when the partial positive end of hydrogen is bonded with the partial negative end of another molecule like, oxygen, nitrogen, etc.
- Dipole-dipole attraction : When the partial positively charged part of the molecule is interact with the partial negatively charged part of the molecule. For example : In case of HCl.
- London-dispersion force : This force is present in all type of molecule whether it is a polar or non-polar, ionic or covalent. For example : In case of Br-Br , F-F, etc
Hydrogen-bonding is present between the oxygen and hydrogen molecule.
Dipole-dipole forces is present between the carbon and oxygen molecule.
London-dispersion forces is present between the carbon and carbon molecule.
Answer:
The answer is
Explanation:
- A nonpolar organic compound that is insoluble in water
Explanation:
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