To begin the experiment, 1.11g of methane CH4is burned in a bomb calorimeter containing 1000 grams of water. The initial tempera
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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:
Fe =52.2%, O = 44.9%, H = 2.81%
Explanation:
Percentage composition is also known as percentage by mass.
First, find the Mr of the compound.
Mr of Fe(OH)₃ = 55.8 + (16 × 3) + (1 × 3)
= 106.8
Now Divide the mass of the element, as per the compound, by the Mr of the compound found and times that by 100.
% composition of Fe = × 100 = 52.2%
% composition of O = × 100 = 44.9%
[There are 3 oxygen atoms in the compound Fe(OH)₃, so we will multiply the atomic mass of oxygen with the number of atoms in the compound: 16×3 ]
% composition of H = × 100 = 2.81%
[There are 3 hydrogen atoms in the compound Fe(OH)₃, so we will multiply the atomic mass of hydrogen with the number of atoms in the compound: 1×3 ]