Methane CH4 CH4 1 hexane C6H14 CH3(CH2)4CH3 5
ethane C2H6 CH3CH3 1 heptane C7H16 CH3(CH2)5CH3 9
propane C3H8 CH3CH2CH3 1 octane C8H18 CH3(CH2)6CH3 18
butane C4H10 CH3CH2CH2CH3 2 nonane C9H20 CH3(CH2)7CH3 35
pentane C5H12 CH3(CH2)3CH3 3 decane C10H22 CH3(CH2)8CH3 75
Answer:
64J of energy must have been released.
Explanation:
Step 1: Data given
One reactant contains 346 J of chemical energy, the other reactant contains 153 J of chemical energy.
The product contains 435 J of chemical energy.
Step 2:
Since the energy is conserved
Sum of energy of Reactants = Energy of Products
Sum of energy of Reactants = 346 J + 153 J = 499 J
The energy of the product = 435 J
435 < 499
This means energy must have been lost as heat.
Step 3: Calculate heat released
499 J - 435 J = 64 J
64J of energy must have been released.
Answer : The final pressure will be, 666.2 mmHg
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

or,

where,
= initial pressure = 790 mmHg
= final pressure = ?
= initial volume = 101.2 mL
= final volume = 120 mL
Now put all the given values in the above equation, we get:


Therefore, the final pressure will be, 666.2 mmHg
Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Explanation :
As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.
Let us assume that the radius of Cl⁻ be, (x) pm
So, the radius of Na⁺ = 
In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.
Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.
Given:
Distance between Na⁺ nuclei = 566 pm
Thus, the relation will be:





The radius of Cl⁻ ion = (x) pm = 181 pm
The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm
Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.