The answer is D. And if it is equally shared it is nonpolar covalent bond
Answer: The entropy change of the surroundings will be -17.7 J/K mol.
Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol
Amount of Acetone given = 10.8 g
Number of moles is calculated by using the formula:

Molar mass of acetone = 58 g/mol
Number of moles = 
If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then
0.1862 moles will have = 
To calculate the entropy change for the system, we use the formula:

Temperature = 56.2°C = (273 + 56.2)K = 329.2K
Putting values in above equation, we get
(Conversion Factor: 1 kJ = 1000J)
At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,


A) For balanced chemical equation: 2HgO(s) → 2Hg(l) + O₂(g).
1) Mole ratio 1: n(HgO) : n(Hg) = 2 : 2 (1 : 1).
2) Mole ratio 2: n(HgO) : n(O₂) = 2 : 1.
3) Mole ratio 3: n(Hg) : n(O₂) = 2 : 1.
B) Balanced chemical equation: 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l).
1) Mole ratio 1: n(NH₃) : n(NO) = 4 : 6 (2 : 3).
2) Mole ratio 2: n(NH₃) : n(N₂) = 4 : 5.
3) Mole ratio 3: n(NH₃) : n(H₂O) = 4 : 6 (2 : 3).
4) Mole ratio 4: n(NO) : n(N₂) = 6 : 5.
5) Mole ratio 5: n(NO) : n(H₂O) = 6 : 6 (1 :1).
6) Mole ratio 6: n(N₂) : n(H₂O) = 5 : 6.
Number 4 is
-Oxidation occurs at the anode, while reduction occurs at the cathode. Recharging a battery involves the conversion of electrical energy to chemical energy. During recharging, there is movement of electrons from an external power source to the anode, and on the other side electrons are removed from the cathode.