In the unbalanced equation shown below how many electrons are needed in the bala
ID: 908647 • Letter: I
Question
In the unbalanced equation shown below how many electrons are needed in the balanced oxidation half-reaction?
9)
C2O42– (aq) + MnO4–(aq) CO2(aq) + Mn2+(aq)
Select one:
A. 10
B. 1
C. 2
D. 5
11)
What are the coefficients in front of NO3-(aq) and Cu(s) when the following redox equation is balanced in an acidic solution:
___ NO3-(aq) + ___ Cu(s) ___ NO(g) + ___ Cu2+(aq)?
Select one:
A. 3, 6
B. 2, 6
C. 2, 3
D. 3, 4
25)
Calculate the total quantity of heat required to convert 25.0 g of liquid CCl4(l) from 25.0°C to gaseous CCl4 at 76.8°C (the normal boiling point for CCl4)? The specific heat of CCl4(l) is 0.857J its heat of fusion is 3.27kj/mol and its heat of vaporization is 29.82kj/mol.
Select one:
A. 6.49 kJ
B. 1.11 kJ
C. 1.64 kJ
D. 5.96 kJ
Explanation / Answer
9) Answer : option A) 10
2 MnO + 6 H + 5 HOOC–COOH(aq) --------------------> 2 Mn² + 8 HO + 10 CO(g)
MnO4- oxidation state = +7
Mn+2 oxidation state = +2
5e- are transferred. for 2 MnO4 total 10 e- are reqired.
10) answer : option C) 2,3
3 Cu + 2 NO3- + 8H+ ----------------------> 3Cu2+ + 2NO + 4H2O
11) answer : option D) 5.96
heat required Q = m Cp dT
= 25 x 0.857 x (76.8 - 25)
= 1.11 kJ
moles of CCl4 = mass / molar mass of CCl4
= 25 / 153.8
= 0.1625
1 mol ------------------ 29.82 kJ/mol
0.1625 mol -----------------??
the amount of heat required to vapourize CCl4 = 29.82 x 0.1625
= 4.846 kJ
so total heat = 1.11 + 4.846
= 5.96 kJ