<u>Answer:</u> The potential of electrode is -0.79 V
<u>Explanation:</u>
When zinc is dipped in zinc sulfate solution, the electrode formed is 
Reduction reaction follows: 
To calculate the potential of electrode, we use the equation given by Nernst equation:
![E_{(Zn^{2+}/Zn)}=E^o_{(Zn^{2+}/Zn)}-\frac{0.059}{n}\log \frac{[Zn]}{[Zn^{2+}]}](https://tex.z-dn.net/?f=E_%7B%28Zn%5E%7B2%2B%7D%2FZn%29%7D%3DE%5Eo_%7B%28Zn%5E%7B2%2B%7D%2FZn%29%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BZn%5D%7D%7B%5BZn%5E%7B2%2B%7D%5D%7D)
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = -0.76 V
n = number of electrons exchanged = 2
(concentration of pure solids are taken as 1)
![[Zn^{2+}]=0.1M](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D%3D0.1M)
Putting values in above equation, we get:

Hence, the potential of electrode is -0.79 V
Diamond. It can only be cut with another diamond.
Answer:
a) both substances are insoluble in water
b) both substances are soluble in ligroin
c) both substances suffer combustion, octane produces more CO₂ than hexene.
d) both substances are less dense than waterl, with hexene having the lowest density.
e) only hexene would react with bromine
f) only hexene would react with permanganate
Explanation:
a) both substances are non-polar and water is polar
b) both substances are non-polar and lingroin is non-polar
c) C₈H₁₈ + 17.5O₂ → 8CO₂ + 9H₂O
C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O
d) water = 997 kg/m³
ocatne = 703 kg/m³
hexene = 673 kg/m³
e) bromine test is used to detect unsaturations
f) permanganate test is used to detect unsaturations
<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>
Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.
Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3
To find for the theoretical yield, we first determine the limiting reactant.
100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2
Therefore, the limiting reactant is O2.
Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3
Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%