The half-reaction are:
Cd ---> Cd(OH)₂
The oxidation number of Cd changed from 0 to +2. So, the number of mol electron transferred here is 2.
NiO(OH) --> Ni(OH)₂
The oxidation number of Cd changed from +3 to +2. So, the number of mol electron transferred here is 1.
Now, the greatest common factor would be 2. So, we use n=2 for the formula for ΔG°. F is Faraday's constant equal to 96,485 J/mol e.
ΔG° = nFE° = (2)(96,485)(1.5) =<em> 289,455 J</em>
Properties of a compound is completely different from their elements.
Water is composed by hydrogen and oxygen.
For example, the boiling point of oxygen is - 183 °C and hydrogen is - 253 °C, meanwhile, water has a boiling point of 100°C
Another example is when you put a burning wooden splint into oxygen, it burns more brightly. Put it in hydrogen, you may hear a "pop" sound, or even explode when large amount of hydrogen. But if u put a burning splint in water, it goes off.
Answer:
ok ............bu am not so sure
As we have the balanced reaction equation is:
N2O4 (g) ↔ 2NO2(g)
from this balanced equation, we can get the equilibrium constant expression
KC = [NO2]^2[N2O4]^1
from this expression, we can see that [NO2 ] is with 2 exponent of the stoichiometric and we can see that from the balanced equation as NO2
is 2NO2 in the balanced equation.
and [N2O4] is with 1 exponent of the stoichiometric and we can see that from the balanced equation as N2O4 is 1 N2O4 in the balanced equation.
∴ the correct exponent for N2O4 in the equilibrium constant expression is 1
Explanation:
1 literThe total of water is equal to 1000.0 g of water
we need to find the molality of a solution containing 10.0 g of dissolved in Na₂S0₄1000.0 g of water
1. For that find the molar mass
Na: 2 x 22.99= 45.98
S: 32.07
O: 4 x 16= 64
The total molar mass is 142.05
We have to find the number of moles, y
To find the number of moles divide 10.0g by 142.05 g/mol.
So the number of moles is 0.0704 moles.
For the molarity, you need the number of moles divided by the volume. So, 0.0704 mol/1 L.
The molarity would end up being 0.0704 M
The molality of a solution containing 10.0 g of Na2SO4 dissolved in 1000.0 g of water is 0.0704 Mliter
