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
A electrochemical reaction is said to be spontaneous, if
Answer 1:
Consider reaction: <span>Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s)
The cell representation of above reaction is given by;
</span>

Hence,

we know that,

and

Therefore,

= - 0.25 - (-0.47) = 0.22 v
Since,
is positive, hence cell reaction is spontaneous
.....................................................................................................................
Answer 2: Consider reaction: <span>Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)
</span>
The cell representation of above reaction is given by;

Hence,

we know that,

and

Therefore,

= - 0.126 - 0 = -0.126 v
Since,
is negative, hence cell reaction is non-spontaneous.....................................................................................................................
Answer 3:
Consider reaction: <span>2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)
</span>
The cell representation of above reaction is given by;

Hence,

we know that,

and

Therefore,

= - 0.22 - (-0.913) = 0.693 v
Since,
is positive, hence cell reaction is spontaneous
B. Share Electrons.
Hope this helps.
Answer: 55.84L
Explanation: Please see attachment for explanation.
1) concentration or partial pressure of species
involved. 2) temperature • 3) presence of catalyst
4) nature of reactants.