Solution :
A cell that is concentrated is constructed by the same half reaction for the anode as well as he cathode.
We know,
In a standard cell,
the reduction half cell reaction is :

The oxidation half ell reaction :

Thus the complete reaction of the cell is :

cell = 
Answer:
V2~0.4839M
Explanation:
We're going to use Boyles law to answer the question.
Boyle's law:
P1V1=P2V2
P1=151mmHg
P2=166mmHg
V1=0.532L
V2=?
V2=(P1 x V1)/P2
V2=(151 x 0.532)/166
V2~0.4839M
Hope it helps:)
Answer:
Explanation:
Molar mass of sodium hydroxide is = 100
any compund with its molar mass dissovled in 1L itres solution gives rise to 1M solution
40 g ------> 1L ---->1M
Xg ------>1L------>2M
X g= 40*1*2/1*1 =80 gram