Answer:
<h2>
The no. of moles present in one liter solution is 0.843</h2>
Explanation:
Mass of ammonium chloride = 11.5 gm
Molar mass of ammonium chloride = 53.491 gm
No. of moles
N = 0.215 moles
M = 0.843 M
Thus the no. of moles present in one liter solution is 0.843
Answer:
a. Minimum 1.70 V
b. There is no maximum.
Explanation:
We can solve this question by remembering that the cell potential is given by the formula
ε⁰ cell = ε⁰ reduction - ε⁰ oxidation
Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the oxidized species 0.80 V, thus
ε⁰ reduction - ε⁰ oxidation ≥ ε⁰ cell
Since ε⁰ oxidation is by definition the negative of ε⁰ reduction , we have
ε⁰ reduction - ( 0.80 V ) ≥ 0.90 V
⇒ ε⁰ reduction ≥ 1.70 V
Therefore,
(a) The minimum standard reduction potential is 1.70 V
(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V
Answer:
on each side of the salt bridge, which is represented by a double vertical line
Explanation:
While writing a cell notation, the general convention is; anode || cathode. The anode and the cathode are separated by a double line. The anode is written on the lefthand side while the cathode is written on the righthand side.
The cell notation is a shorthand representation of a cell, hence any electrochemical cell can easily be produced based on its cell diagram.
Hey there!:
Molar mass:
CHCl3 = ( 12.01 * 1 )+ (1.008 * 1 ) + ( 35.45 * 3 ) => 119.37 g/mol
C% = ( atomic mass C / molar mass CHCl3 ) * 100
For C :
C % = (12.01 / 119.37 ) * 100
C% = ( 0.1006 * 100 )
C% = 10.06 %
For H :
H% = ( atomic mass H / molar mass CHCl3 ) * 100
H% = ( 1.008 / 119.37 ) * 100
H% = 0.008444 * 100
H% = 0.8444 %
For Cl :
Cl % ( molar mass Cl3 / molar mass CHCl3 ):
Cl% = ( 3 * 35.45 / 119.37 ) * 100
Cl% = ( 106.35 / 119.37 ) * 100
Cl% = 0.8909 * 100
Cl% = 89.9%
Hope that helps!
