Answer:
Explanation:
If the fission is induced by two neutrons, the unbalanced equation is
The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation .
Balancing the superscripts, we get
2 + 239 = 89 + 149 + x
241 = 238 + x
x = 3
The balanced nuclear equation is
Answer:
You can find in the given attachemnet
Answer:
a) [Ag+]dilute = 6.363 × 10⁻¹⁶ M
b) 1.273 × 10⁻¹⁶
c) 2.629×10⁻¹⁹ M Thus; the value for [Ag+ ]dilute will be too low
Explanation:
In an Ag | Ag+ concentration cell ,
The anode reaction can be written as :
Ag ----> Ag+(dilute) + e- &:
The cathode reaction can be written as:
Ag+(concentrated) + e- ----> Ag
The Overall Reaction : is
Ag+(concentrated) -----> Ag+(dilute)
However, the Standard Reduction potential of cell = E°cell = 0
( since both cathode and anode have same Ag+║Ag )
Also , given that the theoretical slope is - 0.0591 V
Therefore; the reduction potential of cell ; i.e
Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )
0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )
log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963
[Ag+]dilute = × 1.0 × 10⁻¹ M
[Ag+]dilute = 6.363 × 10⁻¹⁶ M
b)
AgI ----> Ag + (dilute) + I⁻
So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]
= 6.363 × 10⁻¹⁶ M × 0.20 M
= 1.273 × 10⁻¹⁶
c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :
Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )
1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )
log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804
[Ag+]dilute = × 1.0 × 10⁻¹ M
[Ag+]dilute = 2.629×10⁻¹⁹ M
Thus, the value for [Ag+ ]dilute will be too low