For a comparison of the nucleus 5626fe, the density of the nucleus 112 48cd is mathematically given as the same.
n(Cd) / n(Fe)=1
<h3>What is the density of the nucleus 112 48cd?</h3>
Generally, the equation for the density  is mathematically given as
d=\frac{A}{4/3}\piR^3
Therefore
n(Cd) / n(Fe) = [A (Cd) / (A Fe) ] * [ R (Fe) / R (Cd)]^3
n(Cd) / n(Fe)= (112 / 56 ) * (1/1.26)3
n(Cd) / n(Fe)=1
In conclusion, The ratio of n(Cd) = n(Fe) is 1, hence same
Read more about density
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Answer:
<u>a nonspontanenous reaction is forced to occur</u>
<em>Hope</em><em> this</em><em> helps</em><em> </em><em>:</em><em>)</em>
 
        
             
        
        
        
Balanced Half reactions are:
At anode 2
  ==> Cl₂+ 
 + H₂O ==> 
+ 2
 + 
 
At Cathode:  2
 + 
 ==> H₂ 
Since the question states that you are using an aqueous solution of MnCl₂,  so ions will have present are, H₂O, 
, 
 and 
Now at Anode reaction will occur as given:
 2
  ==> Cl₂+ 
 + H₂O ==> 
+ 2
 + 
  (will occur)
At Cathode: 
2
 + 
 ==> H₂ (will occur)
At Cathode: 
 +  
==> Mn (This reaction will not occur)
The deposition of solid Mn will not occur because in aqueous solution, 
will be reduced before 
 .
The reduction potentials for 
 is zero whereas reduction potential for 
 is - 1.18V.
The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.
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Explanation:
pls, refer to the above picture, i hope you will find it helpful.