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SVETLANKA909090 [29]
3 years ago
9

Which missing item would complete this beta decay reaction?

Chemistry
2 answers:
Karo-lina-s [1.5K]3 years ago
7 0
The question is incomplete. Complete question is attached below:
.................................................................................................................

Correct Answer is : 54 131Xe

Reason: 
Emission of beta particle from radioactive material increases atomic number by one, but atomic mass number donot change.

In present case, the reactant is 53 131 I. When, iodine emits beta particle i.e. electron, it atomic number increases by one. Hence, it gets converted into Xe. Atomic number of Xenon is 54. 

Zinaida [17]3 years ago
5 0
Since you did not write down the eaction,

The Beta Decay Reaction NEVER change the total numbers of the nucleons,

for example :

131                                  131            0          0
        I      >>>>>>>>>>          Xe  +     e    +     v
   53                                    54           -1          0 

(see that the total proton and electron is still 131 and 53)
  
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The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
Tanggung jawab kabinet​
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Answer:

djjdjd

Explanation:

jrjrndjdjjdjjdjdjdjdjdjrjjrjr

8 0
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I am a chemical ph of 1
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If your pH is 1, then you're a very strong Acid.

Hope this helps!
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