Question

please help with question 25 a and b. showing all work

25. a. (3pts) A radioactive sample of substance X has initial mass = 3.25mg. It decays by a first- order reaction. If its half-life = 5.6 days, calculate its mass after 11.2 days. Given k = 0.0490. k= b. (4pts) Compound Z decomposes by a second-order reaction. We are told the initial concentration is 0.660M. If the value of k = 7.12 x 10 sec"!, what will the concentration be after 2 minutes? Ans.

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Questions: in ...typed format...

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$\Rightarrow$Part - A:

$\Rightarrow$Answer:

Mass of 'X' remaining: after 11.2 days = 0.8125 mg ( milligram)

$\Rightarrow$Explanation:

• Given:

1. Initial mass of X =  3.25 mg
2. Half Life t_1/2 = 5.6 days
3. Time passed: =  11.2 days = 2.0 half lifes ( since, 5.6 x 2 = 11.2 days )

• ​​​​​​​Step - 1:

​​​​​​​Thus, mass of compound: X remaining: after two half lifes:

$\Rightarrow$ Mass of X  remaining: = ( initial mass ) x ( 1/2 ) ^{number of half lifes} = ( 3.25 mg ) x ( 0.5 ) ^2.0 =   0.8125 mg ( milligram)

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$\Rightarrow$ Part - B:

$\Rightarrow$Answer:

Concentration after two ( 2 ) minutes =  [Z]_t = 0.422 M ( mol/L )​​​​​​​ ....Answer...

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above: Answer.

• Given:

1. ​​​​​​​The reaction is of second order:
2. Initial concentration of 'Z' : [ Z ]_o = 0.660 M ( mol/L )
3. Dissociation constant: k = 7.12 x 10^-3 sec^-1
4. Let the concentration of 'Z' , after time ' t ' be : [ Z ]_t
5. Time passed:  t = 2.0 minutes = ( 2.0 min ) x ( 60 sec. / min. ) = 120 sec

• ​​​​​​​Step - 1:

​​​​​​​We know the following: integrated rate law: for second order equation:

$\Rightarrow$1/ [Z]_t = kt + 1/[Z]_o   ------------------Equation - 1

$\Rightarrow$1/ [Z]_t =   ( 7.12 x 10^-3 sec^-1 ) x ( 2 min x 60 sec/ min ) + 1.0 / ( 0.660 M ) =  2.37   ( approx. )

$\Rightarrow$   [Z]_t = 1/ ( 2.37 ) =  0.422 M ( mol/L )