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professor190 [17]
3 years ago
9

223Ra decays by alpha emission with a half-life of 11.43 days. For a 1.0g sample of 223Ra, after one half-life, what mass of 223

Ra remains? What has happened to the remaining mass?
Chemistry
1 answer:
Ray Of Light [21]3 years ago
6 0

Explanation:

The reaction that shows the alpha decay of ^{223}_{88}Ra is:

^{223}_{88}Ra\rightarrow ^{219}_{86}Rn+^{4}_{2}He

Half life is the time at which the concentration of the reactant reduced to half. So,

Mass\ of\ ^{223}_{88}Ra\ remained=\frac {Initial\ concentration}{2}=\frac {1.0\ g}{2}=0.5\ g

The remaining mass is converted to ^{219}_{86}Rn.

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1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

where; a = 78.2 ; b = 1 ; c= - 0.0530

= \frac{-b+\sqrt{b^2-4ac} }{2a}   or \frac{-b-\sqrt{b^2-4ac} }{2a}

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= 0.0204  or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

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Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

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if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

where;

P (pressure) = 1 atm

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R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K  (fixed constant temperature )

V (volume) = ???

V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

V = 1.64604

V ≅ 1.65 L

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