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Hunter-Best [27]
3 years ago
8

Iodine-131, t1/2 = 8.0 days, is used in the diagnosis and treatment of thyroid gland diseases. If a laboratory sample of iodine-

131 initially emits 9.95 × 1018 β particles per day, how long will it take for the activity to drop to 6.22 × 1017 β particles per day?
Chemistry
1 answer:
SashulF [63]3 years ago
7 0

Answer:

32,0 days.

Explanation:

The radioactive decay follows:

N_{t} = N_{0}e^{\frac{-0.693t}{t_{1/2}}

Where Nt is the concentration in a time t (6,22x10¹⁷), N₀ is the initial concentration (9,95x10¹⁸) Half life time is 8,0 days and t is the time it take to drop this concentration. Replacing:

6.22x10^{17} = 9,95x10^{18}e^{\frac{-0.693t}{8 days}

0,0625 = e^{\frac{-0.693t}{8days}

ln 0,0625 = {\frac{-0.693t}{8days}

-2,77*8days = -0.693t

-22,2days = -0.693t

32,0days = t

It take <em>32,0 days</em>

<em></em>

I hope it helps!

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An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
djverab [1.8K]

Answer:

The new partial pressures after equilibrium is reestablished for PCl_3:

P_1'=6.798 Torr

The new partial pressures after equilibrium is reestablished Cl_2:

P_2'=26.398 Torr

The new partial pressures after equilibrium is reestablished for PCl_5:

P_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=P_1=13.2 Torr

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K_p=\frac{P_1}{P_1\times P_2}

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Partial pressure of the PCl_3=P_1'=13.2 Torr

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Partial pressure of the PCl_5=P_3'=217.0 Torr

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P=P_1'+P_2'+P_3'

263.0Torr=13.2 Torr+P_2'+217.0 Torr

P_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{P_3'}{P_1'\times P_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

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The new partial pressures after equilibrium is reestablished for PCl_3:

P_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

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P_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

The new partial pressures after equilibrium is reestablished for PCl_5:

P_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

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