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4 years ago

3.) A 208 g sample of sodium-24 decays to 13.0 g of sodium-24 within 60.0 hours.

Chemistry

1 answer:

Ulleksa [173]4 years ago

3 0

Answer:

Half life = 12 hrs

Explanation:

Given data:

Total amount of sample = 208 g

Amount left after 60 hr = 13.0 g

Half life = ?

Solution:

Step one:

amount half life

208 g time 0

208/2 1 half life

104/2 2 half life

52/2 3 half life

26/2 4 half life

13 5 half life

Step 2:

half life = elapsed time/ total half life when 13 g is left

Half life = 60 hrs/ 5

Half life = 12 hrs

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Answer : The initial concentration of HI and concentration of $HI$ at equilibrium is, 0.27 M and 0.386 M respectively.

Solution : Given,

Initial concentration of $H_2$ and $I_2$ = 0.11 M

Concentration of $H_2$ and $I_2$ at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

Initially 0.11 0.11 C

At equilibrium (0.11-x) (0.11-x) (C+2x)

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The expression of $K_c$ will be,

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By solving the terms, we get:

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