Answer:
(a) The concentration of
after 5.00 min is 0.9672 mol/L
(b) The fraction of
decomposed after 5.00 min is 0.568
The problem can be solved by using first order integrated reaction.
Explanation:
(a)
Rate constant of reaction = K =

Initial concentration of
= 2.24 mol/L
Assuming final concentration of
to be x mol/L
time (t) = 5.00 min
The first order integrated equation is shown below

Concentration of
decomposed = 0.9672 mol/L
(b)
Concentration of
decomposed = 
Fraction of
decomposed = 
<em>We know<u> One mole of Argon occupy the volume of 22.4 L
</u></em>
<em>Therefore , 22.4 L of Argon weighs 40 g.</em>
<em>∴ 1 L of Argon weighs , 40/22.4 </em>
<em>∴ 100 L of Argon weighs (40/22.4 ) × 100</em>
<em>= 178.57 g.</em>
<em>Hence <u>moles of Argon is 178.57 g.
</u>Hope it Helps :-)</em>
Answer:
<em><u>0.0365 is your answer</u></em>
Explanation:
<h3>hope it will help u</h3>