I would say the first three. But I'm not 100% sure. I'm truly sorry if it's wrong
Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
Answer:
[A]²
Explanation:
Since the formation is independent of D, D is 0 order.
Since a quadruples when it is doubled it can be written as
2A^X= 4
To find the unknown power we can assume A= 1 to make the math simple. So When a = 2 (Because you doubled it) raised to X power it will equal 4
so the unknown power is 2
Making the rate law
[a]²[b]⁰
or simply just
[A]²
Answer:
Please refer to the attachment for answers.
Explanation:
Please refer to the attachment for explanation
Probably C hope this helps