Answer:
2.858g
Step-by-step explanation:
First order reaction.
Half life (t1/2) = 688
Specific rate constant, K = ?
The relationship between t1/2 and K id given as;
t1/2 = 0.693 / k
k = 0.693 / t1/2
k = 0.693 / 688 = 0.0010073 hr-1
Amount of 3.0 g sample of CS2 that remains after 48 hours.
Integrated rate law for a first order reaction is given as;
ln[A] = ln[A]o - kt
ln[A] = ln[3] - 0.0010073(48)
ln[A] = 1.05026
[A] = 2.858g