Gen Chem Questions?

[Philippines03j]

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Alright here are a few questions for gen chem that I have been struggling to figure out.

1. The activity of a radioisotope is 3000 counts per minute at one time and 2736 countes per minute 48 hours later. What is the half-life of the radioisotope?

A) 831hr B) 521hr C) 361hr D) 1.44 hr

Answer: C

2. The half-life of the first order conversion of cyclobutane to thylene is 22.7 s at a particular temperature. How many seconds are need for the partial pressure of cyclobutan to decrease from 100mmHg to 10mmHg?

c4h8-->2c2h4

A) 52 s B) 75.4 s C) 90 s D) 227 s

Answer: B

 

[UCLAzy]

hey, so for the first one i didnt get exactly the same answer, but its relatively close so i guess i rounded too early. anyways the way i did it was by using the formula R(amount remaining) = A(initial amount) * .5^n(number of half lives)

so you have 2736 = 3000(.5^n). solve for n using logs and you get n = .132

since that represents the number of half lives that occurred in 48 hours, i just used proportions to find how long one half live is

.132/48 = 1/x, or x = 363 (there might be a different or more accurate way, i'm not sure)


for the second one there's two ways. since the numbers are so spread apart, its easy to eyeball it and get the right answer. to go from 100 to 10, there are 3 half lives and change. if you look at 3 half lives, it gets you down to 12.5 mmHg. 3 half lives is 3*22.7 seconds, or 68.1 seconds. 4 half lives is 4*22.7 seconds, or 90.8 seconds. since 10 is closer to 3 half lives than 4, you can knock out all the answers but 75. this way is less accurate, but saves you a lot of time since i did it in my head in about 20 seconds.

if the numbers are really close together, the way to do this problem is using that formula for half life i used earlier. here you would say

10 = 100(.5^n), solve for n and get 3.32 for n

multiply 3.32*22.7 and you get 75.4

hope this helps