Question

After decaying for 48 hours, 1/16 of the original
mass of a radioisotope sample remains unchanged.

What is the half-life of this radioisotope?
(1) 3.0 h (3) 12 h
(2) 9.6 h (4) 24 h

Answer is (3) 12 h, but how would you get that as an answer?

Answer 1

1/16 * 48.

Answer 2

Nope, 1/16 * 48 = 3. 3 is the wrong answer.

Answer 3

No, it's correct. :)

Answer 4

1/16 = 0.0625
0.0625 * 48 = 3

Answer 5

This is chemistry, you are aware of that right? 12 h is the right answer, I have the answers in front of me. I need the correct steps so I know how to do the problem.

Answer 6

This doesn't belong in math then, it belongs in Chemistry.

Answer 7

This is chemistry, not math.

Answer 8

Look at the topic.

Answer 9

omg lol I was switched! xD My appologies

Answer 10

All good.

Answer 11

If there's only 1/16th of the initial concentration of a radioisotope sample remaining, then the sample must have undergone 4 half lives (b/c 1/2 x 1/2 x 1/2 x 1/2 = 1/16). We know that the total time these 4 half lives took is 48 hours, so each half life took 48hr total / 4 half lives = 12 hours per half life

Answer 12

So do you understand it @Albert0898 ?

Answer 13

yes, yes, thank you!

What about;

A 10.0-gram sample of H2O(ℓ) at 23.0°C absorbs 209 joules of heat. What is the final temperature of the H2O(ℓ) sample?
(1) 5.0°C (3) 28.0°C
(2) 18.0°C (4) 50.0°C

and

A 10.0-gram sample of H2O(ℓ) at 23.0°C absorbs 209 joules of heat. What is the final
temperature of the H2O(ℓ) sample?
(1) 5.0°C (3) 28.0°C
(2) 18.0°C (4) 50.0°C

Answer 14

@aprehan

Answer 15

Sorry, the 2nd question is; A 10.0 milliliter sample of NaOH (aq) is neutralized by 40.0 milliliters of 0.50 M HCl. What is the molarity of the NaOH (aq)?

1. 1.0 M
2. 2.0 M
3. 0.25 M
4. 0.50 M

Answer 16

Q = mCpdeltaT
209 = 10 x 1 x (Tfinal - 23 degrees celsius)
Tfinal = 439 degrees celsius, thus it increased which makes sense b/c the sample ABSORBED energy so the temp should go up

Answer 17

I had a brain fart for the first one, then I figured it out. Thank you.

Answer 18

wait someting went wrong on my part

Answer 19

actually the temperature I got was 43.9 degrees Celsius, which isn't an answer choice...

Answer 20

OOOh wait the specific heat of water is 4.184 J/g celsius lol not 1 which has untis of cal/g cel

Answer 21

so if you plug that into the equation
209 = 10 x 4.184 x (Tfinal - 23 degrees Celsius)
you get Tfinal = 28 degrees Celsius

Answer 22

that's your answer

Answer 23

he 2nd question is; A 10.0 milliliter sample of NaOH (aq) is neutralized by 40.0 milliliters of 0.50 M HCl. What is the molarity of the NaOH (aq)?

As for your second question, we just use M1V1 = M2V2,

M of NaOH x 10mililiter NaOH = 0.5M HCl x 40 mililiter of HCl
M of NaOH = 2M, that's your answer