anonymous
  • anonymous
HAlf Life problem Radon-222 has a half-life of 4.0 d (abbreviation for “diem”—Latin for “day”). If the initial mass of the sample of this isotope is 6.8 g, calculate the mass of radon-222 remaining in the sample after: a) 8 d b) 16 d c) 32d the awnsers are : a) 1.7g b)0.62g c) 0.026g i can't figure out how to do it
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
You mean 0.42 for b?
anonymous
  • anonymous
no it is 0.62 for 6 this question is from a text book with given awnsers
anonymous
  • anonymous
Check it again.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
i will show you a picture one second
anonymous
  • anonymous
Kk.
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
its a word file
anonymous
  • anonymous
Is that the only you cant figure or is it all of them?
anonymous
  • anonymous
It is probably a typo as I know I got a and c right. If b is between the ones I got right, then I am sure it is a mistake
anonymous
  • anonymous
Is that really a textbook? If so, what publisher is it?
anonymous
  • anonymous
i am not sure about the publisher i had a pdf version
anonymous
  • anonymous
and i can't figure out how to do all 3
anonymous
  • anonymous
i think it is called nature of matter 11 thre could have possible been a mistake, they are common in textbooks
anonymous
  • anonymous
\[Half life period=\frac{ \ln2 }{ k }\] Mass after some days is given as y(t) where t is the time in days y0 would be initial mass. Formula for mass after some days is \[y(t)=y _{0}e ^{-kt}\]
anonymous
  • anonymous
Try it and would get a and c right but not b. Strange.
anonymous
  • anonymous
okay ty i will try it and see if i get the awnser right
anonymous
  • anonymous
KK, I can wait.
anonymous
  • anonymous
in the formula you put -kt what does that represent?
anonymous
  • anonymous
as the expoenet
anonymous
  • anonymous
There would be a negative sign because it is decaying. k is the constant from knowing the half life period. t is the time in days
anonymous
  • anonymous
with the formula i am doing y(8)=6.8 and im am confused for what i do after
anonymous
  • anonymous
because i am still getting the wrong answer
anonymous
  • anonymous
can you do a question so i know how its done
anonymous
  • anonymous
you forgot yo use the e^-kt part.
anonymous
  • anonymous
to*
anonymous
  • anonymous
You done e before in Algebra right?
anonymous
  • anonymous
yeah it's the exponent right? i didnt know what values would go for the k
anonymous
  • anonymous
Read my explanation above for the formula.
anonymous
  • anonymous
lol i did but i still dont understand
anonymous
  • anonymous
we briefly learned this in class so i am very confused
anonymous
  • anonymous
Half life period is the period it takes for an element to lose half its mass
anonymous
  • anonymous
which was four days
anonymous
  • anonymous
so is the k value supposed to be 4
anonymous
  • anonymous
Yes! Now solve for k
anonymous
  • anonymous
No. Read my 2 formulas above.
anonymous
  • anonymous
what does the IN mean?
anonymous
  • anonymous
Oh man, I thought algebra 2 was needed to take chemistry. ln is natural logarithm. But dont worry about its meaning since you probably havent took algebra 2 yet. It is in a scientific calculator. Do you see it. Its actually lowercase of L then n, not in btw.
anonymous
  • anonymous
yeah i found it now i can solve for k using this
anonymous
  • anonymous
btw this is highschool chemistry fir grade 11 and you only need grade 10 scince to be able to take it :P
anonymous
  • anonymous
is k value 1.45?
anonymous
  • anonymous
How?
anonymous
  • anonymous
beacuse in*2 *1.45=4
anonymous
  • anonymous
\[4=\frac{ \ln2 }{ k} \] Cross multiply \[4k=\ln2\] Divide both sides by 4 \[4=\frac{ \ln2 }{ k }\]
anonymous
  • anonymous
Dont forget to close the parentheses inside the ln
anonymous
  • anonymous
i am getting 0.17
anonymous
  • anonymous
if this isnt correct i will ask my chemistry teacher at school tom
anonymous
  • anonymous
im very confused right now
anonymous
  • anonymous
Yea you are correct. But makes but includes the other decimals. I got 0.1732
anonymous
  • anonymous
yeah i rounded
anonymous
  • anonymous
Now do a).
anonymous
  • anonymous
yeah i fiannly got the right awnser
anonymous
  • anonymous
tysm for helping me
anonymous
  • anonymous
No problem! Just remember the formula and its meanings and you're all good.
anonymous
  • anonymous
Okay thank you

Looking for something else?

Not the answer you are looking for? Search for more explanations.