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

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

You mean 0.42 for b?

- anonymous

no it is 0.62 for 6
this question is from a text book with given awnsers

- anonymous

Check it again.

Looking for something else?

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

## More answers

- anonymous

i will show you a picture one second

- anonymous

Kk.

- anonymous

##### 1 Attachment

- anonymous

its a word file

- anonymous

Is that the only you cant figure or is it all of them?

- 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

Is that really a textbook?
If so, what publisher is it?

- anonymous

i am not sure about the publisher i had a pdf version

- anonymous

and i can't figure out how to do all 3

- anonymous

i think it is called nature of matter 11
thre could have possible been a mistake, they are common in textbooks

- 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

Try it and would get a and c right but not b.
Strange.

- anonymous

okay ty i will try it and see if i get the awnser right

- anonymous

KK, I can wait.

- anonymous

in the formula you put -kt what does that represent?

- anonymous

as the expoenet

- 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

with the formula i am doing y(8)=6.8
and im am confused for what i do after

- anonymous

because i am still getting the wrong answer

- anonymous

can you do a question so i know how its done

- anonymous

you forgot yo use the e^-kt part.

- anonymous

to*

- anonymous

You done e before in Algebra right?

- anonymous

yeah
it's the exponent right?
i didnt know what values would go for the k

- anonymous

Read my explanation above for the formula.

- anonymous

lol i did but i still dont understand

- anonymous

we briefly learned this in class so i am very confused

- anonymous

Half life period is the period it takes for an element to lose half its mass

- anonymous

which was four days

- anonymous

so is the k value supposed to be 4

- anonymous

Yes!
Now solve for k

- anonymous

No.
Read my 2 formulas above.

- anonymous

what does the IN mean?

- 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

yeah i found it
now i can solve for k using this

- anonymous

btw this is highschool chemistry fir grade 11 and you only need grade 10 scince to be able to take it :P

- anonymous

is k value 1.45?

- anonymous

How?

- anonymous

beacuse in*2 *1.45=4

- anonymous

\[4=\frac{ \ln2 }{ k} \]
Cross multiply
\[4k=\ln2\]
Divide both sides by 4
\[4=\frac{ \ln2 }{ k }\]

- anonymous

Dont forget to close the parentheses inside the ln

- anonymous

i am getting 0.17

- anonymous

if this isnt correct i will ask my chemistry teacher at school tom

- anonymous

im very confused right now

- anonymous

Yea you are correct.
But makes but includes the other decimals.
I got 0.1732

- anonymous

yeah i rounded

- anonymous

Now do a).

- anonymous

yeah i fiannly got the right awnser

- anonymous

tysm for helping me

- anonymous

No problem!
Just remember the formula and its meanings and you're all good.

- anonymous

Okay thank you

Looking for something else?

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