Question

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

Answer 1

You mean 0.42 for b?

Answer 2

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

Answer 3

Check it again.

Answer 4

i will show you a picture one second

Answer 5

Kk.

Answer 6

its a word file

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

\[Half life period=\frac{ \ln2 }{ k }\]
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
\[y(t)=y _{0}e ^{-kt}\]

Answer 14

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

Answer 15

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

Answer 16

KK, I can wait.

Answer 17

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

Answer 18

as the expoenet

Answer 19

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

Answer 20

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

Answer 21

because i am still getting the wrong answer

Answer 22

can you do a question so i know how its done

Answer 23

you forgot yo use the e^-kt part.

Answer 24

to*

Answer 25

You done e before in Algebra right?

Answer 26

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

Answer 27

Read my explanation above for the formula.

Answer 28

lol i did but i still dont understand

Answer 29

we briefly learned this in class so i am very confused

Answer 30

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

Answer 31

which was four days

Answer 32

so is the k value supposed to be 4

Answer 33

Yes!
Now solve for k

Answer 34

No.
Read my 2 formulas above.

Answer 35

what does the IN mean?

Answer 36

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.

Answer 37

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

Answer 38

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

Answer 39

is k value 1.45?

Answer 40

How?

Answer 41

beacuse in*2 *1.45=4

Answer 42

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

Answer 43

Dont forget to close the parentheses inside the ln

Answer 44

i am getting 0.17

Answer 45

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

Answer 46

im very confused right now

Answer 47

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

Answer 48

yeah i rounded

Answer 49

Now do a).

Answer 50

yeah i fiannly got the right awnser

Answer 51

tysm for helping me

Answer 52

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

Answer 53

Okay thank you