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anonymous
 one year ago
HAlf Life problem
Radon222 has a halflife 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 radon222 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
 one year ago
HAlf Life problem Radon222 has a halflife 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 radon222 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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You mean 0.42 for b?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no it is 0.62 for 6 this question is from a text book with given awnsers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i will show you a picture one second

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that the only you cant figure or is it all of them?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It 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
 one year ago
Best ResponseYou've already chosen the best response.0Is that really a textbook? If so, what publisher is it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am not sure about the publisher i had a pdf version

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and i can't figure out how to do all 3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think it is called nature of matter 11 thre could have possible been a mistake, they are common in textbooks

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[Half life period=\frac{ \ln2 }{ k }\] 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
 one year ago
Best ResponseYou've already chosen the best response.0Try it and would get a and c right but not b. Strange.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay ty i will try it and see if i get the awnser right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in the formula you put kt what does that represent?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There 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
 one year ago
Best ResponseYou've already chosen the best response.0with the formula i am doing y(8)=6.8 and im am confused for what i do after

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because i am still getting the wrong answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you do a question so i know how its done

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you forgot yo use the e^kt part.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You done e before in Algebra right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah it's the exponent right? i didnt know what values would go for the k

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Read my explanation above for the formula.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol i did but i still dont understand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we briefly learned this in class so i am very confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Half life period is the period it takes for an element to lose half its mass

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so is the k value supposed to be 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes! Now solve for k

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No. Read my 2 formulas above.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what does the IN mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah i found it now i can solve for k using this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw this is highschool chemistry fir grade 11 and you only need grade 10 scince to be able to take it :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0beacuse in*2 *1.45=4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[4=\frac{ \ln2 }{ k} \] Cross multiply \[4k=\ln2\] Divide both sides by 4 \[4=\frac{ \ln2 }{ k }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Dont forget to close the parentheses inside the ln

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if this isnt correct i will ask my chemistry teacher at school tom

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im very confused right now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea you are correct. But makes but includes the other decimals. I got 0.1732

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i fiannly got the right awnser

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No problem! Just remember the formula and its meanings and you're all good.
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