Confused about Chemical Q Decay Calculation?

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Discussion Overview

The discussion revolves around the calculation of the original mass of Chemical Q based on its decay rate, specifically addressing the mathematical formulation of the decay process over time. Participants are examining the implications of the decay model and the correct application of the formula in a practical context.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for the original mass of Chemical Q, using the formula based on a decay of one-third every 7 years.
  • Another participant suggests that the correct approach should involve using two-thirds of the mass remaining after each decay period, rather than one-third.
  • A participant expresses confusion regarding the transition from one-third loss to two-thirds remaining, seeking clarification on the reasoning behind this adjustment.
  • Further clarification is provided that if one-third is lost, then two-thirds remains, leading to a revised equation for the decay process.
  • A later reply confirms that the original calculation should indeed use two-thirds instead of one-third, indicating a consensus on this point.

Areas of Agreement / Disagreement

Participants generally agree that the decay model should use two-thirds of the mass remaining after each period, but there is initial disagreement on the correct formulation of the equation. The discussion reflects a process of clarification and correction rather than a definitive resolution of the original calculation.

Contextual Notes

The discussion highlights the importance of understanding the relationship between mass loss and remaining mass in decay calculations, as well as the potential for confusion in applying these concepts mathematically.

helpm3pl3ase
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I have a question Iam stuck on. I was wondering if i did the right calculations. Please let me know:

Through decay, Chemical Q loses one-third of its mass every 7 years.
Twenty-five years ago, a container of Chemical Q was buried. The EPA has
just dug it up and found 1,000 grams remaining. How much Chemical Q was
originally buried?

1000=(Q)(1/3)^(x/7) when x = 25

1000=(Q)(1/3)^(25/7)

Q = 1000/(3)^(3/7)/81

Q = 50583?

I am not sure if I did this correct??
 
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If it loses 1/3 of its mass every 7 years, then 2/3 of its mass is left at the end of 7 years, and (2/3)^2 at the end of 14 years, etc. I think you used 1/3 where you should have used 2/3...?
 
hmm I don't really understand that?? The question called for one third. Iam not understanding wher eyou get 2/3
 
If it loses 1/3 then you have 2/3 left. so every 7 years the new amount would equal 2/3 the old amount. that's why the equation is
(Q)(2/3)^(x/7) - you're doing (Q)(2/3 * 2/3 * 2/3 * ... x/7 times)
in other words, to take away 1/3 each time you multiply by 2/3
 
O alright.. I get it now. Thank you

So everything thing else is correct, but instead of 1/3 use 2/3?
 
Yes, use 2/3 instead of 1/3.
 

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