The initial mass of a radioactive goo is 180 grams. After 195 minutes, the sample has decayed to 2.8125 grams. Find the half-life of the substance in minutes? Find a formula for the amount, G(t), remaining at time t.? Find grams remain after 35 min?

Answer 1

Forms of the decay formula that we are aware of are:

#G(t) = G(0)(1/2)^(t/t_(1/2))" [1]"#
Where #G(0)# is the initial amount and #t_(1/2)# is the half-life.
Given #G(0) = 180" g" and G(195" min") = 2.8125" g"#

Enter the previous value in equation [1]:

#2.8125" g" = (180" g")(1/2)^((195" min")/t_(1/2))#
Solve for half-life, #t_(1/2)#.
Divide both sides by #180" g"#:
#(2.8125" g")/(180" g") = (1/2)^((195" min")/t_(1/2))#

Calculate both sides' natural logarithms:

#ln((2.8125" g")/(180" g")) = ln((1/2)^((195" min")/t_(1/2)))#

Make use of the logarithmic property, which permits the exponent to be moved to the outside as a coefficient:

#ln((2.8125" g")/(180" g")) = ((195" min")/t_(1/2))ln(1/2)#
Multiply both sides by #t_(1/2)#
#t_(1/2)ln((2.8125" g")/(180" g")) = (195" min")ln(1/2)#
Divide both sides by #ln((2.8125" g")/(180" g"))#:
#t_(1/2) = (195" min")ln(1/2)/ln((2.8125" g")/(180" g"))#
#t_(1/2) = 32.5" min"#
The formula for #G(t)# is:
#G(t) = G(0)(1/2)^(t/(32.5" min")#

The 180-gram sample's remaining quantity after 35 minutes:

#G(35" min") = (180" g")(1/2)^((35" min")/(32.5" min"))#
#G(35" min") = 85.327" g"#
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Answer 2

Half-life ((t_{1/2})): (t_{1/2} = \frac{\ln(2)}{\text{slope}})

Exponential decay formula: (G(t) = G_0 \cdot e^{-kt})

Given dataHalf-life ((t_{1/2})): (t_{1/2} = \frac{\ln(2)}{\text{slope}})

Exponential decay formula: (G(t) = G_0 \cdot e^{-kt})

Given Initial mass ((G_0)): 180 g Final mass after 195 min: 2.8125 g

Solve for (k): (2.8125 = 180 \cdot e^{-195k})

Formula for remaining mass: (G(t) = 180 \cdot e^{-kt})

Grams remaining after 35 min: (G(35) = 180 \cdot e^{-35k})

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Answer 3

Half-life = ( t_{\frac{1}{2}} = \frac{t \times \ln(2)}{\ln(\frac{G_0}{G_t})} )

Formula for the remaining amount at time ( t ): ( G(t) = G_0 \times e^{-\frac{t}{t_{\frac{1}{2}}}} )

Grams remaining after 35 minutes: ( G(35) = 180 \times e^{-\frac{35}{t_{\frac{1}{2}}}} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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