Question

A 45 gram sample of a substance that's used to preserve fruit and vegetables has a k-value of 0.1088

Answer 1

Answer:

The substance's half-life is 6.4 days

Step-by-step explanation:

Recall that the half life of a substance is given by the time it takes for the substance to reduce to half of its initial amount. So in this case, where they give you the constant k (0.1088) in the exponential form:

$N=N_0\,e^{-k\,*\,t}$

we can replace k by its value, and solve for the time "t" needed for the initial amount $N_0$ to reduce to half of its value ($N_0/2$). Since the unknown resides in the exponent, to solve the equation we need to apply the natural logarithm:

$N=N_0\,e^{-k\,*\,t}\\\frac{N_0}{2} =N_0\,e^{-0.1088\,*\,t}\\\frac{N_0}{2\,*N_0} =e^{-0.1088\,*\,t}\\\frac{1}{2} =e^{-0.1088\,*\,t}\\ln(\frac{1}{2} )=-0.1088\,t\\t=\frac{ln(\frac{1}{2} )}{-0.1088} \\t=6.37\,\,days$

which rounded to the nearest tenth is: 6.4 days

Answer 2

Answer:

6.4

Step-by-step explanation:

I did it on the same site and got it correct