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Strontium-90, a radioactive isotope, is a major product of an atomic bomb explosion. It has a half-life of 28.1 yr. (a) Calculate the first-order rate constant for the nuclear decay. (b) Calculate the fraction of $^{90}$ Sr that remains after 10 half-lives. (c) Calculate the number of years required for 99.0 percent of 90 Sr to disappear.

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(a) $0.0247 \mathrm{yr}^{-1}$(b) $\frac{1}{1024}$(c) 186 years

Chemistry 102

Chapter 6

Chemical Kinetics

Kinetics

Lionel G.

September 17, 2021

A 2.5 gram sample of an isotope of strontium-90 was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific Test Site. The half-life of strontium-90 is 28 years. In what year will only 0.625 grams of this strontium-90 remain?

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University of Maryland - University College

University of Kentucky

Brown University

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22:42

In probability theory, the conditional probability of an event A given that another event B has occurred is defined as the probability of A given B, written as P(A|B). It is a function of the probability of B, the probability of A given B, and the probability of B.

04:55

In chemistry, kinetics is the study of the rates of chemical reactions. The rate of a reaction is the change in concentration of a reactant over time. The rate of reaction is dependent on the concentration of the reactants, temperature, and the activation energy of the reaction.

01:12

Strontium-90 is a radioact…

01:30

Radioactive decay exhibits…

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Radioactive Decay Strontiu…

Well, today we're going to be looking at strong team 90 which has the half life shown here 28.1 years. So we know the half life and this is a first order reaction. So let's figure out the great constant if we remember for a first order reaction, the half life is equal to the natural log of to over there a constant so they're a constant would just be the natural log of to over the half life. So let's take the natural log of two and divided by 28.1, and we see that our rate constant is zero 0.0 to 47 in verse years. So we found the rate constant. Now let's think about, say, 10 half lifes pass what fraction of strong Tim will remain. Well, half life for every half life we are time. We are dividing the fraction in half, so we start off with one and we have 10 half lives. So if there was one half life, it would be half. It was to half life, so it would be a quarter. And so let's just take half to the 10th power and we will see that our fraction will be 0.0 nine eat, which is 0.98%. So this is less than 0.1 point percent. This is very, very small, as we can see So now. But we want Teoh. Let's say we want to find out how many years have to pass for 99% of the strong, too young to disappear. Well, how would we figure that out? Well, we know that for first order reaction, the natural log of the concentration of the reactant in this key strong team at a certain time is equal to the natural log of the industrial concentration. It's retracting three constant times time we can rearrange that. So it looks like this. So we know the rate constant already, and we're given that 99% of it has really of the strong team has reacted. That means we're going to have zero point zero one left. That's 1%. Well, the well, it's one over 100 and this is going to be equal to negative. We know our K is 0.2 for seven get times time. And so if we take the natural log of 0.1 We see that it will be negative 4.6. So let's divide that by negative 0.247 and we see that we need ah, 186 years to pass, which is a lot, and so there.

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