The half-life of cesium-137 is 30 years. suppose we have a 40 mg sample. exercise (a) find the mass that remains after t years. step 1 let y(t) be the mass (in mg) remaining after t years. then we know the following. y(t) = y(0)ekt = · ekt exercise (b) how much of the sample remains after 20 years? step 1 after 20 years we have the following. y(20) = 40 · 2 = mg (round your answer to two decimal places.)

1) y(t) = (40.0 mg)(e∧(-(0.0231 year⁻¹)t).

2) 25.2 mg.

Explanation:

(a) Find the mass that remains after t years. Step 1 Let y(t) be the mass (in mg) remaining after t years. Then we know the following. y(t) = y(0)ekt

For first order reactions: y(t) = y(0)(e∧-kt)

where, y(t) is the mass of the substance at any time (t).

y(0) is the initial concentration of the substance at (t = 0) (y(0) = 40.0 mg).

k is the rate constant of the reaction.

t is the time of the reaction.

For first order reactions: k = ln2/(t1/2) = 0.693/(30 years) = 0.0231 year⁻¹.

∴ y(t) = y(0)(e∧-kt)

y(t) = (40.0 mg)(e∧(-(0.0231 year⁻¹)t).

Exercise (b) How much of the sample remains after 20 years?

∵ y(t) = y(0)(e∧kt)

k = 0.03465 year⁻¹, t = 20.0 years, y(0) = 40.0 mg.

∴ y(t) = y(0)(e∧-kt) = (40.0 mg)e∧-(0.0231 year⁻¹)(20.0 years) = 25.2 mg.

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.