The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1×1010 133ba atoms. how many are left after (a) 6 years, (b) 10 years, and (c) 200 years?

(a) 7.4 x 10⁹ atoms.

(b)

(c)

Explanation:

k = ln(2)/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(10.5 years) = 0.066 year⁻¹.

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 0.066 year⁻¹).

t is the time of the reaction.

[A₀] is the initial concentration of (133-Ba) ([A₀] = 1.1 x 10¹⁰ atoms).

[A] is the remaining concentration of (133-Ba) ([A] = ??? g).

(a) 6 years:

t = 6.0 years.

∵ kt = ln([A₀]/[A])

∴ (0.066 year⁻¹)(6.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])

∴ 0.396 = ln((1.1 x 10¹⁰ atoms)/[A]).

Taking exponential for both sides:

∴ 1.486 = ((1.1 x 10¹⁰ atoms)/[A]).

∴ [A] = (1.1 x 10¹⁰ atoms)/(1.486) = 7.4 x 10⁹ atoms.

(b) 10 years

t = 10.0 years.:

∵ kt = ln([A₀]/[A])

∴ (0.066 year⁻¹)(10.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])

∴ 0.66 = ln((1.1 x 10¹⁰ atoms)/[A]).

Taking exponential for both sides:

∴ 1.935 = ((1.1 x 10¹⁰ atoms)/[A]).

∴ [A] = (1.1 x 10¹⁰ atoms)/(1.935) = 5.685 x 10⁹ atoms.

(c) 200 years:

t = 200.0 years.

∵ kt = ln([A₀]/[A])

∴ (0.066 year⁻¹)(200.0 year) = ln((1.1 x 10¹⁰ atoms)/[A])

∴ 13.2 = ln((1.1 x 10¹⁰ atoms)/[A]).

Taking exponential for both sides:

∴ 5.4 x 10⁵ = ((1.1 x 10¹⁰ atoms)/[A]).

∴ [A] = (1.1 x 10¹⁰ atoms)/(5.4 x 10⁵) = 2.035 x 10⁴ atoms.

answer is a

explanation:

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.