k = 0.693/t1/2 = 0.693/2.43 = ??
No is what we want to start with and we want N to be 5.00 mg at the end of 4.50 days. .
ln(No/N) = kt
ln(No/5.00) = k(4.50 days)
solve for No which will be mg 128Ba. Convert that to BaCl2.
Check my work.
A researcher needs 5.00 mg of 128Ba for an experiment. If the half-life of 128Ba is 2.43 days, how many milligrams of 128BaCl2 must she order from the manufacturer if it takes 4.50 days to ship the material from the manufacturer to the university? (Assume the molar mass of 128Ba is 128 g/mol.)
4 answers
No=18.0432
but im not sure what you mean by convert to BaCl2
but im not sure what you mean by convert to BaCl2
I don't think your prof would like for you not to include units. That is 18.043 mg 128Ba (technically that is too many place--we are allowed only 3 from the 5.00 mg and 4.50 days but I like to keep extra places, usually just keep them in the calculator, and round at the very end.) The conversion part--note that the problem asks for BaCl2, not Ba. You know the 128Ba so you must convert that to BaCl2. You can't look up the molar mass of BaCl2 because the numbers in the periodic table are AVERAGES of all of the isotopes of naturally occurring Ba which may or may not include 128Ba. So you look up the mass of Cl, that x 2 for the Cl2 part, then add 128 for the 128Ba to obtain the molar mass of 128BaCl2. Then 18.043 mg Ba x (1 mol BaCl2/1 mol Ba) x (molar mass BaCl2/1 mol BaCl2) = ? mg, then round to three significant figures. That is the amount of 128BaCl2 that should be bought so that it can be shipped and 4.50 days latter it still will contain 5.00 mg Ba.
As a passing note of interest, I hope, is I worked with an isotope of Cu (64Cu) for about a year and it had a half-life of 12.9 hours. It was made in Oak Ridge, Tennessee and shipped to the lab in Missouri and by the time I received it the stuff had gone through four half-lives (only 1/16 of the original activity was there). Of course it took a day before I could use it and get around to counting it. By then I had almost nothing left in the form of radioactive Cu.But I persevered.
As a passing note of interest, I hope, is I worked with an isotope of Cu (64Cu) for about a year and it had a half-life of 12.9 hours. It was made in Oak Ridge, Tennessee and shipped to the lab in Missouri and by the time I received it the stuff had gone through four half-lives (only 1/16 of the original activity was there). Of course it took a day before I could use it and get around to counting it. By then I had almost nothing left in the form of radioactive Cu.But I persevered.
oh ok thank you very much, is an online class and have to teach myself so somethings get confusing
1 answer