Thulium-167 has a half life of 9.0 days. if the initial amount of thulium is 50.0 grams. how many grams are left after 36.0 days?

3.125 grams.

Explanation:

∵ Thulium-167 has a half life of 9.0 days.

∴ The time is needed to calculate the grams are left after (36.0 days) represents (36.0 days / 9.0 days) = 4.0 half-lives.

50.0 grams → (first half life) 25.0 grams → (second half life) 12.5 grams → (third half life) 6.25 grams → (fourth half life) 3.125 grams.

So, the grams are left after 36.0 days = 3.125 grams.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

answer: you could do some invislibe ink.

explanation: i will tell you how to if u answer my question.