There is a 30 g sample of Be-11. It has a half-life of about 14 seconds. How much will be left after 28 seconds?A.3.75 gB.7.5 gC.15 gD.30 g

After 28 sec 7.5 grams of Be will be left

Explanation:

Given data:

Total amount of Be = 30 g

Half life of Be = 14 sec

Mass left after 28 sec = ?

Solution:

First of all we will calculate the number of half lives in 28 sec.

Number of half lives = T elapsed / half life

Number of half lives = 28 sec / 14 sec

Number of half lives = 2

AT time zero = 30 g

At first half life = 30 g/2 = 15 g

At second half life = 15 g/ 2 = 7.5 g

So after 28 sec 7.5 grams of Be will be left.

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.

the answer is na (sodium)