3. An experiment was conducted to determine how various drop heights of ball bearings affects the diameter of the indentation. The ball bearings were 5 cm in
diameter and dropped into 250 grams of sand. The sand was contained in a pie
tin 25 cm across. Students measured the width (cm) of the crater caused by the
ball bearings falling into the sand. The height of the ball bearings was increased
by 10 cm cach time.

Volume of Box / volume of tin = number of tins required

Volume of box =50*40*20= 40000 cu.cm

Volume of Tin = pi*r^2*h=pi*4^2*15 = 754 cu.cm

4000/754= 53 tins (rounded off)

i'm sorry if this is a bit complicated but i hope it answers your question. Take care.

Explanation:

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.

i don't see a text, but i am gonna assume this *orbit* the curved path, usually elliptical, described by a planet, satellite, spaceship, etc., around a celestial body, as the sun.