Chemistry help. Mole introduction.?

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AnnaKedavrat

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I had activities about moles that I had to do in class, and answer questions about it. I am so confused about it all. So I'll just give you three that I can use to check my other stuff(so you know I'm not trying to get you to do my homework).

1. How any drops of H2O can fit on a penny (3 trials).
Trial 1- 53 drops.
Trial 2- 27 drops.
Trial 3- 38 drops.
Average=40 drops.

Show how to determine the number of molecules:
HINT: 20 drops= 1 mL


2. How many moles of grassblades are in an average football field?
REMEMBER: 150 blades per square inches, a football field is 360ft by 160ft.

I did all the work to find out how many blades of grass and got 10,370,000 (sig figs) BUT THIS IS NOT GUARENTEED NUMBERS, SO CAN YOU PLEASE CHECK THIS?


3. How many molecules of carbon dioxide are in the balloon (if perfectly sphericle) The circumfrence, I measured wasa 77 cm.
(1 mole CO2= 44.0 g, 1 mole CO2=22.4 L, 1cm^3=1 mL, 1 in=2.54 cm.)


10 points to anyone that can help me =/ I'm so lost.
Please help me...I'm desperate.
 
1. 40 drops of H2O can fit on a penny
Show how to determine the number of molecules:
40 drops @ 1 ml / 20 drops = 2ml

2ml of water @ density of 1 g/ml = 2 grams

2 grams of water @ molar mass of 1 mole / 18 grams = 0.111 moles of water

avagadro's number:
0.111 moles @ 6.02 e23 molecules / mole = 6.69 e22 molecules

which should at least be rounded to 6.7 e22 molecules
==========================================================


2. How many moles of grassblades are in an average football field?

a football field is 360ft by 160ft = 57,600 square feet
57,600 square feet @ 144 sq inches / ft2 = 8,294,400 in2
8,294,400 in2 @ 150 blades / in2 = 1.24 e9 blades

using avagadro's number:
1.24 e9 blades @ 1mole / 6.02 e23 blades = 2.07 e-15 moles
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3. How many molecules of carbon dioxide are in the balloon (if perfectly sphericle) The circumfrence, I measured wasa 77 cm.
circumference = 2 pi r
77 = 2 (3.14) r
r = 12.26 cm in radius

volume of a sphere = 4/3 pi r3
vol = (4/3) (3.14)(77)^3
vol = 7717 cm3

7717cm3 @ 1 litre / 1000cm3 = 7.717 litres

7.717 litres @ 1 mole / 22.4 litres = 0.3445 moles

0.3445 moles @ 6.02e23 molecules / mole = 2.07 e23 molecules

which likely should be rounded off to 2.1 e23 molecules
(1 mole CO2= 44.0 g, 1 mole CO2=22.4 L, 1cm^3=1 mL, 1 in=2.54 cm.)
 
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