For complete proper combustion of Propane:
C3H8 + 5O2 = 3CO2 + 4H2O
The relative atomic weights of a molecule of Propane and Oxygen are:
Propane: 3 × C + 8 × H = 3 × 12 + 8 × 1 = 44
Oxygen: 2 × O = 2 × 16 = 32
Thus a molecule of propane is 44/32 = 1⅜ times heavier than a molecule of oxygen; and the same amount (number of molecules) of propane as 24 g of oxygen is 24g × 1⅜ = 33g
Each propane molecule takes 5 oxygen molecules, thus:
33 g ÷ 5 = 6 3/5 g = 6.6 g
If the combustion produces the poisonous carbon monoxide instead of carbon dioxide:
2C3H8 + 7O2 = 6CO + 8H2O
→ propane = 33g × 2/7 = 9 3/7 g ≈ 9.4 g
A complete answer is thus between 6 3/5 g (6.6g) and 9 3/7 g (9.4 g) depending upon how much carbon monoxide relative to carbon dioxide is produced by the burning - the safe amount is 6.6 g.
1 gram = 1000 milligrams so 24 grams = 24*1000 = 2400 milligrams. Simple!
3/8 of 24 grams is 9 grams.
12 x 14/24 = 7 grams
1 kg = 1000 grams because that is the definition of a kilogram. You have 24 lots of 1 kg so you have 24 lots of 1000 grams. And, 24 lots of 1000 grams is 1000*24 grams.
24.
The answer is 6,61 g.
Let's see. C3H8 + 5O2 -> 3CO2 + 4H2O For every one molecule of propane burned there is four molecules of water produced. Or, this is the actuality. 1 molecule propane (1 mole C3H8/6.022 X 1023)(4 mole H2O/ 1mole C3O8) = 6.64 X 10 -24 molecules water
There are 10,886.2169 grams in 24 pounds.
There are 1000 grams in one kilogram. Therefore, 24 kilograms is equal to 24 x 1000 = 24000 grams.
how many grams is 39cc?
24 oz = 680.4 g
15 grams of water in 1 tablespoon. So 24 is 360 grams
There are 1000 milligrams in one gram. Therefore, 24 milligrams is equal to 24/1000 = 0.024 grams.
120 grams of water is 24 teaspoons.
24 grams.
1 gram = 1000 milligrams so 24 grams = 24*1000 mg - 24,000 mg. Simple!
1 gram = 1000 milligrams so 24 grams = 24*1000 = 2400 milligrams. Simple!