Well, honey, sucrose is made up of carbon, hydrogen, and oxygen. Carbon makes up about 40% of the molecular weight of sucrose, so in 100.0g of sucrose, you'd have about 40.0g of carbon. Just remember, sugar might be sweet, but chemistry sure ain't.
11 grams because all is reacted and there is no reactant left over, although if there were only 3 grams of carbon there would have to be 6 grams of oxygen for this to be viable as carbon dioxide is CO2 so the question asked was itself wrong.
5 moles carbon? 5 grams carbon? Both done. 5 moles carbon (6.022 X 1023/1 mole C) = 3.0 X 1024 atoms carbon =================== 5 grams carbon (1 mole C/12.01 grams)(6.022 X 1023/1 mole C) = 2.5 X 1023 atoms carbon ==================== your choice
Ideal gas equation: pV=nRT STP is 298K (temperature) and 1 ATM (pressure). Use the ideal gas equation to calculate the number of moles of oxygen in 2L at STP. Complete oxidation means that the molecule is oxidised to carbon dioxide and water. Sucrose is C12H22O11 meaning is will be oxidised to 12 carbon dioxide and 11 water molecules (it has 12 carbon atoms and 22 hydrogen atoms in). To produce these, it will need 12x2 + 11 oxygen atoms. Since 11 atoms are already present in the sucrose, it requires an additional 24 atoms of oxygen per molecule of sucrose. One mole is a fixed number of atoms, so the ratio will stay the same. Hence 1 mole of sucrose requires 24 moles of oxygen to oxidise it fully. Divide the number of moles of oxygen you have calculated are present in the 2L by the number of moles of oxygen required to fully oxidise the sucrose (24). This tells you how many moles of sucrose you can fully oxidise with your oxygen. m=n x Mr The mass is equal to the number of moles multiplied by the relative molecular mass. You can therefore multiply your result by the molecular mass of sucrose to find how many grams of sucrose you can fully oxidise. Molecular mass is the addition of the atomic masses of each of the atoms in a molecule. Sucrose has a molecular mass of 342.3 grams/mole.
Balanced equation. C + 2Cl2 -> CCl4 10 grams carbon (1 mole C/12.01 grams)(1 mole CCl4/1 mole C)(153.81 grams/1 mole CCl4) 128 grams carbon tetrachloride produced -----------------------------------------------------
To calculate the number of moles of carbon dioxide in 19 grams, divide the given mass by the molar mass of carbon dioxide, which is approximately 44 grams/mol. Therefore, 19 grams of carbon dioxide is equal to 19/44 ≈ 0.43 moles.
1000g
1kg = 1000g 6.9kg x 1000g/kg = 6900g
1kg = 1000g 1.73kg x 1000g/kg = 1730g
1kg = 1000g 20005000g x (1kg/1000g) = 20005kg
1kg = 1000g 57.2g x 1kg/1000g = 0.0572kg
1kg = 1000g 400g x 1kg/1000g = 0.400kg
1kg = 1000g 1.4kg x 1000g/kg = 1400g
1kg is equal to 1000g
1000g
1kg = 1000g, so 4.97kg = 4970 grams.
There are 1000g in 1 kg So 14 x 1000g = 14000g
There are approximately 28.35 grams in one ounce.