6.02 ten to the power of 23
1 mole CO2 has about 44 grams, so half a mole of CO2 equals 22 grams
atomic weight of carbon dioxide is 2 * 16 + 12 = 44 1 kg = 1000 g 1 kg of co2 has 1000/44 = 22.7 moles yeh i think that's wrong lol isn't it 3.37E25?
When 1 mole of C8H18 is burned, it forms 8 moles of CO2. Therefore, when 451 moles of C8H18 is burned, it will form 8 * 451 = 3608 moles of CO2.
CaCO3 +2HCl ------------> CaCl2 + CO2 + H2O number of moles of CO2 in .44 grams = .44/ 44 = .01 From equation it is clear that 1 mole of CO2 is produced from CaCO3 = 1 mole .01 mole of CO2 is formed from CaCO3 = .01 mole Weight of .01 mole of CaCO3 is = .01mole *100 g/mole = 1 gram weight % of CaCO3 is = 1*100/ 1.25 = 80 % w/w I've post my answer, so why don't you show that answer here with the question. It's fare. I must be informed about my answer weather it is right or wrong. Please inform me at amitmahalwar@yahoo.com
carbon= 12.01 oxygen=16*2 12.01+32=44.01 CO2
6.02 ten to the power of 23
1 mole CO2 has about 44 grams, so half a mole of CO2 equals 22 grams
1 mole of CO2 has 1 mole of carbon atoms and 2 moles of oxygen atoms.
Sodium Bicarbonate come from absorption of CO2 into NaOH Total reaction is 2NaOH + CO2 -> NaHCO3 + H2O Each mole of NaHCO3 come from absorption of 1 mole CO2 Molecular weight of NaHCO3 is 84 g/mol and CO2 is 44 g/mol It is that 84 g of NaHCO3 had 44 g of CO2 By weight ratio it is 52% CO2 in Sodium Bicarbonate.
1 mole of CO2 has 1 mole of carbon atoms and 2 moles of oxygen atoms. So, 0.000831 mole of CO2 will have 0.000831 mole of carbon atoms.
Assuming you mean gaseous CO2. You can roughly approximate by PV=nRT, where P and T are ambient pressure and temperature and V is the volume of the Lorry. Solve for # of moles, n. (n=PV/(RT)) For weight of CO2, each mole = atomic weight of Carbon plus 2x atomic weight of Oxygen. (44.01 grams / mole)
atomic weight of carbon dioxide is 2 * 16 + 12 = 44 1 kg = 1000 g 1 kg of co2 has 1000/44 = 22.7 moles yeh i think that's wrong lol isn't it 3.37E25?
Yes. One mole of anything contains 6.02x10^23 "particles". In the case of the element uranium, it would be 6.02x10^23 atoms of uranium in 1 mole. In the case of CO2, it would be 6.02x10^23 molecules of CO2 in 1 mole.
When 1 mole of C8H18 is burned, it forms 8 moles of CO2. Therefore, when 451 moles of C8H18 is burned, it will form 8 * 451 = 3608 moles of CO2.
CaCO3 +2HCl ------------> CaCl2 + CO2 + H2O number of moles of CO2 in .44 grams = .44/ 44 = .01 From equation it is clear that 1 mole of CO2 is produced from CaCO3 = 1 mole .01 mole of CO2 is formed from CaCO3 = .01 mole Weight of .01 mole of CaCO3 is = .01mole *100 g/mole = 1 gram weight % of CaCO3 is = 1*100/ 1.25 = 80 % w/w I've post my answer, so why don't you show that answer here with the question. It's fare. I must be informed about my answer weather it is right or wrong. Please inform me at amitmahalwar@yahoo.com
Yes. CO2 has a weight of 44g/mol and O2 has a weight of 32g/mol.