6.02 ten to the power of 23
1 mole CO2 has about 44 grams, so half a mole of CO2 equals 22 grams
3.61 *10^3 mol CO2
1.76 grams CO2 (1 mole CO2/44.01 grams)(2 mole O/1 mole CO2)(6.022 X 10^23/1 mole O2) = 4.82 X 10^22 atoms of oxygen gas
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
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. So, 0.000831 mole of CO2 will have 0.000831 mole of carbon 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.
3.61 *10^3 mol CO2
1.76 grams CO2 (1 mole CO2/44.01 grams)(2 mole O/1 mole CO2)(6.022 X 10^23/1 mole O2) = 4.82 X 10^22 atoms of oxygen gas
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
6.5 grams CO2 divided by 44 grams CO2 per mole CO2 = 6.8 mole CO2 (Molar mass CO2 = 12 + 2*16 = 44 g/mol)
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.
CO2 + 4H2 --> CH4 + 2H2O0.500 moles CO2 (1 mole CH4/1 mole CO2) = 0.500 moles CH40.500 moles CO2 (2 moles H2O/1 mole CO2) = 1.00 moles H2O-------------------------------------------------------------------------------------add= 1.50 moles total product====================