1.50553543e+30 ppm
6.02 ten to the power of 23
3.61 *10^3 mol CO2
-- Take the number of percent. -- Multiply it by 10,000-- The answer is the number of parts per million. . 10 percent = 100,000 parts per million.
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
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)
6.02 ten to the power of 23
3.61 *10^3 mol CO2
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.
-- Take the number of percent. -- Multiply it by 10,000-- The answer is the number of parts per million. . 10 percent = 100,000 parts per million.
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
For every million parts of air there are 400 ppm (parts per million) or 0.04% of CO2(carbon dioxide). This is serious increase from 280 ppm (parts per million) or 0.028% which was the level for thousands of years before the Industrial Revolution, 200 years ago.
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)
1 mole CO2 has about 44 grams, so half a mole of CO2 equals 22 grams
7/2O2+C2H6->2CO2+3H2O Mole ratio of C2H6:CO2=1:2 No.of mole of CO2=2mol
Also 0,1 mole carbon dioxide.
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====================
1.3 mole C2H2 will produce 2.6 mole CO2, weighting 2.6(mol) x 44 (g/mol) = 114 g CO2