0,0235 x 180,16 = 4,238 g
Molarity = moles of solute/Liters of solution Find moles glucose, which is molecular formula------C6H12O6 154 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.8548 moles C6H12O6 Molarity = 0.8548 moles C6H12O6/1 Liter = 0.855 M glucose ---------------------------
147.2 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6 moles O/1 mole C6H12O6)(16.0 grams/1 mole O) = 78.4 grams oxygen ----------------------------
Find moles of glucose. 32.8 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.18206 moles C6H12O6 Molarity = moles of solute/Liters of solution M = 0.18206 moles C6H12O6/1.0 L = 0.18 M C6H12O6
7.11 grams glucose (1 mole C6H12O6/180.156 grams)(6 mole C/1 mole C6H12O6)(6.022 X 10^23/1 mole C) = 1.43 X 10^23 atoms of carbon ------------------------------------------
glucose 1 mole has 180,156 grams and has 6.022 x 1023 atoms carbon 1 mole has 12,01 grams and has 6.022 x 1023 atoms. There are 6 carbon atoms in a glucose molecule so that times six would give you a total of 72,06 grams out of the 180,156. Carbon makes up about 40 percent of the total glucose mass so the final answer would be it would be around 2.4088 x 1023 atoms of carbon in one gram of glucose.
Molarity = moles of solute/Liters of solution Find moles glucose, which is molecular formula------C6H12O6 154 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.8548 moles C6H12O6 Molarity = 0.8548 moles C6H12O6/1 Liter = 0.855 M glucose ---------------------------
147.2 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6 moles O/1 mole C6H12O6)(16.0 grams/1 mole O) = 78.4 grams oxygen ----------------------------
(5.72g c6h12o6)*6.022*10^23atoms c6h12o6/180.156gc6h12o6=1.91*10^22 atoms c6h12o6
18 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6.022 X 10^23/1 mole C6H12O6) = 6.02 X 10^22 atoms
Find moles of glucose. 32.8 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.18206 moles C6H12O6 Molarity = moles of solute/Liters of solution M = 0.18206 moles C6H12O6/1.0 L = 0.18 M C6H12O6
7.11 grams glucose (1 mole C6H12O6/180.156 grams)(6 mole C/1 mole C6H12O6)(6.022 X 10^23/1 mole C) = 1.43 X 10^23 atoms of carbon ------------------------------------------
0.67 moles of C6H12O6
C6H12O6 + 6O2 --> 6CO2 + 6H2O 45 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6 moles CO2/1 mole C6H12O6)(44.01 grams/1 mole CO2) = 66 grams carbon dioxide produced ==========================
15.8 grams of C6H12O6 is equivalent to 0.087 moles.
glucose 1 mole has 180,156 grams and has 6.022 x 1023 atoms carbon 1 mole has 12,01 grams and has 6.022 x 1023 atoms. There are 6 carbon atoms in a glucose molecule so that times six would give you a total of 72,06 grams out of the 180,156. Carbon makes up about 40 percent of the total glucose mass so the final answer would be it would be around 2.4088 x 1023 atoms of carbon in one gram of glucose.
C6H1206 has 6 Carbon atoms, 12 Hydrogen atoms, and 6 Oxygen atoms, otherwise known as Glucose, a plant food made for long-term storage. Sorry if that's not what you're looking for! Added: This is what you are looking for. 300 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6 mole C/1 mole C6H12O6)(6.022 X 10^23/1 mole C)(1 mole C/6.022 X 10^23) = 9.99 moles Carbon atoms in that mass glucose 300 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(12 mole H/1 mole C6H12O6)(6.022 X 10^23/1 mole H)(1 mole H/6.022 X 10^23) = 19.98 moles Hydrogen atoms in that mass glucose Now, you have seen two examples of this procedure, so you do the oxygen number crunching.
Molarity = moles of solute/Liters of solution ( 450 ml = 0.450 liters) 5M C6H12O6 = moles C6H12O6/0.450 liters = 2.25 moles C6H12O6 (180.156 grams/1 mole C6H12O6) = 405.351 grams of glucose ( you do significant figures )