18 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6.022 X 10^23/1 mole C6H12O6)
= 6.02 X 10^22 atoms
the answer is 21!
The best way of doing this is by invoking the concept of moles and avagadro's number. Avagadros number is 6.022*10^23. One mole is an avagadro's number of atoms. So to solve this problem a periodic table can be used to find the molar mass (the mass of one mole) of C6H12O6. This comes out to be: (6*12.01)+(12*1)+(6*16)=180.06 g/mole.Then we use dimensional analysis to find the number of C, H, and O atoms:Carbon atoms:3.00g*(1 mole C6H12O6/180.06g)*(6 mole C/1 mole C6H12O6)*(6.022*10^23 atoms /mole C)=6.02*10^22 atoms of carbonHydrogen atoms:3.00g*(1 mole C6H12O6/180.06g)*(12 mole H/1 mole C6H12O6)*(6.022*10^23 atoms /mole H)=1.20*10^23 atoms of hydrogenOxygen atoms:3.00g*(1 mole C6H12O6/180.06g)*(6 mole H/1 mole C6H12O6)*(6.022*10^23 atoms /mole H)=6.02*10^22 atoms of oxygen
There are 58 atoms in C18H32O18. This is calculated by adding up the total number of each element: 18 carbon atoms + 32 hydrogen atoms + 18 oxygen atoms.
Yes
Glucose has a molar mass of 180. There are thus 0.1 moles present in 18 g. 1 mole contains 6.022 x 1023 molecules. 0.1 moles contains 6.022 x 1022 molecules and so glucose with 24 atoms per molecule contains 24 x 6.022 x 1022 = 1.44528 x 1024 atoms.
To create three glucose molecules, it would require a total of 72 carbon atoms because each glucose molecule contains 6 carbon atoms. So, 6 carbon atoms x 3 glucose molecules = 18 carbon atoms for each individual glucose molecule, and 18 carbon atoms x 3 = 72 carbon atoms in total.
the answer is 21!
18 carbon atoms (6 in each)
Of the 54 carbon atoms in a triglyceride, 18 can potentially be converted into glucose. Each glucose molecule contains 6 carbon atoms.
The best way of doing this is by invoking the concept of moles and avagadro's number. Avagadros number is 6.022*10^23. One mole is an avagadro's number of atoms. So to solve this problem a periodic table can be used to find the molar mass (the mass of one mole) of C6H12O6. This comes out to be: (6*12.01)+(12*1)+(6*16)=180.06 g/mole.Then we use dimensional analysis to find the number of C, H, and O atoms:Carbon atoms:3.00g*(1 mole C6H12O6/180.06g)*(6 mole C/1 mole C6H12O6)*(6.022*10^23 atoms /mole C)=6.02*10^22 atoms of carbonHydrogen atoms:3.00g*(1 mole C6H12O6/180.06g)*(12 mole H/1 mole C6H12O6)*(6.022*10^23 atoms /mole H)=1.20*10^23 atoms of hydrogenOxygen atoms:3.00g*(1 mole C6H12O6/180.06g)*(6 mole H/1 mole C6H12O6)*(6.022*10^23 atoms /mole H)=6.02*10^22 atoms of oxygen
There are 58 atoms in C18H32O18. This is calculated by adding up the total number of each element: 18 carbon atoms + 32 hydrogen atoms + 18 oxygen atoms.
Each glucose molecule contains 6 carbon atoms. Thus we need 18 molecules of CO2 to make 3 molecules of glucose.
Yes
Glucose has a molar mass of 180. There are thus 0.1 moles present in 18 g. 1 mole contains 6.022 x 1023 molecules. 0.1 moles contains 6.022 x 1022 molecules and so glucose with 24 atoms per molecule contains 24 x 6.022 x 1022 = 1.44528 x 1024 atoms.
They will contain 18 atoms total in all the products.
Yes, all atoms of argon have the same atomic number, which is 18. Atomic number refers to the number of protons in the nucleus of an atom, and for argon, it always remains constant at 18.
There are 18 hydrogen atoms in 6NH3 molecules. Each NH3 molecule contains 3 hydrogen atoms, so 6NH3 would have 6 x 3 = 18 hydrogen atoms.