Will have to make some assumptions with the little info given.
Solid sucrose is 1.587 g/ml in density and has a mass of 342.30 grams/mole
Density = grams/milliliters
1.587 g/ml = grams/250 ml
= 396.75 grams/342.30 grams
= 1.159 moles of sucrose
6
10*(1/6)=1.67
There are 6 atoms of oxygen in a molecule of glucose (C6H12O6).
C6H12O6 Glucose has twelve hydrogen atoms
The number of atoms is 2,167 970 708 52.10e25.
The number of moles is 2,997.
0.67 moles of C6H12O6
6
Do you mean this reaction? C6H12O6 + 6O2 -> 6CO2 + 6H2O 6 moles oxygen required. --------------------------------
To determine the number of carbon atoms in 7.11g of glucose, you first need to calculate the moles of glucose using its molar mass. The molar mass of glucose (C6H12O6) is 180.16 g/mol. Next, calculate the number of moles of carbon in one mole of glucose (6 moles). Finally, multiply the moles of glucose by the number of moles of carbon to find the total number of carbon atoms in 7.11g of glucose.
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.
10*(1/6)=1.67
There are 6 atoms of oxygen in a molecule of glucose (C6H12O6).
C6H12O6 Glucose has twelve hydrogen atoms
The number of atoms is 2,167 970 708 52.10e25.
The balanced chemical equation for the combustion of glucose (C6H12O6) is: C6H12O6 + 6O2 -> 6CO2 + 6H2O. From the equation, we can see that 1 mol of C6H12O6 produces 6 mol of CO2. First, calculate the number of moles of C6H12O6 in 45 g. Then use the mole ratio to find the moles of CO2 produced, and finally convert that to grams.
To find the number of molecules in 936 g of glucose (C₆H₁₂O₆), first calculate the molar mass of glucose, which is approximately 180.18 g/mol. Next, use the formula: number of moles = mass (g) / molar mass (g/mol). Thus, 936 g of glucose corresponds to about 5.19 moles. Finally, multiplying the moles by Avogadro's number (approximately (6.022 \times 10^{23}) molecules/mol) gives roughly (3.12 \times 10^{24}) molecules of glucose.