There are 6 oxygen atoms in one molecule of glucose (C6H12O6). To find the number of moles of glucose in 5.23 grams, you divide the mass by the molar mass of glucose. Then multiply the number of moles by the number of oxygen atoms in one glucose molecule to find the total number of oxygen atoms in 5.23 grams of glucose.
To calculate the number of oxygen atoms in 2.50 grams of oxygen gas (O2), you will first need to convert grams to moles using the molar mass of O2 (32 g/mol). Then, use Avogadro's number (6.022 x 10^23) to determine the number of oxygen molecules in moles. Since there are 2 oxygen atoms in each O2 molecule, you will multiply the number of moles by 2 to find the total number of oxygen atoms.
To determine the number of oxygen atoms present in 1.659 grams of ozone gas, you first need to calculate the number of moles of ozone using its molar mass. Then, since each molecule of ozone (O3) contains 3 oxygen atoms, you can use Avogadro's number (6.022 x 10^23) to find the total number of oxygen atoms in the given mass of ozone.
The chemical equation for the combustion of glucose is C6H12O6 + 6O2 -> 6CO2 + 6H2O. From this, we can see that 1 mole of glucose produces 6 moles of CO2. Therefore, 45 grams of glucose (which is approximately 0.25 moles) would produce 6 times that amount of CO2, which is about 90 grams.
It depends on the sugar. The monosaccharides glucose, fructose, and galactose have the molecular formula C6H12O6, and therefore have 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms per molecule. The disaccharides sucrose and maltose have the molecular formula C12H22O11, and therefore have 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms per molecules.
In 2 moles of potassium dichromate, there are 16 moles of oxygen atoms (from the two oxygen atoms in each formula unit). The molar mass of oxygen is 16 g/mol, so in 2 moles of potassium dichromate, there are 32 grams of oxygen.
There are 6 atoms of oxygen in a molecule of glucose (C6H12O6).
There are three oxygen atoms in 3.45 grams of water (H2O).
There are 6 atoms of both hydrogen and oxygen in one molecule of glucose (C6H12O6).
glucose is C6H12O6. So there are three types of atoms - carbon, hydrogen and oxygen
There are 24 atoms in a molecule of glucose (C6H12O6), consisting of 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
There are 12.
Yes, the amount of hydrogen atoms in glucose is 12, and the amount of oxygen atoms is 6, therefore, there are two times as many hydrogen atoms than oxygen atoms in a molecule of glucose. C6-H12-O6 is the formula.
Glucose has three elements: carbon, hydrogen, and oxygen. It consists of 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen.
Glucose (C6H12O6) is a monosaccharide that contains twelve hydrogen atoms, six carbon atoms and six oxygen atoms. A glucose and fructose molecule combine to create a sucrose molecule.
The first step is calculating how many molecules of glucose are in 3.00 grams. To do this, you need the molecular mass of the compound (glucose), which is found by adding up the weights of the elements involved in C6H12O6. C6: 12.0 × 6 = 72.0 H12: 1.0 × 12 = 12.0 O6: 16.0 × 6 = 96.0 72.0 + 12.0 + 96.0 = 180.0 grams/mol With this and Avogadro's constant (6.02 × 1023), we can then convert 3.00 grams of glucose to number of molecules. 3.00 grams ÷ 180.0 grams/mol × (6.02 × 1023) = 1.00 × 1022 molecules of glucose So now we know how many molecules of glucose there are. We also know from the formula that in one molecule of glucose, there are 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen. Number of atoms in one molecule × number of molecules = number of atoms in given amount 6 × (1.00 × 1022) = 6.00 × 1022 atoms of carbon 12 × (1.00 × 1022) = 1.20 × 1023 atoms of hydrogen 6 × (1.00 × 1022) = 6.00 × 1022 atoms of oxygen
Glucose is an example of a carbohydrate which is commonly encountered. It is also known as blood sugar, and dextrose.Its chemical formula is C6H12O6,Which Means it has total 24 atoms which include 6 carbon, 12 hydrogen and 6 oxygen atoms
6The formula is C6H12O6