This mass is 244,459.10e23 g.
To determine the mass of oxygen in 147.2 grams of glucose (C6H12O6), we need to consider the molecular formula of glucose. For each mole of glucose, there are 6 moles of oxygen atoms. The molar mass of glucose is approximately 180.16 g/mol. Thus, the mass of oxygen in 147.2 grams of glucose would be (6/180.16) * 147.2 = approximately 4.88 grams.
The molecular formula of ozone is O3 because an ozone molecule consists of three oxygen atoms bonded together. The presence of three oxygen atoms in each molecule gives rise to the chemical formula O3.
The percentage of oxygen is 54,84 %.
320 grams of oxygen is the equivalent of 10 moles.
16 grams of oxygen how many moles is 0,5 moles.
0.758 moles of NH3 is the amount of moles in 50 grams of NH42SO4.
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.
To determine the mass of oxygen in 147.2 grams of glucose (C6H12O6), we need to consider the molecular formula of glucose. For each mole of glucose, there are 6 moles of oxygen atoms. The molar mass of glucose is approximately 180.16 g/mol. Thus, the mass of oxygen in 147.2 grams of glucose would be (6/180.16) * 147.2 = approximately 4.88 grams.
Wallastonite is a silicate mineral is comprised of calcium, silicon, and oxygen. The molecular formula is CaSi03, with a molecular weight of 116.16 grams per mole.
36 grams of water is equal to 2 moles. Therefore, to find the quantity of oxygen molecules that contain the same number of molecules as 36 grams of water, you would need 4 moles of oxygen since the molecular formula of water is H2O.
The formula with this disaccharide is, C12H22O11 17.1 grams C12H22O11 (1 mole C12H22O11/342.296 grams)(11 mole O/1 mole C12H22O11)(6.022 X 10^23/1 mole O) = 3.31 X 10^23 atoms of oxygen ------------------------------------------
If 12 grams of carbon were used to form the 22 grams of carbon dioxide, this implies that 12 grams of oxygen were consumed in the reaction. Since 20 grams of oxygen were initially available, only 8 grams of oxygen are left unused.
A sample of a compound contain 1.52 g of Nitrogen and 3.47 g of Oxygen. The molar mass of this compound is between 90 grams and 95 grams. The molecular formula and the accurate molar mass would be N14O35.
The percentage of oxygen is 54,84 %.
For every 40 grams of calcium (Ca), 32 grams of oxygen (O) will be needed to react. This is based on the chemical formula for calcium oxide (CaO), where one calcium atom reacts with one oxygen atom to form one molecule of CaO.
The molecular formula of ozone is O3 because an ozone molecule consists of three oxygen atoms bonded together. The presence of three oxygen atoms in each molecule gives rise to the chemical formula O3.
We assume 100 grams total and turn those percentages into grams. Get moles elements. 29.6 grams O (1 mole O/16.0 grams) = 1.85 mole O 70.4 grams F (1 mole F/19.0 grams) = 3.71 mole F now divide both moles by smallest number to get mole ratio for empirical formula 1.85/1.85 = 1 for O 3.71/1.85 = 2 for F so the empirical formula for this compound is........ OF2