Boron trifluoride is BF3. So each mole of BF3 contains 1 moles of boron (B) and 3 moles of fluorine (F). Thus, 3 moles of BF3 contains NINE moles of fluorine.
The answer is 3 moles of Cu)NO3)2..
When 3 moles Cu react 3 moles of copper nitrate are obtained.
6 moles
3 moles of O, so that must be 6 moles of O2
6,1x10-3 moles of carbon 36,7350592277.1020 atoms.
To determine the number of moles of fluorine present in 2.67 grams of nitrogen trifluoride (NF3), we first need to calculate the molar mass of NF3, which is 71.0 g/mol. Then we find the moles of NF3 in 2.67 g by dividing the mass by the molar mass: 2.67 g / 71.0 g/mol ≈ 0.038 moles. As there are three fluorine atoms in each molecule of NF3, there are 3 times more moles of fluorine: 0.038 moles NF3 x 3 = 0.114 moles of fluorine.
One: The formula for magnesium fluoride is MgF2. Since each mole of fluorine molecules, which have the formula F2, contains two moles of fluorine atoms, one mole of each is the right ratio.
The answer is 3 moles of Cu)NO3)2..
When 3 moles Cu react 3 moles of copper nitrate are obtained.
The answer is 3 moles.
6 moles
3 moles of calcium is equivalent to 120,234 g.
Moles = Mass/Molar Mass.Using this equation, we can take 25/(1.0 + 19) and find that it is equal to 1.25 moles.Keep in mind that very few significant figures were used for this example (2), and that if accuracy is to be expected, as many as possible should be used.
3 moles of O, so that must be 6 moles of O2
6,1x10-3 moles of carbon 36,7350592277.1020 atoms.
Since ammonia has a chemical formula of NH3, it contains one mole of nitrogen and three moles of hydrogen per mole of ammonia. Therefore, 3 moles of ammonia contain 3 moles of nitrogen and 9 moles of hydrogen atoms.
There are 6 moles of sulfur present in 3 moles of aluminum sulfate, because aluminum sulfate has a 2:3 ratio of aluminum to sulfur.