10.4 g NaF x 1 mol NaF/42 g x 1 mol F/mol NaF = 0.248 moles F- ions (3 significant figures).
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∙ 7y agoTo calculate the number of moles of fluoride ions present in 10.4 g of NaF, you first need to determine the molar mass of NaF, which is 41 g/mol. Then, divide the given mass by the molar mass to find the number of moles of NaF (0.253 moles). Since there is one fluoride ion per molecule of NaF, there are also 0.253 moles of fluoride ions in 10.4 g of NaF.
There would be 4.38 moles of fluoride ions in 1.46 moles of aluminum fluoride, as the formula for aluminum fluoride is AlF3 with three fluoride ions per molecule of aluminum fluoride.
In a compound such as carbon fluoride, the number of ions present depends on the ionic form of the elements. For example, carbon typically forms covalent bonds and does not usually exist as an ion, while fluoride ions have a charge of -1. So, in a compound like carbon fluoride (CF₄ or CF₂), there are no ions of carbon but four fluoride ions for CF₄ and two fluoride ions for CF₂.
There are 3 moles of sulfate ions (SO4^2-) present in 1 mole of Al2(SO4)3. Therefore, in 1.7 moles of Al2(SO4)3, there would be 3 * 1.7 = 5.1 moles of sulfate ions.
To find the number of moles of H ions in the solution, first calculate the moles of HNO3 using the given concentration and volume. Since each mole of HNO3 yields 1 mole of H ions in solution, the number of moles of H ions is the same as the moles of HNO3. Therefore, in this case, there are 0.4512 moles of H ions present in the solution.
In 1 mole of PO4-3 ions, there are 3 ions present. Therefore, in 1.33 moles, there would be 3 ions/mol x 1.33 mol = 3.99 ions, which rounds down to 3 ions in 1.33 moles of PO4-3 ions.
There would be 4.38 moles of fluoride ions in 1.46 moles of aluminum fluoride, as the formula for aluminum fluoride is AlF3 with three fluoride ions per molecule of aluminum fluoride.
To determine the number of fluoride ions in 175 g of barium fluoride, first calculate the number of moles of barium fluoride using its molar mass. Then, use the ratio of fluoride ions to barium fluoride in the formula BaF\u2082 to find the number of fluoride ions. Finally, multiply this by Avogadro's number (6.022 x 10^23) to get the total number of fluoride ions.
There are 4.5 moles of sodium fluoride in 4.5 moles of sodium fluoride.
In a compound such as carbon fluoride, the number of ions present depends on the ionic form of the elements. For example, carbon typically forms covalent bonds and does not usually exist as an ion, while fluoride ions have a charge of -1. So, in a compound like carbon fluoride (CF₄ or CF₂), there are no ions of carbon but four fluoride ions for CF₄ and two fluoride ions for CF₂.
5,7 moles (SO4)3-.
There are 3 moles of sulfate ions (SO4^2-) present in 1 mole of Al2(SO4)3. Therefore, in 1.7 moles of Al2(SO4)3, there would be 3 * 1.7 = 5.1 moles of sulfate ions.
There are two fluoride ions in magnesium fluoride (MgF2), as the formula indicates the ratio of magnesium ions (Mg2+) to fluoride ions (F-) is 1:2.
To find the number of moles of H ions in the solution, first calculate the moles of HNO3 using the given concentration and volume. Since each mole of HNO3 yields 1 mole of H ions in solution, the number of moles of H ions is the same as the moles of HNO3. Therefore, in this case, there are 0.4512 moles of H ions present in the solution.
In 1 mole of PO4-3 ions, there are 3 ions present. Therefore, in 1.33 moles, there would be 3 ions/mol x 1.33 mol = 3.99 ions, which rounds down to 3 ions in 1.33 moles of PO4-3 ions.
1,46 moles of aluminum fluoride contain 35,16848.10e23 atoms.
In CaF2, there is one calcium atom for every two fluoride atoms. This means that in 0.18 moles of CaF2, there are 0.18 moles of calcium atoms.
The individual ions for calcium fluoride have the formulas Ca+2 and F-1 respectively. That means that in any sample of calcium fluoride, there must be twice as many of the fluoride ions.