The answer is 5,978 moles.
The molar mass of MgCl2 = 95.211 g/mol
0,75 moles of AlCl3 (anhydrous) is equivalent to 100,005 g.
To find the molarity of the substance, divide the moles of the substance by the volume of the solution in liters. First, calculate the moles of magnesium bromide: Convert grams to moles by dividing the given mass by the molar mass of magnesium bromide (184.1 g/mol). Moles of magnesium bromide = 4.13 g / 184.1 g/mol = 0.0224 mol. Next, divide the moles by the volume of the solution in liters: Molarity = 0.0224 mol / 845 L ≈ 0.00003 M (rounded to 3 significant figures).
0.688 moles*6.02x1023=4.14x1023 Formula units
The formula means, among other things, that there are 7 atoms of oxygen in each mole of the compound. Therefore, in 4.00 moles of the compound, there are 28.00 moles of oxygen atoms. Elemental oxygen usually is diatomic, so that there would be the equivalent of 14 moles of diatomic elemental oxygen.
One formula unit of MgBr2 has three ions; one Mg2+ ion and two Br- ions. One mole of MgBr2 formula units has one mole of Mg2+ ions and two moles of Br- ions, for a total of three moles of ions.
2.50 g MgBr2 x 1 mole MgBr2/184 g x 2 mole Br/mole MgBr2 = 0.0272 moles Br^-0.0272 moles Br^- x 6.02x10^23 anions/mole = 1.6x10^22 anions
The formula is: number of moles = g Be/9,012.
The molar mass of MgCl2 = 95.211 g/mol
2.03mol
0,75 moles of AlCl3 (anhydrous) is equivalent to 100,005 g.
the formula does not show how many moles are in a compound, but it tells how many atoms of an element that are in the compound...
The term molecule is not adequate for an ionic compound; correct is formula unit.60 g NaCl contain the equivalent of 1,026 formula units.
0,665 moles NaCl is equivalent to 38,86 g.
To find the molarity of the substance, divide the moles of the substance by the volume of the solution in liters. First, calculate the moles of magnesium bromide: Convert grams to moles by dividing the given mass by the molar mass of magnesium bromide (184.1 g/mol). Moles of magnesium bromide = 4.13 g / 184.1 g/mol = 0.0224 mol. Next, divide the moles by the volume of the solution in liters: Molarity = 0.0224 mol / 845 L ≈ 0.00003 M (rounded to 3 significant figures).
0.688 moles*6.02x1023=4.14x1023 Formula units
Since both the acid and the base have equivalent weights equal to their formula weights, 2 moles of KOH are needed to neutralize 2 moles of nitric acid.