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 determine the empirical formula, first calculate the moles of each element present by dividing the given amount by their respective molar mass. For aluminum (Al), 1.0 moles of Al is equivalent to 1.0 moles, while 1.5 moles of oxygen (O) equal 1.5 moles. The ratio of Al to O is 1:1, so the empirical formula is Al2O3 (Aluminum oxide).
Molarity= mols/ L So, take 4.13g MgBr2 ( because Mg has a +2 charge and Bromide has a +1 charge) and divide it by 184.1 g MgBr2 which gives you .022 mols of MgBr2 then you divide .022 by 845 which gives you 2.65 M Molarity = 2.65M
0.688 moles*6.02x1023=4.14x1023 Formula units
There are a total of 2 moles of anions in 2.50 g of MgBr2. Each formula unit of MgBr2 contains 2 moles of anions (Br-). The molar mass of MgBr2 is 184.113 g/mol, so 2.50 g is equivalent to 0.0136 moles, and therefore 0.0136 moles * 2 moles = 0.0272 moles of anions.
There are 0.021 moles of MgBr2 in 3.50 g. Since MgBr2 contains 2 bromide ions per formula unit, there are 0.042 moles of Br- ions. So, there are 0.042 * 6.022 x 10^23 = 2.53 x 10^22 Br- ions in 3.50 g of MgBr2.
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.
There are 1.35 moles of MgBr2 in 1 L of solution, which corresponds to 2 moles of bromide ions. Therefore, in 750.0 mL of 1.35 M MgBr2 solution, there will be 1.0125 moles of bromide ions.
The molar mass of MgCl2 = 95.211 g/mol
The formula is: number of moles = g Be/9,012.
0,75 moles of AlCl3 (anhydrous) is equivalent to 100,005 g.
The molar mass of each element gives 0.0134g of iron is equivalent to 0.00024 moles, 0.00769g of sulfur is equivalent to 0.00024 moles, and 0.0115g of oxygen is equivalent to 0.00071 moles. Simplifying the ratio of the moles of each element gives the empirical formula FeS2O4.
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 determine the number of formula units in 2.45 moles of potassium chloride, you first need to find the molar mass of KCl, which is approximately 74.55 g/mol. Then, divide the number of moles by the molar mass to get the number of formula units. Therefore, 2.45 moles of KCl is equivalent to about 163 formula units.
To determine the empirical formula, first calculate the moles of each element present by dividing the given amount by their respective molar mass. For aluminum (Al), 1.0 moles of Al is equivalent to 1.0 moles, while 1.5 moles of oxygen (O) equal 1.5 moles. The ratio of Al to O is 1:1, so the empirical formula is Al2O3 (Aluminum oxide).