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
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.
To find the number of moles of anions in 37.4 g of AlF3, we first need to calculate the molar mass of AlF3. Aluminum (Al) has a molar mass of 26.98 g/mol, and fluorine (F) has a molar mass of 19.00 g/mol. So, the molar mass of AlF3 is 26.98 + (3 * 19.00) = 83.98 g/mol. Therefore, the number of moles of AlF3 in 37.4 g is 37.4 g / 83.98 g/mol = 0.445 moles. Since there are 3 fluoride ions (anions) in each formula unit of AlF3, the number of moles of anions in 37.4 g of AlF3 is 0.445 moles * 3 = 1.34 moles.
Total mass of MgBr2- 36.9 g Mass of Br 32.0 / total mass 36.9 = .867 = 86.7% Mass of Mg 4.9 / 36.9 = .133 = 13.3% There is 86.7% of Bromine and 13.3% of Magnesium in the compound MgBr2
250 grams CaCO3 (1 mole CaCO3/100.09 grams) = 2.50 moles of calcium carbonate
0.25 g is equal to 250 mg, as there are 1000 milligrams (mg) in one gram.
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.
The molar mass of MgCl2 = 95.211 g/mol
1.39 moles
250 g = about 0.55116 pounds.
To find out how many 250 g packets make 1 kg, you can divide 1000 g by 250 g. This calculation shows that 1 kg is equal to 4 packets of 250 g. Therefore, you need 4 packets of 250 g to make 1 kg.
250 g = 0.25 kgTo convert from g to kg, divide by 1000.
250 mg = 0.25 g
1 tonne = 1,000,000 g ⇒ 1 tonne ÷ 250 g = 1,000,000 g ÷ 250 g = 4,000
250 g iron (III) oxide is equal to 1,565 moles.
250 grams = 8.81849049 ounces
250 g = 0.25 kgTo convert from g to kg, divide by 1000.
0.55lbs