0,75 moles of AlCl3 (anhydrous) is equivalent to 100,005 g.
10 formula units
The answer is 5,978 moles.
The answer is 32,4.10e23.
The answer is 2,09 moles.
6.022 x 1023 formula units of K2O are equal to one mole of K2O.
How many formula units of sodium acetate are in 0.87 moles of sodium acetat
10 formula units
The answer is 3,33 moles.
The answer is 5,978 moles.
The answer is 32,4.10e23.
.21 moles/liter * .0655 L = 0.013755 moles AlCl3 .013755 * 3 moles Cl / 1 mole AlCl3 = 0.041265 moles Cl * avagadro's number = number of chloride ions
There is 1.42 x 10^-2 moles of formula units that are in 5.67 g of iron III sulfate.
The result is 0,921.10e14 moles.
The answer is 2,09 moles.
mom
1 mole CaCl2 = 6.022 x 1023 formula units CaCl2 1.26 x 1024 formula units CaCl2 x 1mol CaCl2/6.022 x 1023 formula units CaCl2 = 2.09 moles CaCl2
42.394 grams.