The answer is 32,4.10e23.
0.688 moles*6.02x1023=4.14x1023 Formula units
10 formula units
The answer is 5,978 moles.
0,75 moles of AlCl3 (anhydrous) is equivalent to 100,005 g.
The answer is 2,09 moles.
There is 1.42 x 10^-2 moles of formula units that are in 5.67 g of iron III sulfate.
1.42 x 10^-2 mole
0.0482 moles Na2SO4 x 6.02x10^23 form.units/mole = 2.91x10^22 formula units
2.45 x 6.02 x 1023.4749 x 1025
0.688 moles*6.02x1023=4.14x1023 Formula units
10 formula units
The answer is 3,33 moles.
The answer is 5,978 moles.
How many formula units of sodium acetate are in 0.87 moles of sodium acetat
1 mole Fe2(SO4)3 = 6.022 x 1023 formula units Fe2(SO4)3 0.671mol Fe2(SO4)3 x 6.022 x 1023 formula units Fe2(SO4)3/1mol Fe2(SO4)3 = 4.04 x 1023 formula units Fe2(SO4)3
K2O is potassium oxide. Formula mass = 94g32.6 g x 1 FU/94 g = 0.35 formula units
K2O is potassium oxide. Formula mass = 94g32.6 g x 1 FU/94 g = 0.35 formula units