To find the number of moles in 16.4 g of calcium nitrate, Ca(NO3)2, you first need to calculate its molar mass. The molar mass of Ca(NO3)2 is approximately 164.1 g/mol. Using the formula for moles (moles = mass / molar mass), you can calculate the moles: ( \text{moles} = \frac{16.4 , \text{g}}{164.1 , \text{g/mol}} \approx 0.100 , \text{moles} ). Therefore, there are about 0.100 moles of Ca(NO3)2 in 16.4 g.
735 g of Ca3(PO4)2 are obtained.
978 g calcium contain 24,4 moles.
29,0 g of calcium is equal to 0,723 moles.
67,4 g HCl is equivalent to 1,85 moles.
14,84 g magnesium are equivalent to 0,61 moles.
735 g of Ca3(PO4)2 are obtained.
The formula is: number of moles = g Be/9,012.
14,84 g magnesium are equivalent to 0,61 moles.
97,5 g of oxygen is equal to 5,416 moles.
978 g calcium contain 24,4 moles.
573,28 of g of AgCI is equivalent to 4 moles.
67,4 g HCl is equivalent to 1,85 moles.
29,0 g of calcium is equal to 0,723 moles.
27.4 g H2O x 1 mole/18 g = 1.52 moles
156 g calcium is equivalent to 3,89 moles.
231 g of Fe2O3 are equal to 0,69 moles.
1 g of sodium sulfite is equivalent to 0,0079 moles.