A properly written formula for the compound, using the subscript fonts available for answers but not for questions, is Al2(SO4)3. This formula shows that there are 2 aluminum atoms, three sulfur atoms, and twelve oxygen atoms in each formula unit, for a total of 342.14 grams per formula unit. 647 of these formula units will have a mass of 647 X 342.14 gramsor 221 kilograms, to the justified number of significant digits.
Aluminum sulfate is formulated by Al2(SO4)3
To find the mass of 0.25 moles of aluminum sulfate, you need to know the molar mass of aluminum sulfate. The molar mass of aluminum sulfate (Al2(SO4)3) is approximately 342.15 g/mol. Therefore, the mass of 0.25 moles of aluminum sulfate would be around 85.54 grams.
Well...Since Aluminum is insoluble in water as an element I assume you mean that an 0.12 mol Aluminum ions react with 0.12 mol Sulfate ions to form x mol of Aluminum Sulfate. Balanced equation (Kinda since since the other portions of the compounds are unknown): 2 Al3+(aq) + 3 SO42-(aq) <----->Al2(SO4)3 (aq) 0.12mol*(1mol Al2(SO4)3 (aq)/2mol Al3+)=0.06mol Al2(SO4)3 (aq) 0.12mol*(1molAl2(SO4)3/3mol SO42-) =0.04mol Al2(SO4)3 (aq) *Answer*
The empirical formula for aluminum sulfate is Al2(SO4)3.
Aluminium Sulphate= Al2(SO4)3
Al2 SO4(wrong) this is how it should be written Al2(SO4)3This how you do it3.5 moles Al2(SO4)3 x 3 mole Sulfur / 1 moles Al2(SO4)3 = put that in ur calculator it will give u the answer
There are 3 moles of sulfate ions (SO4^2-) present in 1 mole of Al2(SO4)3. Therefore, in 1.7 moles of Al2(SO4)3, there would be 3 * 1.7 = 5.1 moles of sulfate ions.
5,7 moles (SO4)3-.
Aluminum sulfate is formulated by Al2(SO4)3
To find the mass of 0.25 moles of aluminum sulfate, you need to know the molar mass of aluminum sulfate. The molar mass of aluminum sulfate (Al2(SO4)3) is approximately 342.15 g/mol. Therefore, the mass of 0.25 moles of aluminum sulfate would be around 85.54 grams.
The name of Al2(SO4)3 is aluminum sulfate.
How many moles are in 4.70g of NH3?
To convert grams of aluminum sulfate to moles, you first need to determine the molar mass of aluminum sulfate (Al2(SO4)3), which is approximately 342.15 g/mol. Then, divide the given mass by the molar mass to obtain the number of moles. In this case, 6.7 grams of aluminum sulfate is approximately 0.02 moles.
Well...Since Aluminum is insoluble in water as an element I assume you mean that an 0.12 mol Aluminum ions react with 0.12 mol Sulfate ions to form x mol of Aluminum Sulfate. Balanced equation (Kinda since since the other portions of the compounds are unknown): 2 Al3+(aq) + 3 SO42-(aq) <----->Al2(SO4)3 (aq) 0.12mol*(1mol Al2(SO4)3 (aq)/2mol Al3+)=0.06mol Al2(SO4)3 (aq) 0.12mol*(1molAl2(SO4)3/3mol SO42-) =0.04mol Al2(SO4)3 (aq) *Answer*
To find the grams in 2.4 moles of SO4, you need to multiply the number of moles by the molar mass of SO4 (sulfate). The molar mass of SO4 is 96.06 g/mol. Therefore, 2.4 moles of SO4 would be 2.4 moles * 96.06 g/mol = 230.544 grams.
To calculate the formula mass of Al2(SO4)3, we need to determine the molar masses of each element and multiply them by the respective subscripts in the formula. The molar mass of aluminum (Al) is 26.98 g/mol, sulfur (S) is 32.06 g/mol, and oxygen (O) is 16.00 g/mol. Therefore, the formula mass of Al2(SO4)3 is 2(26.98) + 3(32.06 + 4(16.00)) = 342.15 g/mol.
The empirical formula for aluminum sulfate is Al2(SO4)3.