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In chemistry what is ScBr3?
Scandium Bromide
How many moles of ions are present in aqueous solutions prepared by dissolving 10.00g of the f scandium III bromide in water to make a 1.25 L solution?
First figure out the formula of Scandium(III) Bromide: its ScBr3. Next ignore the 1.25 L solution part -- it's just extra information. Use the 10.00g to convert it to moles. To do so first figure out the molar mass of ScBr3: its 285g/mol. Now you do: 10.00g x 1 mol/285g = .0350877193... mol. Next round it to 3 significant figures: .0351. Now you find out the number of ions in the formula and use it as a conversion factor (instead of Avagrado's number, 6.02 x 1023): it has 4 ions (Scandium = 1; Bromide = 3). Finally you do: .0351 mol x 4 mol ions/1 mol = .1404 mol ions. This is your answer! By Geovonni Bell ;-)
What is the chemical formula for the compound formed between scandium(lll) and the bromate ion?
The chemical formula for the compound formed between scandium(III) and the bromate ion is Sc(BrO3)3. Each scandium(III) ion will combine with three bromate ions to achieve a neutral compound with a balanced charge.
