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How many Li ions are needed to combine with one bromide ion?
Natalieramos
Bromine has -1 charge and Lithium has +1 charge. Therefore,only one lithium ion is required to react with a bromine ion.
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∙ 14y ago
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What mass of barium bromide needed to precipitate the silver ions from 100 ml of 52 M silver nitrate solution?
The equation of the reaction is BaBr2 + 2 AgNO3 -> 2 AgBr + Ba(NO3)2. Therefore, exactly as many bromide ions from barium bromide must be supplied to precipitate any particular number of silver ion from silver nitrate. From the definition of molarity, 100 ml of 52 M solution contains 5.2 moles (preferably called "gram formula units") of silver nitrate. The gram formula unit mass of silver nitrate is 169.87, and each gram formula mass contains equal numbers of silver and of nitrate ions. Therefore, 5.2 gram elemental masses of bromide ions will be required for the precipitation. This amount of bromide ions can be supplied by 5.2/2* or 2.6 gram formula masses of barium bromide, and the gram formula unit mass of barium bromide is 297.14. Multiplying this number by 2.6 shows that 7.7 X 102 grams of barium bromide, to the justified number of significant digits, will be needed.
How many grams of o2 are needed to combine with 6.85g of P?
8.85
How do you know how many ammoniums will combine with how many nitrate ions?
Ammonium ions have a 1+ charge, and nitrate ions have a 1- charge. One ammonium ion combined with one nitrate ion will produce an ionic compound with no overall charge. NH4+ + NO3- ---> NH4NO3
How many moles of bromide are there in Iron Bromide?
FeBr3 (Iron III Bromide) has three moles of bromide for every mole of iron. FeBr2 (Iron II Bromide) has two moles of bromide ion per mole of Iron.
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 ;-)
How many bromide ions can combine with one ion of barium to form an ionic compound?
Two bromide ions can combine with one barium cation to form an ionic compound, because a barium cation has an electrical charge of +2, while a bromide anion has an electrical charge of -1.
How many ions of bromide will combine with one of aluminum?
Three (3) Here is the equation Al^3+ + 3Br^- = AlBr3
How many moles of Cl ions are needed to completely combine with 0.25 moles of Mg 2 ions?
Balanced equation first.Mg + Cl2 -> MgCl2All is one to one, so,0.25 moles of chlorine ions are needed here.================================
How many moles of Cl- ions are needed to completely combine with 0.25 moles of Mg plus 2ions?
0,5 moles Cl-
How many moles of Cl- ions are needed to completely combine with 0.25 moles of Mg 2 ions?
Mg2+ + 2 Cl- are in 1 : 2 ratio (of ions) so also 0.25 : 0.50 mole ratio
What mass of barium bromide needed to precipitate the silver ions from 100 ml of 52 M silver nitrate solution?
The equation of the reaction is BaBr2 + 2 AgNO3 -> 2 AgBr + Ba(NO3)2. Therefore, exactly as many bromide ions from barium bromide must be supplied to precipitate any particular number of silver ion from silver nitrate. From the definition of molarity, 100 ml of 52 M solution contains 5.2 moles (preferably called "gram formula units") of silver nitrate. The gram formula unit mass of silver nitrate is 169.87, and each gram formula mass contains equal numbers of silver and of nitrate ions. Therefore, 5.2 gram elemental masses of bromide ions will be required for the precipitation. This amount of bromide ions can be supplied by 5.2/2* or 2.6 gram formula masses of barium bromide, and the gram formula unit mass of barium bromide is 297.14. Multiplying this number by 2.6 shows that 7.7 X 102 grams of barium bromide, to the justified number of significant digits, will be needed.
How many moles of bromide ions are present in 750.0 mL of 1.35 M MgBr2?
2.03mol
How many potassium ions are needed to bond with phosphate ions?
Three Potassium Ions are needed to Bond with one Phosphate Ion.
How many chloride ions combine with a single sodium ion?
It is a lattice. There are 6 cl- ions around a sodium ion.
How many potassium ions are needed to bond with a phosphate ion?
This process requires three potassium ions.
How many sodium ions are needed to balance the charge of one phosphate ion?
three sodium ions
How many Na plus ions will combine with an O2- to form a neutral salt?
Two Na plus ions will combine with an O2- to form the basic compound sodium oxide. When water is further added, the neutral salt, sodium hydroxide is formed.
