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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 moles of bromide are there in Iron Bromide?
To determine how many moles of bromide are in iron (III) bromide (FeBr3), you can use the chemical formula FeBr3 to see that there are three moles of bromide ions for every mole of iron (III) bromide. So, the number of moles of bromide ions is equal to the number of moles of FeBr3.
How many ions does MgBr2 have?
One formula unit of MgBr2 has three ions; one Mg2+ ion and two Br- ions. One mole of MgBr2 formula units has one mole of Mg2+ ions and two moles of Br- ions, for a total of three moles of ions.
How many types of chemical bonds are there in ammonium bromide?
Two. covalent bonds in the ammonium ion NH4+ and ionic bond between the ammonium ion and the bromide ion, Br-
How many ions in copper II bromide?
copper has 29 protons, when dealing with Cu^2+ all that means is it lost two electrons. so now the element has 29 protons and 27 electrons. Protons are positive and electrons are negative and neutrons are neutral. So say you had an element X^2- then you have gained two more electrons so the element has an overall negative charge. hope that helps
How many ions of bromide will combine with one of aluminum?
Three ions of bromide will combine with one ion of aluminum to form aluminum bromide (AlBr3). This is because aluminum has a +3 charge and bromide ions each have a -1 charge, so three bromide ions are needed to balance the +3 charge of aluminum.
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.
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 are there in Iron Bromide?
To determine how many moles of bromide are in iron (III) bromide (FeBr3), you can use the chemical formula FeBr3 to see that there are three moles of bromide ions for every mole of iron (III) bromide. So, the number of moles of bromide ions is equal to the number of moles of FeBr3.
How many moles of bromide ions are present in 750.0 mL of 1.35 M MgBr2?
There are 1.35 moles of MgBr2 in 1 L of solution, which corresponds to 2 moles of bromide ions. Therefore, in 750.0 mL of 1.35 M MgBr2 solution, there will be 1.0125 moles of bromide ions.
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
How many moles of Cl ions are needed to completely combine with 0.25 moles of Mg 2 ions?
Since the charge of Mg^2+ ion is 2+ and Cl^- ion has a charge of 1-, each Mg^2+ ion will combine with 2 Cl^- ions to form MgCl2. Therefore, for 0.25 moles of Mg^2+ ions, you will need 0.50 moles of Cl^- ions for complete combination.
How many ions does MgBr2 have?
One formula unit of MgBr2 has three ions; one Mg2+ ion and two Br- ions. One mole of MgBr2 formula units has one mole of Mg2+ ions and two moles of Br- ions, for a total of three moles of ions.
How many types of chemical bonds are there in ammonium bromide?
Two. covalent bonds in the ammonium ion NH4+ and ionic bond between the ammonium ion and the bromide ion, Br-
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 mol of bromide ions are in 0.500 L of a 0.300 M solution of AlBr3?
Molarity is mols of solute/volume of solution. We will do this in two parts, though you could use one conversion string. 0.300 M AlBr3 = mols/0.500L = 0.15 mols of AlBr3 (3mol Br/1 mol AlBr3 )( 6.022 X 10^23/1mol Br )(1mol Br/6.022 X 10^23 ) = 0.45 moles of Br
