This is an incomplete question for several reasons. Iron bromide can be FeBr2 or FeBr3 and this will influence the answer. Also, in order to precipitate silver bromide, one would add silver nitrate, but it isn't stated how much silver nitrate is used. Assuming FeBr2 is used, and there is excess AgNO3, then ....
2AgNO3 + FeBr2 ----> 2AgBr(s) + Fe(NO3)
moles FeBr2 = 2.96 g x 1 mol/216 g = 0.0137 moles
moles AgBr produced = 2 x 0.0137 = 0.0274 moles
mass AgBr produced = 0.0274 moles x 188 g/mole = 5.15 grams <---answer
The number of Neutrons and Protons in the nucleus.So it would be 26 Protons and 30 Neutrons so...26+30=56<so that is the mass number.
The statue has a mass of about 42,660.44 g or 42.66044 kg.
10 grams. There is no weight lost or gained in this reaction.
That could be ANY substance whose density is greater than 3 grams per cm3. Examples include but are not limited to mercury, copper, gold, iron, lead, platinum, uranium, brass, silver, and steel.
no because it is a metal single Ag is an atom
The atomic mass of silver is 107.868 and the atomic mass of chlorine is 35.453. Therefore, the fraction by mass of silver in silver chloride is 107.868/(107.868 + 35.453) or 0.7526. The precipitated silver chloride therefore contains 0.7526 X 6.21 or 4.674 grams of silver from the coin. The mass percent silver in the coin therefore is 100(4.674/6.80) or 68.7 % silver, to the justified number of significant digits.
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.
Yes, silver is heavier than iron. One atom of iron has a mass of 55.85 amu (atomic mass units). One atom of silver has a mass of 107.87 amu. All of this information is easily found on a periodic table.
The molar mass of calcium bromide is........99,88.
The molar mass of magnesium bromide is approximately 203 g/mol.
The atomic mass of lithium is 7. The atomic mass of the bromide ion is 80. Therefore the molecular mass is 87u.
From the definition of molar concentration, 5.00 mL of 0.100M silver nitrate solution contains 0.100 X 5.00/1000 or 5.00 X 10-4 gram formula masses of silver nitrate and 5.00 mL of 0.200 X 5.00/1000 or 1.00 X 10-3 gram formula masses of sodium bromide. Because all of the ions involve in this question are monovalent ones, that means that the limiting reagent for the precipitation reaction is silver, and the very low solubility product constant for silver bromide means that the amount of unreacted silver can be disregarded in calculating the answer within the number of significant digits justified by the precision of specifying the amounts of materials reacted. Therefore, 5.00 x 10-4 gram formula masses of silver bromide are formed. The gram formula mass of silver bromide is 187.77, and the mass formed is therefore 0.939 grams to the justified number of significant digits.
gold, just look at the atomic mass in the periodic table.
Potassium bromide is KBr, the atomic mass of this compound is ca. 119.1. no.moles = mass/relitive molecular mass, so in this case that's 245/119.1 = 2.057 moles of KBr.
The atomic weight of bromine is 79,904 amu.
sodium iodide
Copper (I) bromide. Unlike with a zinc compound question I just answered, the (I) here is pretty important; both copper (I) bromide and copper (II) bromide exist and are commercially available.