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How many formula units of AgNO3 are present in 147g of this compound?
The answer is 5,21131981201352.10e23 formula units of silver nitrate.
What mass of silver chloride can be produced from 1.88 L of a 0.139 M solution of silver nitrate?
Since the four named compounds are the only reactants and products, this question can be answered from the law of conservation of mass: The amount of silver nitrate must be 14.35 + 8.5 - 5.85 or 17.0 grams.
How many moles of CaCl2 are in 2.00 1024 formula units?
The answer is 3,33 moles.
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 formula units would there be in a 54.3 gram sample of sodium nitrate?
3
How many moles for silver chloride are produced from 7 mol of silver nitrate?
Since both chloride anions and nitrate anions have a charge of -1, there will be the same number of moles of silver chloride produced as the moles of silver nitrate reacted. (Since both silver nitrate and silver chloride are ionic compounds, it would be preferable to call their "moles" "formula units" instead.)
How many formula units of AgNO3 are present in 147g of this compound?
The answer is 5,21131981201352.10e23 formula units of silver nitrate.
What mass of silver chloride can be produced from 1.88 L of a 0.139 M solution of silver nitrate?
Since the four named compounds are the only reactants and products, this question can be answered from the law of conservation of mass: The amount of silver nitrate must be 14.35 + 8.5 - 5.85 or 17.0 grams.
Determine the number of formula units that are in 0.688 moles of AgNO3?
0.688 moles*6.02x1023=4.14x1023 Formula units
How many formula units of salt make up 10 moles?
10 formula units
How many moles of CaCl2 are in 2.00 1024 formula units?
The answer is 3,33 moles.
How many moles are equivalent to 3.6x1024 formula units of MgBr2?
The answer is 5,978 moles.
How many formula units of sodium acetate are in 0.87 moles of sodium acetate?
How many formula units of sodium acetate are in 0.87 moles of sodium acetat
How many formla units are present in 85.6g of AgNO3?
The molar mass of silver nitrate is 169,87 g: approx. two formula units.
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.
What is the number of formula units that are in 5.68 moles of magnesium oxide?
5.68 mol MgO x 6.02E23 formula units / 1 mol = 3.42E24 Formula Units
How many moles of formula units are in 5.67 g of iron III sulfate?
There is 1.42 x 10^-2 moles of formula units that are in 5.67 g of iron III sulfate.
