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How many moles of silver nitrate do 2.8881015 formula untied equal?
To find the number of moles in 2.8881015 formula units of silver nitrate, you first need to know the molar mass of silver nitrate (AgNO3), which is 169.87 g/mol. Then, you can use the formula: moles = formula units / Avogadro's number. Therefore, moles = 2.8881015 / 6.022 x 10^23 = 4.79 x 10^-24 moles.
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 formula units of AgNO3 are present in 147g of this compound?
To determine the number of formula units of AgNO3 in 147g of the compound, you first need to calculate the molar mass of AgNO3. The molar mass of AgNO3 is 169.87 g/mol. Next, divide the given mass (147g) by the molar mass to find the number of moles present in the sample. Finally, use Avogadro's number (6.022 x 10^23) to convert moles to formula units.
How many formula units would there be in a 54.3 gram sample of sodium nitrate?
To determine the number of formula units in a sample of a compound, you first need to calculate the molar mass of the compound. Sodium nitrate has a molar mass of 85 grams/mol. Next, you convert the given mass of the sample (54.3 grams) to moles by dividing by the molar mass. This gives you approximately 0.639 moles of sodium nitrate. Since one mole of a compound contains Avogadro's number of formula units (6.022 x 10^23), the 54.3 gram sample of sodium nitrate would contain approximately 3.85 x 10^23 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.
How many moles of silver nitrate do 2.8881015 formula untied equal?
To find the number of moles in 2.8881015 formula units of silver nitrate, you first need to know the molar mass of silver nitrate (AgNO3), which is 169.87 g/mol. Then, you can use the formula: moles = formula units / Avogadro's number. Therefore, moles = 2.8881015 / 6.022 x 10^23 = 4.79 x 10^-24 moles.
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.)
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 formula units of AgNO3 are present in 147g of this compound?
To determine the number of formula units of AgNO3 in 147g of the compound, you first need to calculate the molar mass of AgNO3. The molar mass of AgNO3 is 169.87 g/mol. Next, divide the given mass (147g) by the molar mass to find the number of moles present in the sample. Finally, use Avogadro's number (6.022 x 10^23) to convert moles to formula units.
How many formula units would there be in a 54.3 gram sample of sodium nitrate?
To determine the number of formula units in a sample of a compound, you first need to calculate the molar mass of the compound. Sodium nitrate has a molar mass of 85 grams/mol. Next, you convert the given mass of the sample (54.3 grams) to moles by dividing by the molar mass. This gives you approximately 0.639 moles of sodium nitrate. Since one mole of a compound contains Avogadro's number of formula units (6.022 x 10^23), the 54.3 gram sample of sodium nitrate would contain approximately 3.85 x 10^23 formula units.
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 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
What is the number of formula units that are in 5.68 moles of magnesium oxide?
To find the number of formula units of magnesium oxide in 5.68 moles, you first need to determine the formula of magnesium oxide (MgO). Then use Avogadro's number (6.022 x 10^23) to convert moles to formula units. So, in 5.68 moles of MgO, there are approximately 3.43 x 10^24 formula units.
How many formula units of salt make up 10 moles?
10 formula units
How many moles are equivalent to 3.6x1024 formula units of MgBr2?
The answer is 5,978 moles.
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.
