0
Molar mass is the mass per amount substance .What you need to do is just add the mass of everything .
For ex -mass of siver -107
mass of nitrogen -14
mass of oxygen -16 x 3=48(there are three atoms of oxygen )
Molar Mass of AgNO3 - 107+14+48=169 mols
You can include decimals also as per the Periodic Table .
What else can I help you with?
Mass of silver in 3.4g AgNO3?
To find the mass of silver in 3.4g of AgNO3, you need to consider the molar mass of silver nitrate (AgNO3). The molar mass of AgNO3 is 169.87 g/mol. Since the molar ratio of Ag to AgNO3 is 1:1, the mass of silver in 3.4g of AgNO3 would be 3.4g * (1/169.87) ≈ 0.02g.
What is the mass of silver in 3.4g AgNo3?
To find the mass of silver in 3.4g of AgNO3, you need to consider the molar mass of AgNO3. The molar mass of AgNO3 is 169.87 g/mol. From this, you can calculate the mass of silver (Ag) in AgNO3, which is 107.87 g/mol. Therefore, the mass of silver in 3.4g of AgNO3 is (107.87/169.87) * 3.4g.
How many moles in 4.50 grams of silver nitrate?
To find the number of moles in 4.50 grams of silver nitrate (AgNO3), you first need to calculate the molar mass of AgNO3. The molar mass of AgNO3 is 169.87 g/mol. Then, use the formula: moles = mass/molar mass. So, 4.50 grams of AgNO3 is equal to 0.0265 moles.
What is the molar mass of 2AgNO3?
The molar mass of silver nitrate (AgNO3) is approximately 169.87 g/mol. Multiplying this by 2 gives a molar mass of 339.74 g/mol for 2 moles of AgNO3.
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.
Mass of silver in 3.4g AgNO3?
To find the mass of silver in 3.4g of AgNO3, you need to consider the molar mass of silver nitrate (AgNO3). The molar mass of AgNO3 is 169.87 g/mol. Since the molar ratio of Ag to AgNO3 is 1:1, the mass of silver in 3.4g of AgNO3 would be 3.4g * (1/169.87) ≈ 0.02g.
What is the mass of silver in 3.4g AgNo3?
To find the mass of silver in 3.4g of AgNO3, you need to consider the molar mass of AgNO3. The molar mass of AgNO3 is 169.87 g/mol. From this, you can calculate the mass of silver (Ag) in AgNO3, which is 107.87 g/mol. Therefore, the mass of silver in 3.4g of AgNO3 is (107.87/169.87) * 3.4g.
How many moles in 4.50 grams of silver nitrate?
To find the number of moles in 4.50 grams of silver nitrate (AgNO3), you first need to calculate the molar mass of AgNO3. The molar mass of AgNO3 is 169.87 g/mol. Then, use the formula: moles = mass/molar mass. So, 4.50 grams of AgNO3 is equal to 0.0265 moles.
What is the molar mass of 2AgNO3?
The molar mass of silver nitrate (AgNO3) is approximately 169.87 g/mol. Multiplying this by 2 gives a molar mass of 339.74 g/mol for 2 moles of AgNO3.
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 moles of AgNO3 does 85 grams of AgNO3 represents?
To find the number of moles, you need to divide the given mass (85 grams) by the molar mass of AgNO3 (169.87 g/mol). 85 grams of AgNO3 represents 0.500 moles.
How many moles of silver are present in 32.46g of AgNO3?
To find the number of moles of silver in 32.46g of AgNO3, first calculate the molar mass of AgNO3 (169.87 g/mol). Then, divide the given mass by the molar mass to find the number of moles (32.46g / 169.87 g/mol ≈ 0.191 moles). Since there is one mole of Ag in one mole of AgNO3, there are 0.191 moles of silver present.
How do you convert 22.6g AgNO3 to moles?
To convert grams to moles, you need to divide the given mass by the molar mass of the substance. The molar mass of AgNO3 is approximately 169.87 g/mol. Therefore, to convert 22.6g of AgNO3 to moles, you would divide 22.6g by 169.87 g/mol to get approximately 0.133 moles of AgNO3.
What is the mass of 0.118 moles of AgNO3?
Mole / Molar mass = mass = 0.118 [mol] /169.87 [g mol−1] = 20.05 grams
What mass of the precipitate could be produced by adding 100.0ml of 0.887 m agno3 to a na3po4 solution assume that the sodium phosphate reactant is present in excess?
To find the mass of the precipitate that forms when 100.0mL of 0.887M AgNO3 is added to a Na3PO4 solution, you need to determine the limiting reactant. Since Na3PO4 is in excess, AgNO3 is the limiting reactant. Calculate the moles of AgNO3 using its molarity and volume, then use the mole ratio between AgNO3 and the precipitate to find the moles of the precipitate. Finally, convert the moles of the precipitate to mass using its molar mass.
How many moles are in 4.5 g of AgNO3?
To find the number of moles in 4.5 g of AgNO3, you first need to determine the molar mass of AgNO3 which is 169.87 g/mol. Then you can use the formula: moles = mass / molar mass. Therefore, moles = 4.5 g / 169.87 g/mol ≈ 0.0265 moles.
What mass of solid agcl is obtained when 25 ml of 0.068m agno3 reacts with excess of aqueous hcl?
By definition, No. of moles = given mass/molecular mass; and also by definition, molar concentration of a solute means the number of moles of solute per liter of solution. Therefore, 25 mL of 0.068 M AgNO3 contains AgNO3 = (0.068 * 25) / 1000 = 0.0017 mol of AgNO3.The equation for the reaction is AgNO3 + HCl -> AgCl + HNO3, showing that 0.0017 mol of AgNO3 gives 0.0017 mol of AgCl. The molecular mass of AgCl = 107+35.5 = 143.5 gTherefore, the mass of AgCl produced by the reacion = No. of moles*molecular mass = .0017*143.5 = 0.24g, to the justified number of significant digits.100 mL of 0.068 M AgNO3 contains AgNO3 = 0.068 molSo,Therefore,
