0
What else can I help you with?
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 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.
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 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.
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 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,
What is the molarity of a solution if 255 grams AgNO3 is dissolved in 1500 mL of solution?
Get moles silver nitrate. 255 grams AgNO3 (1 mole AgNO3/169.91 grams) = 1.5008 moles AgCO3 --------------------------------Now; Molarity = moles of solute/Liters of solution ( 1500 ml = 1.5 Liters ) Molarity = 1.5008 moles AgNO3/1.5 Liters = 1.00 M AgNO3 ---------------------
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.
What mass of solid agcl is obtained when 25 ml of 0.068 m agno3 reacts with excess of aqueous hcl?
See it's an easy one..!! AgNO3 + HCl -> AgCl + HNO3 100 mL of 0.068 M AgNO3 contains AgNO3 = 0.068 mol So, 25 mL of 0.068 M AgNO3 contains AgNO3 = (0.068 * 25) / 1000 = 0.0017 mol From the equation, we can see 1 mol of AgNO3 gives 1 mol of AgCl 0.0017 mol of AgNO3 gives 0.0017 mol of AgCl Amount of AgCl can be found this way.! No. of moles = given mass/ molecular mass molecular mass of AgCl = 107+35.5 = 143.5 g Therefore, Given mass = No. of moles*molecular mass = 0.0017*143.5 = 0.244g Note : In your question, you have written 0.068 "m" .. (small) m represents for Molality and (capital) M represents for Molarity..! Hope I helped.. :)
How many moles of AgCl will be produced from 83.0 g of AgNO3 assuming NaCl is available in excess?
The balanced chemical equation for this reaction is: AgNO3 + NaCl -> AgCl + NaNO3 From this equation, we can see that 1 mole of AgNO3 produces 1 mole of AgCl. Since the molar mass of AgNO3 is 169.87 g/mol, 83.0 g of AgNO3 is equivalent to 0.488 moles. Therefore, 0.488 moles of AgCl will be produced.
Calculate the mass of AgBr is formed when 35.5mL of 0.184 M AgNO3 is treated with an excess of aqueous hydrobromic acid HBr?
To find the mass of AgBr formed, first calculate the moles of AgNO3 in 35.5 mL of 0.184 M solution. Then, use the mole ratio from the balanced chemical equation between AgNO3 and AgBr to find the moles of AgBr formed. Finally, multiply the moles of AgBr by its molar mass to get the mass. Note that since HBr is in excess, AgNO3 will be the limiting reagent.
How many grams of AgNO3 are required to make 500.0 mL of a 0.10 M solution?
Ah, what a lovely question! To make a 0.10 M solution of AgNO3 in 500.0 mL, we can use the formula: moles = molarity x volume (in liters). First, convert 500.0 mL to liters by dividing by 1000. Then, multiply the molarity (0.10 M) by the volume in liters to find the moles of AgNO3 needed. Finally, convert moles to grams using the molar mass of AgNO3. Happy calculating!
