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AgNO3 is 169.87 g/mol. Ag is 107.87 g/mol meaning that Ag is 63.5% of AgNO3. 63.5% of 32.46g is 20.61g. 20.61g / 107.87 g/mol = .191 moles.

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Q: How many moles of silver are present in 32.46g of AgNO3?
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How many moles of Ag are produced when starting with 6.2 moles of AgNO3?

6,2 moles of silver


How many moles of silver of ions are presented in 32.46 g of AgNO3?

The number of moles is 0,19.


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 ---------------------


How many moles of silver chromate (Ag2CrO4) will be produced from 4 mol of silver nitrate (AgNO3)?

6


How many moles of silver nitrate do 2.8881015 formula untied equal?

2,888.10e15 molecules of AgNO3 equal 0,48.10e-8 moles.


How many moles of AgNO3 does 85 grams of AgNO3 represents?

85 grams of AgNO3 represents 0,.5 moles.


How many silver atoms would be in 4.55 moles of AgNO3?

4.55 moles AgNO3 (1 mole Ag/1 mole NO3)(6.022 X 10^23/1 mole Ag) = 2.74 X 10^24 atoms of silver


How many moles of silver ions are present in 32.46g of AgNO3?

First, get the molar mass of silver nitrate (AgNO3): Ag-108, N-14, O-16(3)=48; total=170g/mol. Now, find out how many moles that is: 32.46/170=0.19mol. Now, look at the formula: there's 1 silver ion per formula unit. So, 0.19(6.02x1023)=approx. 1.14x1023 silver ions.


How many moles of silver chloride are produced from 7 moles of silver nitrate?

AgNO3 + NaCl ===> AgCl(s) + NaNO37 moles silver nitrate will produce 7 moles of silver chloride provided there is sufficient (at least 7 moles) of sodium chloride.


How many mL of .117M AgNO3 solution would be required to react exactly with 3.82 moles of NaCl?

Balanced equation first! AgNO3 + NaCl -> AgCl + NaNO3 all one to one, get moles AgNO3 3.82 moles NaCl (1 mole AgNO3/1 mole NaCl) = 3.82 moles AgNO3 ------------------------------- Molarity = moles of solute/Liters of solution 0.117 M AgNO3 = 3.82 moles AgNO3/Liters Liters = 3.82/0.117 = 32.6 Liters which is 32600 milliliters which is unreasonable; check answer if you can


How many moles of agno3 are needed to prepare 0.50 l of a 4.0 m solution?

Molarity = moles of solute/liters of solution or, for our purposes moles of solute = liters of solution * Molarity moles of AgNO3 = 0,50 liters * 4.0 M = 2.0 moles of AgNO3 needed --------------------------------------


How many moles are in 680g of AgNO3?

Roughly 4 moles.