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Mole / Molar mass = mass = 0.118 [mol] /169.87 [g mol−1] = 20.05 grams
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How many moles of AgNO3 does 85 grams of AgNO3 represents?
85 grams of AgNO3 represents 0,.5 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,
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.
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 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 Ag are produced when starting with 6.2 moles of AgNO3?
6,2 moles of silver
Calculate the number of moles in 5.34 10 21 atoms of boron?
Divide it by Avogadro constant.There are 0.0088 moles
How many grams are there in 2.5 x 1023 molecules of AgNO2?
7.4x10^23 x ( 1 mol AgNO3/ 6.02x10^23) x (169.88g AgNO3/ 1 mol AgNO3)=2.1x10^2g AgNO3no of molecules=7.4x10^23Mr of AgNO3=169.88we can find the no of moles, thereforein 1mol there are 6.02x10^23 molecules7.4x10^23 molecules represent (7.4x10^23/6.02x10^23)molesin theory when we multiply molar mass (Mr) with the no of moles, we obtain the mass of the substance in gram:Mr x no of Mol= Mass in grams(7.4x10^23/6.02x10^23)x169.88= 2.1x10^2
How many moles of silver are present in 32.46g of AgNO3?
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.
How many moles of silver of ions are presented in 32.46 g of AgNO3?
The number of moles is 0,19.
