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

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Roughly 4 moles.

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

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

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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.

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

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

10 grams of cobalt are in 2 moles

36 grams of water have 2 moles

2,00 grams of H2O have 0,11 moles.

9,8 moles of calcium have 392,76 grams.

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

no

.75 moles times 64 grams/mols = 48 grams

53 grams Ã· 18.01 grams/mole = 2.94 moles

The answer is 0,2686 moles.

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