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There are 107.9 grams in one mole of pure silver. 107.9 a.m.u. is the average mass of isotopes of the element silver
Answer to another (= not this one) question:
The atomic number is the number of grams in a mole of any element. Hydrogen has an atomic number of 1 so a mole of hydrogen (okay, a half-mole of H2) weighs 1 gram.
[ Silver is not Hydrogen ! and an atom is not a molecule]
Wiki User
∙ 11y ago
Wiki User
∙ 15y agoFor this you need the Atomic Mass of Ag. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel.
1.3 moles Ag × (107.9 grams) = 140.3 grams Ag
Wiki User
∙ 7y ago2,00 moles of silver have a mass of 215,7364 g.
Wiki User
∙ 16y ago107.8682g/mole
Wiki User
∙ 11y ago.016
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How many moles is 3 grams of carbon?
By definition, one mole would be the same as the atomic mass. You take the number of moles and multiply it by the atomic mass. So if you have just 1 mole, the number of grams will be the atomic mass. Carbon's atomic mass is 12.011 grams.
How many moles of atoms are in 9.00g of carbon 13?
Formal set up. ( Avogadro's number appears as form of 1 here ) 9.00 grams 13C (1 mole 13C/13.00 grams)(6.022 X 1023/1 mole 13C)(1 mole atoms 13C/6.022 X 1023) = 0.692 mole of 13C atoms ====================
How many moles of aluminum are present in 856g of aluminum sample?
Moles = Mass/ Relative Molecular Mass Aluminum forms Al2 compounds, so the relative molecular mass is 2 * 13 = 26. 856/26 = 32.9 (3sf)
How many grams of Al2S3 can be formed from the reaction of 108.00 grams of Al with 5.00 grams of S?
The answer i got is 12.33 grams of Al2S3. Below i will try to show the steps i used: n=moles m=mass (grams) M= molecular weight (from periodic table) 2Al + 3S --> Al2S3 nAl= m/M = 9/13 = 0.692307 moles of Al nS= m/M = 8/16 = 0.5 moles of S Limiting Reagent: 1.5 moles of S required for every mole of Al (3:2 ratio) This means that there should be about 1.04 moles of S to completely use up all the Al. Since there is less than this amount of S present (only 0.5 moles), S is the limiting reagent and should be used in the mole calculation of the product Amount of product: 0.5 moles S x (1 mol Al2S3/ 3 mol S) = 0.16666 moles of Al2S3 nAl2S3 = m/M Molecular weight of Al2S3 = (2 x 13) + (3 x 16) = 74 0.16666 moles Al2S3 = m/74 m = 74 x 0.16666 = 12.33 grams of Al2S3 Therefore, 12.33 grams of Al2S3 is formed in this reaction ( hopefully this is right :P )
How many electrons are in 9 g of carbon 13?
The molar mass of 13-carbon is approximately 13g. Hence 9g contains 0.6923 moles.
How many moles is 3 grams of carbon?
By definition, one mole would be the same as the atomic mass. You take the number of moles and multiply it by the atomic mass. So if you have just 1 mole, the number of grams will be the atomic mass. Carbon's atomic mass is 12.011 grams.
How many moles of atoms are in 8 grams of carbon 13?
0,666 moles
How many moles of atoms are in 9.00g of carbon 13?
Formal set up. ( Avogadro's number appears as form of 1 here ) 9.00 grams 13C (1 mole 13C/13.00 grams)(6.022 X 1023/1 mole 13C)(1 mole atoms 13C/6.022 X 1023) = 0.692 mole of 13C atoms ====================
How many moles are present in 500.00 g of H2O?
For this you need the atomic (molecular) mass of H2O. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. H2O= 18.0 grams500.0 grams H2O / (18.0 grams) = 27.8 moles H2O
What is the density of a substance that has a mass of 169 grams and a volume of 13 milliliters?
The density is 169/13= 13 grams/ml.
How much are 13 grams of silver worth?
you tell me
What is the number of kilograms of solvent in a 0.70 molal solution containing 5.0 grams of solute (molar mass of solute 30 g)?
molarity = moles of solute/ kilogram of solvent 0.30=13/x 0.30x=13 x=13/0.30 x=4.333333333333333333
How many grams of oxygen are required to burn 4.8 mol of butane?
The balanced equation for the reaction is 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O. This shows that 13 moles of diatomic oxygen are required to burn 2 moles of butane. By proportionality, (4.8/2)13 or 31.2 moles of oxygen are required to burn 4.8 moles of butane. This corresponds to 31.2(32) or 1.0 X 103 grams of oxygen.
If you have 4 grams of sand and 13 millilters of water what is the mass of the solution?
The mass is 17 g.
How many moles of aluminum are present in 856g of aluminum sample?
Moles = Mass/ Relative Molecular Mass Aluminum forms Al2 compounds, so the relative molecular mass is 2 * 13 = 26. 856/26 = 32.9 (3sf)
If the mass of one dozen cookie of uniform size shape and mass is 156 grams calculate the mass of a single cookie?
i think you work divide 156 by 12 and get 13 grams per cookie i think you would divide 156 by 12 and get 13 grams per cookie
How many grams of Al2S3 can be formed from the reaction of 108.00 grams of Al with 5.00 grams of S?
The answer i got is 12.33 grams of Al2S3. Below i will try to show the steps i used: n=moles m=mass (grams) M= molecular weight (from periodic table) 2Al + 3S --> Al2S3 nAl= m/M = 9/13 = 0.692307 moles of Al nS= m/M = 8/16 = 0.5 moles of S Limiting Reagent: 1.5 moles of S required for every mole of Al (3:2 ratio) This means that there should be about 1.04 moles of S to completely use up all the Al. Since there is less than this amount of S present (only 0.5 moles), S is the limiting reagent and should be used in the mole calculation of the product Amount of product: 0.5 moles S x (1 mol Al2S3/ 3 mol S) = 0.16666 moles of Al2S3 nAl2S3 = m/M Molecular weight of Al2S3 = (2 x 13) + (3 x 16) = 74 0.16666 moles Al2S3 = m/74 m = 74 x 0.16666 = 12.33 grams of Al2S3 Therefore, 12.33 grams of Al2S3 is formed in this reaction ( hopefully this is right :P )
