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1 mol of Ag and 1 mol of Au can be said to contain the same amount of atoms of each element.
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What is common between 1 mole of Ag and 1 mole of Au?
Both 1 mole of silver (Ag) and 1 mole of gold (Au) contain Avogadro's number of atoms, which is approximately 6.022 x 10^23. They also have a molar mass equal to their atomic mass in grams, which is around 107.87 g/mol for Ag and 196.97 g/mol for Au.
How much silver can be produced from 125 grams of Ag2S?
1 mol Ag/ 107.87g Ag ---/---------------------------------------- x2=215.74g Ag / 1mol Ag 1 mol S/32.07g S --------/------------------------------------32.07g S /1 mol S total=247.81g Ag2S 215.47g Ag/247.81=.8706 87.06% Ag .8706 or 87.06% Ag x 125g Ag2S = 108.83g Ag can be produced from Ag2S
What are the two equivalent conversion factors for the molar mass of silver?
The molar mass of silver is 107.87 g/mol. Therefore, the conversion factors are: 1 mol Ag = 107.87 g Ag and 107.87 g Ag = 1 mol Ag.
How many moles of Ag are produced from 155g Ag2O?
[155 (g Ag2O) / 231.8 (g Ag2O/mol Ag2O)] * 2 (mol Ag/mol Ag2O) == 310/231.8 mol Ag = 1.337 == 1.34 mole Ag
How many moles are in 47.0 grams of Ag?
To find the number of moles in 47.0 grams of Ag (silver), divide the mass given by the molar mass of silver (107.87 g/mol). [ \text{Number of moles} = \frac{47.0 , \text{g}}{107.87 , \text{g/mol}} \approx 0.436 , \text{moles} ]
How many moles of Ag contain 4.4910e23 atoms Ag?
To find the number of moles, we first need to calculate the number of moles of Ag atoms using Avogadro's number (6.022 x 10^23 atoms/mol). Number of moles = 4.4910e23 atoms Ag / (6.022 x 10^23 atoms/mol) ≈ 0.746 moles of Ag.
What mass of Ag2s is produced from a mixture of 2.0g Ag and 2.0g S8?
The way to solve these problems is to determine which reactant will produce the fewest moles of product. In order to do this, we first need to convert the mass of each reactant into moles of reactant using the molecular mass of silver, Ag (107.9 g/mol), and molecular sulfur, S8 (256.5) g/mol:2.0 g Ag * 1mol/107.9 g/mol = 0.0185 mol Ag2.0 g S8 * 1mol/256.5 g/mol = 0.00779 mol S8Now using the balanced chemical equation (our recipe) from above:16 Ag + S8 -->8 Ag2SWe can see:16 moles of Ag turns into 8 moles of Ag2S1 mole of S8 turns into 8 moles of Ag2SWhat we need to do know, is use these two statements above to see how many moles of product each reactant turns into:0.0185 mol Ag * 8 mol Ag2S/16 mol Ag = 0.00925 mol Ag2S0.00779 mol S8 * 8mol Ag2S/1 mol S8 = 0.0623 mol Ag2SSince 0.0185 mole of Ag produce less Ag2S (the product), silver (Ag) is the limiting reactant. Now you could convert this to grams of product by multiplying the lesser amount produced, 0.00925 moles of Ag2S, by the molecular mass of Ag2S.To determine the mass of reactant left unreacted, we need to convert moles of the limiting reactant into moles of the other reactant to find out how much of the other reactant is lost--Then subtract from the original amount and convert to grams with the molecular mass.0.0185 mol Ag * 1 mol S8/16 mol Ag = 0.00116 mol S8Originally we had 0.00779 mol of S8, from above, so:0.00779 mol S8 - 0.00116 mol S8 = 0.00663 mol S8Now convert to grams left:0.00663 mol S8 * 256.5 g/mol = 1.70 g S8 left unreacted!
How many grams of Ag2 S could be produced from 1.70g H2 S if sufficient amounts of the other reactants were present?
To determine how many grams of Ag₂S can be produced from 1.70 g of H₂S, we first need to use the balanced chemical equation for the reaction between H₂S and silver nitrate (AgNO₃) to form Ag₂S. The molar mass of H₂S is approximately 34.08 g/mol, and the molar mass of Ag₂S is about 247.8 g/mol. Calculating the moles of H₂S in 1.70 g gives approximately 0.050 mol. According to the stoichiometry of the reaction, 1 mole of H₂S produces 1 mole of Ag₂S, so 0.050 mol of H₂S will produce 0.050 mol of Ag₂S. Multiplying the moles of Ag₂S by its molar mass (247.8 g/mol), we find that approximately 12.39 g of Ag₂S can be produced.
How many atoms of gold are in 0.02 g of Au?
To calculate the number of atoms in 0.02 g of gold (Au), you first need to determine the number of moles of gold in 0.02 g using the molar mass of gold (196.97 g/mol). Then, you use Avogadro's number (6.022 x 10^23 mol^-1) to convert moles to atoms. The calculation would be 0.02 g Au / 196.97 g/mol Au × 6.022 x 10^23 atoms/mol.
How many atoms are in 1.500 kilograms of gold?
The mass of 1 mole of an element is its atomic weight in grams.1 mole of an element is 6.022 x 1023 atoms of that element.Known/Given:1 mol Au = 196.96655g Au (atomic weight in grams)1 mol Au = 6.022 x 1023 atoms Au (Avagadro's number)1000g = 1 kgConvert kilograms to grams.1.500kg Au x (1000g/1kg) = 1500g AuConvert grams to moles.1500gAu x (1mol Au/196.96655g Au) = 7.616mol AuConvert moles to atoms.7.616mol Au x (6.022 x 1023 atoms Au) = 4.586 x 1024 atoms Au
How many atoms are in 3.50 g of gold?
1 mole of gold is 196.97 grams. 7.2 mol Au * (196.97 g Au/1 mol Au) = 1418.18 g There are 1418.18 grams in 7.2 moles of gold.
How many ag atoms are there in 0.0001 grams of Ag.Molar mass of Ag is 108 gram per mole?
To find the number of atoms in 0.0001 grams of silver (Ag), first determine the number of moles in 0.0001 grams using the molar mass of Ag (108 g/mol). Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms. So, 0.0001 g of Ag is equal to 6.94 x 10^16 Ag atoms.
