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How many moles are in 0.525g of AgCl?
To find the number of moles in 0.525g of AgCl, you need to divide the mass by the molar mass of AgCl. The molar mass of AgCl is 143.32 g/mol. moles = mass / molar mass moles = 0.525g / 143.32 g/mol moles ≈ 0.0037 mol
How many moles in 0.0688g Ag CL?
To find the number of moles in 0.0688g AgCl, first calculate the molar mass of AgCl. It is 143.32 g/mol. Then divide the given mass (0.0688g) by the molar mass to get the number of moles. This gives you approximately 0.00048 moles of AgCl.
How many moles is 573.28 of g of AgCI?
To find the number of moles in 573.28 g of AgCl, you need to divide the given mass by the molar mass of AgCl. The molar mass of AgCl is approximately 143.32 g/mol. So, 573.28 g / 143.32 g/mol = approximately 4 moles of AgCl.
How many moles is 573.28 g of AgCl?
To find the number of moles, divide the given mass of AgCl by its molar mass. The molar mass of AgCl is 143.32 g/mol (107.87 g/mol for Ag + 35.45 g/mol for Cl). Therefore, 573.28 g ÷ 143.32 g/mol = 4 moles of AgCl.
How many moles of silver chloride are produced from 7 moles of silver nitrate?
1 mole of silver nitrate produces 1 mole of silver chloride in a 1:1 ratio according to the balanced chemical equation AgNO3 + NaCl -> AgCl + NaNO3. Therefore, 7 moles of silver nitrate will produce 7 moles of silver chloride.
How many moles are in 0.525g of AgCl?
To find the number of moles in 0.525g of AgCl, you need to divide the mass by the molar mass of AgCl. The molar mass of AgCl is 143.32 g/mol. moles = mass / molar mass moles = 0.525g / 143.32 g/mol moles ≈ 0.0037 mol
How many moles in 0.0688g Ag CL?
To find the number of moles in 0.0688g AgCl, first calculate the molar mass of AgCl. It is 143.32 g/mol. Then divide the given mass (0.0688g) by the molar mass to get the number of moles. This gives you approximately 0.00048 moles of AgCl.
How many moles is 573.28 of g of AgCI?
To find the number of moles in 573.28 g of AgCl, you need to divide the given mass by the molar mass of AgCl. The molar mass of AgCl is approximately 143.32 g/mol. So, 573.28 g / 143.32 g/mol = approximately 4 moles of AgCl.
How many moles of AgCl will be produced from 83.0 g of AgNO3 assuming NaCl is available in excess?
The balanced chemical equation for this reaction is: AgNO3 + NaCl -> AgCl + NaNO3 From this equation, we can see that 1 mole of AgNO3 produces 1 mole of AgCl. Since the molar mass of AgNO3 is 169.87 g/mol, 83.0 g of AgNO3 is equivalent to 0.488 moles. Therefore, 0.488 moles of AgCl will be produced.
How many moles is 573.28 g of AgCl?
To find the number of moles, divide the given mass of AgCl by its molar mass. The molar mass of AgCl is 143.32 g/mol (107.87 g/mol for Ag + 35.45 g/mol for Cl). Therefore, 573.28 g ÷ 143.32 g/mol = 4 moles of AgCl.
How many grams of solid AgCl are needed to make 4.0 liters of a 5.0 M solution of AgCl?
To determine the mass of AgCl needed, first calculate the number of moles needed using the molarity equation: moles = molarity x volume (in L). Then, convert moles of AgCl to grams by using the molar mass of AgCl (107.87 g/mol for Ag and 35.45 g/mol for Cl). Finally, perform the calculation to find the grams of AgCl required.
How many moles of AgCl will be precipitated when an excess of AgNo3 solution is added to one molar solution of CrCl(H2O)5Cl2?
When an excess of AgNO3 solution is added to a one molar solution of CrCl(H2O)5Cl2, all the chloride ions (Cl⁻) from CrCl(H2O)5Cl2 will react with Ag⁺ ions to form AgCl precipitate. Since CrCl(H2O)5Cl2 contains 6 moles of Cl⁻ per mole of complex, the reaction will precipitate 6 moles of AgCl. Therefore, 6 moles of AgCl will be formed.
How many moles of silver chloride are produced from 7 moles of silver nitrate?
1 mole of silver nitrate produces 1 mole of silver chloride in a 1:1 ratio according to the balanced chemical equation AgNO3 + NaCl -> AgCl + NaNO3. Therefore, 7 moles of silver nitrate will produce 7 moles of silver chloride.
What is the mole ratio of BaCl2 to AgCl?
The mole ratio of BaCl2 to AgCl is 1:2. This means that for every 1 mole of BaCl2, 2 moles of AgCl are produced in the chemical reaction.
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,
What mass in g of agcl is formed from the reaction of 75.0 ml of a 0.078 m agc2h3o2 solution with 55.0 ml of 0.109 m mgcl2 solution?
To find the mass of AgCl formed, first calculate the moles of AgC2H3O2 and MgCl2 using their respective concentrations and volumes. Then, determine the limiting reactant, which is MgCl2 as it produces less AgCl. Use the stoichiometry of the reaction to find the moles of AgCl formed and convert it to grams.
An impure sample of table salt that weighed 0.8421 g when dissolved in water and treated with excess AgNO3 formed 2.044 g of Ag-Cl what is the percentage of Na-Cl in the impure sample?
2.044 g AgCl / 143.33 g per mole = 0.01426 moles of AgClNaCl + AgNO3 --> AgCl + NaNO31 mole of AgCl = 1 mole of NaClThe sample contained 0.01426 moles of NaCl0.01426 moles NaCl x 58.44 g per mole = 0.8334 g NaCl0.8334 / 0.8421 = 98.97% pureI'm pretty sure this is the correct answer.
