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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.
What mass of silver chloride can be produced from 1.75L of a 0.293M solution of silver nitrate?
To determine the mass of silver chloride produced, we need to know the balanced chemical equation for the reaction between silver nitrate (AgNO3) and sodium chloride (NaCl) that produces silver chloride (AgCl) as a precipitate. Once we have the balanced equation, we can use the stoichiometry of the reaction to determine the number of moles of AgCl produced, and then convert that to mass using the molar mass of AgCl.
How many mph equal one G force?
You can not equate miles per hour to G forces. If you were traveling at 1000 mph, for example, but at a constant speed, you would fell no extra force since you are not accelerating.
How many kilometers is 9.33 miles?
well, i mile is 1.6km, so just multiply the 9.33 by 1.6 and you have the answer! that is 14.928km. G
How many grams of silver chloride are equal to a mol of silver chloride?
Silver chloride - AgClAg (107.89 grams) + Cl (35.45 grams) = 143.34 grams
How many moles is 573.28g of AgCl?
There's 4 moles.
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 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 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 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 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 grams of silver chloride will precipitate by silver nitrate 5.85g of sodium chloride?
The answer is 14,35 g AgCl.
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.
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.
What mass of silver nitrate will react with 5.85 g of sodium chloride to produce 14.35 g of silver chloride and 8.5 g of sodium nitrate if the law of conservation of mass is true?
By stoichiometry, the equation for the reaction is: AgNO3 + NaCl --> AgCl + NaNO3 From the reaction, 1 mole of AgNO3 reacts with 1 mole of NaCl to produce 1 mole of AgCl and 1 mole of NaNO3. Using the given masses, we can calculate the mass of AgNO3 needed for the reaction to occur. (Molar masses: AgNO3 = 169.87 g/mol, NaCl = 58.44 g/mol, AgCl = 143.32 g/mol, NaNO3 = 84.99 g/mol)
How many miles is it from Luna New Mexico to Sandia Park New Mexico?
6,000,000,000 Miles About N i G g A
How much AgCl is produced when 4.22 grams of AgNO3 react with 7.73 grams of AlCl3?
3.56 g 3 AgNO3 + AlCl3 --> 3 AgCl + Al(NO3)3 AgCl3 is the limiting reagent (I checked), so: 4.22 g AgNO3 * (1 mol AgNO3/169.88 g AgNO3) * (3 mol AgNO3/3 mol AgCl3) * (143.32 mol AgCl3/1 mol AgCl3) =3.56 g AgCl3See the Related Questions to the left for more information about solving stoichiometry problems of this nature.
