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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.
Do chemical reactions contain the same number of moles as reaction proceeds?
Not necessarily. Some reactions do have the same number of moles, and some do not.Examples: NaCl + AgNO3 ==> NaNO3 + AgCl same # of moles N2 + 3H2 ==> 2NH3 different # of moles
A volume of 10 ml of a .00600 M solution of Cl- ions are reacted with 0.500 m solution of AgNO3 what is the maximum mass of AgCL that precipitates?
Unfortunately this question is a complicated mathematical equation that can not be completed in 750 characters. There is a complex equation, where the user would in put the volume of the ions and the solution and calculate the solution in that manner.
A volume of 10 ml of a .00600 M solution are reacted with 0.500 m solution of AgNO3 what is the maximum mass of AgCL that precipitates?
This was a fun one.Step 1: You're not given the other reactant solution or the other product solution, so let's make one up that works. If you use sodium chloride on the reactant side and sodium nitrate on the product side, you get this correctly balanced double replacement reaction equation:AgNO3(aq) + NaCl(aq) --> NaNO3(aq) + AgCl(s)This works well because you don't need any molar ratio coefficients to muddy things up any further.Step 2: You're given the volume of the unnamed solution (that we've decided will be NaCl) so let's figure out how many moles are dissolved in that volume. Molarity is moles per liter, and you have 10mL, which is 0.01L, so move the decimal point twice to the left to get the moles per 10mL. You should get 0.006 moles of NaCl.Step 3: You're given the molality (m) of the silver nitrate solution, but not its volume. In a weak solution, molality (m) is approximately the same as molarity (M), but it doesn't matter since you aren't given its volume. You should assume then, in this case, that you have an excess of silver nitrate solution, which means that the sodium chloride solution is your limiting reactant. This means that you should now run the stoichiometry with the mass of NaCl.Step 4: The equation above is balanced as is, with no coefficients necessary, so the moles of the NaCl equal the moles of the AgCl, which we established in step 2 was 0.006 moles.Step 5: The precipitate is silver chloride, AgCl, with a molar mass of 143g/mol. Multiply this by the number of moles, 0.006, to get 0.858g AgCl, which is the answer. This is the maximum amount of silver chloride which can be yielded by this reaction under the conditions given.
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 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 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 573.28g of AgCl?
There's 4 moles.
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 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.
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,
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.
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.
Do chemical reactions contain the same number of moles as reaction proceeds?
Not necessarily. Some reactions do have the same number of moles, and some do not.Examples: NaCl + AgNO3 ==> NaNO3 + AgCl same # of moles N2 + 3H2 ==> 2NH3 different # of moles
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.
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.
What mass of solid agcl is obtained when 25 ml of 0.068 m agno3 reacts with excess of aqueous hcl?
See it's an easy one..!! AgNO3 + HCl -> AgCl + HNO3 100 mL of 0.068 M AgNO3 contains AgNO3 = 0.068 mol So, 25 mL of 0.068 M AgNO3 contains AgNO3 = (0.068 * 25) / 1000 = 0.0017 mol From the equation, we can see 1 mol of AgNO3 gives 1 mol of AgCl 0.0017 mol of AgNO3 gives 0.0017 mol of AgCl Amount of AgCl can be found this way.! No. of moles = given mass/ molecular mass molecular mass of AgCl = 107+35.5 = 143.5 g Therefore, Given mass = No. of moles*molecular mass = 0.0017*143.5 = 0.244g Note : In your question, you have written 0.068 "m" .. (small) m represents for Molality and (capital) M represents for Molarity..! Hope I helped.. :)
