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What is the solubility of AgCl in a 0.35 M solution of NaCl?
(AgCl has Ksp = 1.8 x 10-10)-5.1 x 10-10
How many ml of 0.117 M AgNO3 would be required to react exactly with the NaCl solution?
To determine this, you need the concentration of the NaCl solution. Once you have that, you can use the stoichiometry of the reaction between AgNO3 and NaCl to calculate the volume of 0.117 M AgNO3 needed. The balanced chemical equation for this reaction is: AgNO3 + NaCl → AgCl + NaNO3.
4.02 grams of AgNO3 react with excess NaCl how many grams of silver nitrate are produced?
AgNo3(aq) + NaCl(aq) -------------->AgCl (ppt) + No3(-) + Na(+) no silver nitrates are produced it is all consumed. only silver chloride is produced and precipitate . free nitrate and free sodium ions are produced but do not react with each other 1 Mole AgNo3 ------->169.9 gm 1 Mole Nacl ------->58.4 gm 1 Mole AgCl ------->143.3 gm 4.02 gm AgNO3 = (4.02 / 169.9) = 0.02366 M AgCl produced = 0.02366 M = 3.39 gm
What is the solubility of AuCl in a 0.2 M solution of NaCl?
The solubility of AuCl in a 0.2 M solution of NaCl would depend on the solubility product constant (Ksp) of AuCl in water. If the Ksp of AuCl is exceeded by the presence of NaCl, AuCl would precipitate out of solution. If the Ksp is not exceeded, AuCl would remain in solution. Additional information, such as the Ksp value of AuCl, would be needed to calculate the exact solubility.
What is the solubility of AuCl in a 0.2 M solution of NaCl AuCl has Ksp equals 2.0 x 10-13?
1.0 x 10-12
What is 5 M of NaCl equivalent to?
5 M of NaCl is equivalent to 292,2 grams.
It was found that the molar solubility of baso3 in 01 m is 80 x 10-6 m what is the value of ksp baso3?
solubility = [Ba2+]= 80 x 10-6 M When BaSO3 dissolves it gives equal concentration of Ba2+ and SO32-, so Ksp = [Ba2+][SO32-] = 6.4 x10-9
What is the total dilution factor from 1.00 M NaCl to 1.00 M NaCl solution?
The concentration is the same !
preparation of 0.1 m of NaCL ML SOLUTION AND ITS LAP REPORT?
A solution of NaCl 1 M.
Calculate the isotonic coefficient for NaCl if a 07M of NaCl solution equaled the hemolysis of14M glucose solution?
i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2
What volume of a 0.20 M silver nitrate solution is required to precipitate all the Cl− ion in the solution as AgCl?
You cannot make a solution of AgCl, it is an insoluble salt
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.
