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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.
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.
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 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
How much 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl?
To find the volume of 0.075 M NaCl solution that can be made, you can use the formula C1V1 = C2V2. Plugging in the values, we get (9.0 M)(450 mL) = (0.075 M)(V2). Solving for V2 gives V2 = 1.1 L. Therefore, 1.1 L of 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl.
