1.You have 505 ml of 0.145M HCl solution and you want to dilute it to exactly 0.

Question

1.You have 505 ml of 0.145M HCl solution and you want to dilute it to exactly 0.100M how much water should you add ?

2.what volumen of 0.612 M ammonium sulfate contains 6.70g of ammonium ion?

3.how many milliliters of a 0.283 M HCl solution are needed to know neutralize 231 ml of a 0.0365 M Ba(OH)2 solution?

4. If 30.0 ml of 0.150 M CaCl2 is added to 13.5ml of 0.100 M AgNO3, what is the mass of the AgCl precipitate? 1.You have 505 ml of 0.145M HCl solution and you want to dilute it to exactly 0.100M how much water should you add ?

2.what volumen of 0.612 M ammonium sulfate contains 6.70g of ammonium ion?

3.how many milliliters of a 0.283 M HCl solution are needed to know neutralize 231 ml of a 0.0365 M Ba(OH)2 solution?

4. If 30.0 ml of 0.150 M CaCl2 is added to 13.5ml of 0.100 M AgNO3, what is the mass of the AgCl precipitate?

2.what volumen of 0.612 M ammonium sulfate contains 6.70g of ammonium ion?

3.how many milliliters of a 0.283 M HCl solution are needed to know neutralize 231 ml of a 0.0365 M Ba(OH)2 solution?

4. If 30.0 ml of 0.150 M CaCl2 is added to 13.5ml of 0.100 M AgNO3, what is the mass of the AgCl precipitate?

Explanation / Answer

THis response is for question 1. Questions 2,3,and 4 should be submitted separately as they re different.

FOr dilution, M1V1 = M2V2 so: 505mL x 0.145M = 0.1000M x V2 where V2 is the final solution volume. This works out to a total volume of 732 mL. Since the starting volume was 505, the amount of water needed is (732-505) =227mL

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