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Answers:For all dilution problems use the equation: C1V1 = C2V2 C1* 50 = 1.0*45 Remember that the final volume = 50ml. C1 = 45/50 C1 = 0.9MThe final molarity is 0.9M
Answers:You bubble in more HCl gas. There is no way to evaporate the water since heating the solution will decrease the solubility of the dissolved gas.
Answers:2MnO4^- + 16H^+ + 10Cl^- ==> 2Mn^2+ + 5Cl2 + 8H2O i think this is your question certainly acidic conditions you need to do the experiment to be sure of the products
Answers:The first thing to do is to get some rough idea of the molarity of the hydrochloric acid. The concentration is between 10 and 20%. HCl has a molar mass of 35.453+1.008 = 36.461g.This means that a 1M solution of HCl will have a concentration of approx 3.65% So your HCl solution has a molarity between 2.73M and 5.48M. Ideally in a titration you want both solutions to have about the same molarity. The NaOH is 1M. Now imagine if you used 25ml of the HCl you would have to add something between 68 and 137ml of NaOH. This is very inconvenient - remember most burettes are 50ml capacity and you do not want to introduce errors by having to refill the burette during the titration. These are just practical steps that a good lab. technician would consider. So my advice would be to dilute the HCl solution 4 fold. That is carefully measure out 250ml of the HCl and then dilute up to 1000ml with distilled water in volumetric flasks. Now you can easily titrate 25ml of this diluted solution against the 1M NaOH using the phenolphthalein indicator, as has already been so well explained to you. When you have calculated the molarity, do not forget to multiply the result by 4 to account for the dilution.