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  • Birthday 05/05/1985

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  1. Thank you for the reply. The reason why I’m asking this is because I have to plan an experiment to suggest that the enthalpy change between a metal and a solution of a salt is related to the relative position of two metals in the reactivity series. Firstly I thought that if the reaction takes too long I should use a temperature –time graph and use a temperature correction method to obtain an accurate result. If the reaction is slower than a neutralisation reaction I could simply just note the maximum temp it reaches. Which one is most sensible If I was going to use about 3.30g of zinc with the solutions being approximately 50cm^3 ? Thank you.
  2. Can anyone please tell me how fast these reactions will occur. Haven't done the experiments yet, so I'm mystified Zn(s) + CuSO4(aq) ==> ZnSO4(aq) + Cu(s) Zn(s) + Pb(NO3)2 (aq) ==> Zn(NO3)2 + Pb(s) Any help would be much appreciated. Thank you.
  3. Ahh ... So if I find the mass of the saturated Ca(OH)2 to be 0.3g and I originally added 0.5g to 0.1 dm^3. The solubility would be 0.3g x 1 dm^3/0.1 dm^3 ?
  4. Thanks for the reply. What I can duduct is that there is a method of calculating the solubility by determining the Ksp, although I have no idea how to do that . Also, will the solubility be in g dm^-3 . Unfortunatelly my teacher told me that the titration will have to be done with HCl, he also told me that I have to calculate a reasonable concentration of HCl that needs to be used in the titration. The only way I can think of calculating is by trial and improvement. So it doesn't matter what mass of Ca(OH)2(s) I use in the experiment? Thanks again
  5. Greetings, I was wondering if someone could help me with some of my chemistry difficulty. I have to do an experiment to find the solubility of calcium hydroxide. I have gone as far as to prepare the saturated solution of calcium hydroxide and to plan a titration with HCl. My problem is that I do not know what concentration of HCl I should use, how would I calculate a reasonable amount? Also, correct me if this is wrong. If I measure out 0.5g of calcium hydroxide dissolve it in water, filter it through a funnel and then titrate the saturated solution, I will be able to compare it with the original mass I measure out. To compare the solubility would I just say the mass of calcium hydroxide that was not saturated is evidence to suggest that it is sparingly soluble? Or is there a standard method of doing this? Please, any help would be much appreciated. Thank you in advance.
  6. Sorry I can’t answer your question, but I have on myself. Did you get the Americium from a smoke detector?
  7. Does anyone know what the melting point of pyrex glassware is?
  8. If only I can comprehend, but interesting nonetheless.
  9. Oops, didn’t realise the melting point of NaCl is 800C
  10. Only if I use NaCl dissolved in water, I’ll be using molten NaCl.
  11. That won’t be much fun will it? Another benefit of making HCl out of NaCl is that I get pure Na as a by-product, which also has its uses. My main problem now is setting up a contraption for this.
  12. LOL .... Vomiting would probably be the easiest way to obtain it, I just think it will take a hell of a lot of vomiting and also it wouldn’t be as pure as what is needed. I thought perhaps electrolysis with molten NaCl. The problem is setting up the apparatus, I know if I spilt the ions by means of electrolysis, Cl2 will be given of which could then react with H2O, I think, giving me HCl. Don’t know if this will work though.
  13. Can anyone give me some help on how I could effectively and efficiently make HCl ? Any help would be greatly appreciated, thank you.
  14. Good try. I manage to work it out after spending almost 10 hours on it. (i) MgO + 2HCl ==> MgCl2 + 2H2O (ii) NaOH + HCl ==> NaCl + H2O moles of hydrochloric acid added to the magnesium oxide = 100 ÷ 1000 x 2 = 0.20 mol HCl moles of excess hydrochloric acid titrated = 19.7 ÷ 1000 x 0.2 = 0.00394 mol HCl (mole ratio NaOH:HCl is 1:1 from equation (ii)) moles of hydrochloric acid reacting with the magnesium oxide = 0.20 - 0.00394 = 0.196 mole MgO reacted = 0.196 ÷ 2 = 0.098 (1: 2 in equation (i)), the formula mass of MgO = 40.3, so mass of MgO reacting with acid = 0.098 x 40.3 = 3.95g, % purity = 3.95 ÷ 4.06 x 100 = 97.3%
  15. Local anaesthetic drugs act by causing a reversible block to conduction along nerve fibres.
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