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Hedron123

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Hedron123 last won the day on December 21 2012

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  1. Hey, I believe HL sciences are much harder than SL. I did Chemistry SL and Biology HL and got a 7 in both. However, Biology required DAYS and DAYS of studying + lab reports. I've been told that Chemistry HL is the hardest science (I'm not acquainted with the syllabus so I can't tell) so it's up to you. If you are aiming at an overall high mark I would say stick to SL since if you study well you can surely get a 7. I strongly recommend you to put a lot of effort when doing lab reports due to the fact that they will raise your mark a lot if they are good. Good luck with your choice! Regards. P
  2. Hey Zananok, the procedure to calculate a best-fit line is called least squares / least squares fitting. It can approximate data points to any polynomic function you want. I can introduce you to it via MSN / Skype since it's kind of hard to explain through the forum. If you are interested send me a private message and I'll give you my email. Regards.
  3. Hey there. You could use the enzyme catalase (which is found in the liver of animals). This enzyme catalyses the decomposition of hydrogen peroxide into water and oxygen. By measuring the volume of O2 produced you can determine which conditions are better for the enzyme to work. The reaction is really fast so I believe your issue as regards time would be solved. I hope this helps you! Regards.
  4. I think it's because in CH3F the Hs aren't directly bonded to the F (all 3 Hs and F are bonded to the carbon) so there is no H-bonding between the molecules (they can form H-bonds with water). In D, however, there's an OH group (oxygen directly bonded to hydrogen); hence there is inter-molecular H-bonding. CH3F has a higher boiling point than C2H6 because it is polar. (so in your question, the intermolecular bonding goes: dispersion forces in C2H6 (Van der Waals / instantaneous dipole-dipole), polar in CH3F, H-bonding in CH3OH) Exactly, in CH3F fluorine is not attached to hydrogen so no hydro
  5. I'm hoping your math is right because it's too late for me to do that You should be using c=n/v to find your moles of HCl in each solution and can find the mass from that. But for rates of reaction thats just added work you don't need. HCl and MgCO3 should release MgCl+H2O+CO2 (actually its HCO3 but at your temp and pressure it RAPDILY decomposes to water and CO2). Since CO2 is a gas that will be exiting the system you can record mass of the beaker+HCl+MgCO3 to find your initial mass, then after X seconds after beginning the reaction record the mass again. This should give you experimental da
  6. Hey, I would certainly suggest you to study from the Course Companion book or the summary book by Geoff Neuss. Those are, in my opinion, the most IB-like bibliography you can find. You should practice with some past papers too in order to know how the actual exam is like. If there are exercises you are troubled with, don't hesitate to post them here and we will help you solve them. I wish you the best for your mocks. Regards.
  7. Then you should use my delta H result which is negative, meaning that the reaction is exothermic. If you have to state where you took the values from: Chang, Raymond (2007). Chemistry. China: McGraw-Hill. Appendix 3.
  8. HCl + NaHCO3 ===> NaCl + H2O + CO2 Okay, if you have delta H literature values the calculation would be: ∆H = ∆H products - ∆H reactants (∆H CO2 + ∆H H2O + ∆H NaCl) – (∆H HCl + ∆H NaHCO3) (-393.5kJ/mol + (-285.8kJ/mol) + (-411.0kJ/mol)) – (-92.3kJ/mol - 947.68kJ/mol) ∆H = -50.32kJ/mol K cool i got 46kJ/mol probably just because we have diff values from diff sources. BUT. my experimental values are like 37 JOULES, with a 96% error... The mass of my solution was 3.31, the temp range was 2.0 and the specific heat of water is 4.18J/g/C yea? So q=3.31(4.18)(2) which equals 27.7J..... Edit: I
  9. HCl + NaHCO3 ===> NaCl + H2O + CO2 Okay, if you have delta H literature values the calculation would be: ∆H = ∆H products - ∆H reactants (∆H CO2 + ∆H H2O + ∆H NaCl) – (∆H HCl + ∆H NaHCO3) (-393.5kJ/mol + (-285.8kJ/mol) + (-411.0kJ/mol)) – (-92.3kJ/mol - 947.68kJ/mol) ∆H = -50.32kJ/mol K cool i got 46kJ/mol probably just because we have diff values from diff sources. BUT. my experimental values are like 37 JOULES, with a 96% error... The mass of my solution was 3.31, the temp range was 2.0 and the specific heat of water is 4.18J/g/C yea? So q=3.31(4.18)(2) which equals 27.7J..... Edit: I
  10. HCl + NaHCO3 ===> NaCl + H2O + CO2 Okay, if you have delta H literature values the calculation would be: ∆H = ∆H products - ∆H reactants (∆H CO2 + ∆H H2O + ∆H NaCl) – (∆H HCl + ∆H NaHCO3) (-393.5kJ/mol + (-285.8kJ/mol) + (-411.0kJ/mol)) – (-92.3kJ/mol - 947.68kJ/mol) ∆H = -50.32kJ/mol
  11. Hey Drake, sorry for not answering before. When you are working at a constant pressure Qp (meaning Q at a constant pressure) = ∆H. Demonstration) H = E + PV (2nd law of T.D) so: ∆H = ∆E + ∆PV (I) E is the internal energy of the system which is equal to: Q + W (heat + work) ∆E = Qp - P∆V (II) (we are assuming pressure is constant) ∆H = ∆E + P∆V (I took P as a common factor) By replacing ∆E from eq. II in eq I: ∆H = Qp - P∆V + P∆V => ∆H = Qp This is derived from the 2nd law of Thermodynamics. I'm not sure if I have cleared your thoughts, be sure to ask again if you didn't understand. The uni
  12. Couldn't you just use the volume of water it produced? do some simple conversions and its a 1:1 ratio just like CO2 so its just a 1 step conversion from volume to moles. The fact is you will use a solution of HCl and with the paper on the bottom it will be harder to measure accurate volumes changes. I believe the CO2 production is a more feasible method but I'm open to other suggestions. Besides you cannot predict how much will the volume increase, maybe it's such an insignificant amount that it's hard to account.
  13. I guess you could add excess HCl and measure the volume of CO2 released. Once the volume remains constant for about 2 minutes it means that all the calcium carbonate has been used up. With that volume, use the p x v = n R T equation to calculate the number of moles of CO2 which, in turn, is the same as the initial number of moles of CaCO3. CaCO3 + 2HCl --------> CaCl2 + H2O + CO2 (Sorry for the subindices missing) Weigh the paper before carrying out the experiment and then multiply the number of moles obtained by the molar mass of calcium carbonate and compare both. I think this might hel
  14. Hey people, I've been away for a while. The good thing is that I'm back and I'm willing to help any Chemistry student with their problems or doubts. I'll be glad to help you with questions regarding lab reports, designs, theory or even on how to approach the actual paper. I hope to be useful. Thanks in advance. Regards
  15. The literature value is the result that you should have obtained if the experiment was free of error. For instance: Mg(s) + 2HCl(aq) ====== MgCl2(aq) + H2(g) If this reaction was carried out free of error, you would obtain 1 mole of hydrogen gas. One mole of any gas will have a volume of 22.4dm3. However, if you carried out the experiment under normal conditions in a school lab, you would obtain a lower volume of gas. Hence, the experimental value (the one you obtained) will be lower than the literature value (theoretical value) and the difference between the two is due to the errors of the e
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