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Hedron123

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Hedron123 last won the day on December 21 2012

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  1. Hedron123

    Chemistry HL vs Chemistry SL

    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.S. Remember the IB is not the end of the world!!
  2. Hedron123

    How to calculate best fit line?

    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. Hedron123

    Enzymes Lab

    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. Hedron123

    Quick melting/boiling point question?

    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 hydrogen bonding is formed. Its boiling point is higher than the C2H6 because, although it has no hydrogen bonds, the molecule is polar due to the difference in electronegativity (dipoles-dipoles). C2H6 will only have Van der Waals forces and, as the molar masses of all the compounds is very similar, the strength of the Van der Waals will also be very similar. The compound with oxygen, on the contrary, will be forming hydrogen bonds so its boiling point is the highest of all.
  5. Hedron123

    Chemistry HL/SL help

    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 data that you can find the rate of reaction from. Do this ON the scale so you can watch it drop mass and its just more precise in my opinion if you don't move it around everywhere (since that would actually be the equivalent of stirring which causes more collisions...you get it). Reason I was avoiding the mass from the math is because of just human error, too much HCl or a little less MgCO3 etc etc, measuring isn't perfect Ok I get that thanks, will put that as an explanation. Now on the mouth of the conical flask I added cotton wool to prevent any loss of acid due to acid spray. Then would I need to keep the mass of the cotton wool constant in the different repeats? Yes, stupid question but... yeah Oh one more, is it so necessary to put a diagram to show the experimental setup? EDIT: oh Drake I forgot, since you previously recommended me to also control temperature and pressure (a.k.a keeping them constant), I thought pressure only affects the speed of reaction when at least one of the reactants is in gaseous state? or... not? How to control a pressure though? by doing the experiment in the same place and using the same conical flask? And for temperature, do I need to keep the temperature of the MIXTURE constant, or that of the SURROUNDINGS? Since the reaction might be an exothermic or endothermic reaction, the temperature cannot be constant right? I find it even more troublesome since I now have that cotton wool covering the flask. Where to put the thermometer? Using a water bath is not a good idea since I need to measure the mass and I think a water bath is too heavy to put on an electronic balance but if it is of the surroundings, can I just turn the AC off and close any window, to keep it constant? Hey, I never got to answer your post before. Pressure and temperature can be controlled by carrying out the experiment under the same weather conditions. This will assure constancy throughout the procedure (you don't need to use a water bath, don't worry). As regards the experiment, you could use an indicator to see how much time does it take for all the reactant to be consumed. That might help you calculate the rate of the reaction. P.S. You cannot calculate the mass of a solution by multiplying the number of moles and the molar mass, you need density values in order to do so.
  6. Hedron123

    How to Study.....

    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. Hedron123

    Chemistry HL/SL help

    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. Hedron123

    Chemistry HL/SL help

    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 can't go redo my lab either so w/e data i have is what I have to use, so should i just stick with my values and my giant % error and then explain it in my evaluation? Because I'm pretty sure my calorimeter was iffy, and tons of gas (with the energy) escaped it quite easily which could really mess with my energy values right? It's not surprising that the results diverge from the real value. There are two factors that are determining. First, the literature values for delta H you're using are for the reaction at STP (this is not true under lab conditions). Second, the reactions are done inside a calorimeter which prevents most (if not all) of the heat loss to the surroundings. This a MAJOR error of the experiment since despite the accuracy of the method you're using, there will always be heat loss and so the result for delta H will not be 100% certain. But a 96% error? I mentioned how the calorimeter could be a terrible insulator and let energy out of the system. I also mentioned that the CO2 could also escape through the lid and take energy with it. I've also mentioned that for my last 2 x values, the mass i used includes excess NaHCO3 and not just the solution so its q value is actually kind of high. Oh wait...ok I see what you mean, my reaction is releasing energy (endothermic right?) and the calorimeters insulation is preventing it from releasing and its going back into the solution thus skewing my temperature readings? So scratch the bad insulator thing, my calorimeter's insulation being good was actually an error? o.O I'm not sure about what method you used. What I mean is the literature value you are comparing your result to is defined at STP (standard conditions of temperature and pressure - 273 K and 1 ATM) which are not the conditions you are under in the school's lab. This is one of the reasons for it being different. According to your literature value, the reaction is endothermic (∆H > 0 = endothermic) meaning that is absorbs energy from the surroundings but there isn't much energy to absorb because the calorimeter acts as an insulator and the reaction is (theoretically) not in contact with the surroundings.
  9. Hedron123

    Chemistry HL/SL help

    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 can't go redo my lab either so w/e data i have is what I have to use, so should i just stick with my values and my giant % error and then explain it in my evaluation? Because I'm pretty sure my calorimeter was iffy, and tons of gas (with the energy) escaped it quite easily which could really mess with my energy values right? It's not surprising that the results diverge from the real value. There are two factors that are determining. First, the literature values for delta H you're using are for the reaction at STP (this is not true under lab conditions). Second, the reactions are done inside a calorimeter which prevents most (if not all) of the heat loss to the surroundings. This a MAJOR error of the experiment since despite the accuracy of the method you're using, there will always be heat loss and so the result for delta H will not be 100% certain.
  10. Hedron123

    Chemistry HL/SL help

    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. Hedron123

    Chemistry HL/SL help

    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 units of ∆H will depend upon the heat capacity you use: specific heat capacity will result in joules per units of mass while molar heat capacity will result in joules per moles.
  12. Hedron123

    Calcium Carbonate lab design

    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. Hedron123

    Calcium Carbonate lab design

    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 help you solve your problem. Don't forget to repeat the test to minimise experimental errors. Regards. P.S. You might have to tear the paper into pieces to increase surface area and place the container with the reaction in a water bath to increase speed. (The problem with the water bath is that you won't know at what temperature will CO2 be - an intermediate temperature between the water's temperature and room temperature - so try it first without the water bath and if nothing happens do use it).
  14. Hedron123

    Chemistry HL/SL help

    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 experiment.
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