Msj Chem

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    May 2016
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  1. I have worksheets for this topic on my website. Maybe they will be helpful.
  2. From May 2016 subject report (Copyright IB). Component grade boundaries Grade: 1 2 3 4 5 6 7 Mark range: 0 - 3 4 - 6 7 - 10 11 - 13 14 - 16 17 - 19 20 - 24
  3. You can find two videos on pH curves on my website here: Syllabus details can be found here, (see 18.3 pH curves).
  4. 10.0 mol of C2H3Cl would require 25 mol of O2 to react completely because of the 2:5 molar ratio. 10.0 mol of O2 requires 4 mol of C2H3Cl to react and seeing as we have 10.0 mol, that makes O2 the limiting reactant. The ratio of the limiting reactant O2 to H2O is 5:2, therefore 10.0 mol of O2 will produce 4 mol of H2O.
  5. It all comes down to personal preference, I prefer the Pearson chemistry textbook to the Oxford textbook mainly due to the fact that it comes in separate SL and HL versions and has a good range of practice exam questions. The study guide is good for revising but lacks the detail of a textbook. As far as I know there isn't another study guide available for IB chemistry. If you are looking for chemistry video tutorials, then I have videos that cover almost the whole syllabus. A lot of videos also have worksheets to accompany them.
  6. Each option is assigned 15 hours at SL with an additional 10 hours for HL, giving a total of 25 hours teaching time. So what you are proposing is an extra 75 hours of teaching time to cover all 4 options. I can't speak for your teacher but I don't think that it would be possible for your teacher to teach you all 4 options when there is no requirement to do so. Of course you are free to study all 4 options and some further organic chemistry yourself but that depends on how much free time you have. I have videos that cover almost all of the option topics here:
  7. There are two ways to calculate this. One way is to use the Kw to find the [H+], and then use pH = -log [H+] The second way is to use pOH = -log[OH-] There is a mistake in the mark scheme, it should read -log (3.98 x10-3) = 2.4,
  8. "I don't quite understand the reactions involving ammonia solution, like are both NH4+ and NH3 are present there?" HCl reacts with NH3 to produce NH4Cl HCl + NH3 ==> NH4Cl The NH3 is the excess reactant which is why you have some left over. The NH4+ comes from the dissociation of the salt NH4Cl (assume that the salt fully dissociates). "Why do we have to know the final amount of them both and where did it come from?" You have equal final concentrations of NH3 and NH4+ which acts as a buffer solution. Subtract the amount in mol that reacted from the initial amount in mol. Use the equation C=n/V to get the concentration. Use the Henderson-Hasselbach equation (like in the previous example you posted) to calculate the pH and pOH.
  9. The reaction is: CH3COOH + NaOH ==> NaCH3COO + H2O Use n=CV to calculate amount in mol of each. CH3COOH: n=0.05 mol NaOH n=0.03 mol Now make an ICE box: CH3COOH + NaOH ==> NaCH3COO + H2O initial 0.05 0.03 0 0 eqm 0.02 0 0.03 0.03 This gives 0.02 mol of CH3COOH and 0.03 mol of CH3COO- If you use C=n/V you get 0.200 moldm-3 and 0.300 moldm-3 Then use the Henderson-Hasselbach equation to calculate the pH. But be aware that the use of this equation is only necessary in option B and D (not topic 18).
  10. Convert each mass to amount in mol using the equation n=m/M. Then multiply each of the values by the Avogadro constant 6.02 x 10^23. This will give you the number of molecules of each. You'll find that the compound with the highest molar mass has the lowest number of molecules for the same mass (1g) and vice versa. The answer is A as it has the lowest molar mass and therefore the greatest number of molecules.
  11. By doubling the volume the pressure halves (according to Boyle's law). Then by doubling the temperature, the pressure doubles (according to Gay Lussac's law). So you are halving the pressure and then doubling it so the final pressure is the same as the initial pressure.
  12. You can think of evaporated water molecules (as in steam) as existing as individual water molecules. There are still intermolecular forces between the molecules but their kinetic energy is great enough to overcome these forces. When they lose energy, they will condense back to liquid water. This is why condensation is exothermic.
  13. With regards to question 2, when water evaporates, the water molecules gain enough energy to overcome the intermolecular forces (hydrogen bonds, dipole-dipole attractions and London dispersion forces) between the individual water molecules.
  14. Maybe this video will help:
  15. I have a full range of videos on my website. The majority, but not all, of the videos have worksheets with solutions. If you can get hold of the May 2016 chemistry exams, it would be helpful. I believe they can be ordered directly from the IB web store here: