Higher Chemistry 2018-19

Would be great if someone could explain how 16. is B and 17. is B as well. Had no clue how to even start these questions.

Would be great if someone could explain how 16. is B and 17. is B as well. Had no clue how to even start these questions.

Hi Chemistry people. Can someone please help with this question..? I've done several like these and know how to do them but this one isn't working out. The answer is A.

Sodium and bromine each have a valency of 1, and sulfate has a valency of 2 (since its charge is 2-). That means for each mole of bromide there is one mole of sodium, and for each mole of sulfate, there are 2 moles of sodium.

Given that there are 4 moles of bromide, there are 4 moles of sodium in sodium bromide. There are 10 moles of sodium in total, so there are 6 moles of sodium in sodium sulfate. Since there are 2 moles of sodium for every mole of sulfate, there is half a mole of sulfate for every mole of sodium. Since there are 6 moles of sodium in sodium sulfate, there are 3 moles of sulfate.

For 16 you have to find the limiting reactant. There are 5 moles of N2O4 for every 4 moles of CH3NHNH2, so you need more N2O4, so it's the limiting reactant. It tells you that the enthalpy change for 5 moles of N2O4 is -5116 kJ. You divide that by 5 to get an enthalpy change of -1023 kJ per mole of N2O4. You then multiply that by 2 to get enthalpy change of -2046 kJ. That means that 2046 kJ of energy is released.

For 17, the enthalpy of combustion of a substance is the enthalpy change when one mole of the substance burns completely in oxygen. The energy released when 2 moles of Al are burned completely in oxygen is 1670 kJ, so the energy released when 1 mole of Al is burned completely in oxygen is 835 kJ. Since combustion is exothermic and the energy is being released, the enthalpy change is negative, so it is -835 kJ mol^-1.

I struggled with 16 at first too, but managed to get the answer, I know someone's already answered but thought I'd share anyway... I'll have a go at 17 later tonight

Hi Chemistry people. Can someone please help with this question..? I've done several like these and know how to do them but this one isn't working out. The answer is A.

What does everyone think the open ended questions will be on? Unit-wise

I'll be fine with any unit apart from the Researching Chemistry unit.

Ye i hate thoose open ended questions where it gives you apparutus and chemicals and tells you to investigate something

Hello everyone! How is your chemistry studying going? could someone please if they wouldn’t mind help me with the 2017 paper 2 queatjon 2B paper. I don’t understand why in the answers they have said -9.6 as I thought it is 9.6. Thanks a lot!!

For the forward reaction, the enthalpy of the products is lower than the enthalpy of the reactants, so the enthalpy decreases from the reactants to the products, so the enthalpy change is negative.

Would be great if someone is able to tell me how the answer is D?

Also, is it true that see the cost numeracy questions are they taken out of the higher now?

The chromatogram is based on polarity of molecules so look for the number of OH groups because they're what make the molecule polar since apart from OH groups all 4 molecules are identical. A has the most, D has the second most, C has the third most and B has the fewest. Looking at the chromatogram, the molecule with the highest concentration is the one with the second highest retention time, and therefore the second highest polarity, so you'd go with D because it has the second greatest number of OH groups.