# Higher Chemistry 2018-19Watch

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2 weeks ago
#61
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.
(Original post by Lil Impulse)
Attachment 815126 Attachment 815128Would 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.
1
2 weeks ago
#62
(Original post by Lil Impulse)
Attachment 815126 Attachment 815128Would 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.
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

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2 weeks ago
#63
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.
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2 weeks ago
#64
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.
(Original post by cna1806)
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.
1
2 weeks ago
#65
(Original post by sbneelu)
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.
I see now, thank you!
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2 weeks ago
#66
(Original post by sbneelu)
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.
Thank you that does make sense
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2 weeks ago
#67
(Original post by 536458)
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

That also helps thank you
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2 weeks ago
#68
(Original post by cna1806)
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.
Even though it’s answered, I thought I’ll share my working as well!
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2 weeks ago
#69
What does everyone think the open ended questions will be on? Unit-wise
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2 weeks ago
#70
(Original post by Sb1929)
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.
1
2 weeks ago
#71
(Original post by cna1806)
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
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2 weeks ago
#72
(Original post by Lil Impulse)
Ye i hate thoose open ended questions where it gives you apparutus and chemicals and tells you to investigate something
Me too, I try not to waste my time on them... I do all the other questions and go back to those at the end.
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2 weeks ago
#73
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!!
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2 weeks ago
#74
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.
(Original post by Pearls1)
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!!
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2 weeks ago
#75
Would be great if someone is able to tell me how the answer is D?
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2 weeks ago
#76
Oh okay thank you so much!

(Original post by sbneelu)
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.
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2 weeks ago
#77
Also, is it true that see the cost numeracy questions are they taken out of the higher now?
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2 weeks ago
#78
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.
(Original post by Lil Impulse)
Would be great if someone is able to tell me how the answer is D?
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2 weeks ago
#79
My teacher didn't mention anything about that so I'm not sure, sorry.
(Original post by Pearls1)
Also, is it true that see the cost numeracy questions are they taken out of the higher now?
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2 weeks ago
#80
(Original post by sbneelu)
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.
Cheers think I get it now
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