Need some help with chem q's (1 Viewer)

[hs17]

Can somebody pls explain how to do these?


18) Ans is D but I got B


15) I got D but and is C


Ans is C

 

[hs17]

^

 

[CM_Tutor]

Question 18
Since the combustion heated 200.0 g of water by 19.9 ^oC, we know that

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We also know that the heat of combustion is 3329 kJ mol^-1 which means that the enthalpy of combustion is . From this, we can determine the chemical amount of fuel burned, and thus its molar mass:

​

And, since

• 1-butanol is CH₃CH₂CH₂CH₂OH, with M(1-butanol) ---> NOT "A," does not match the calculated M(fuel).
• butane is CH₃CH₂CH₂CH₃, with M(butane) ---> NOT "B," does not match the calculated M(fuel).
• 1-pentanol is CH₃CH₂CH₂CH₂CH₂OH with M(1-pentanol) ---> "C," does match the calculated M(fuel).
• pentane is CH₃CH₂CH₂CH₂CH₃, with M(pentane) ---> NOT "D," does not match the calculated M(fuel).

I think the answer is C.

EDIT: I just noticed the table in the question! Matching the calculated molar mass of 88.0 to the table, the answer is still 1-pentanol and so is still C! However, it would be D, matching the answer quoted by the OP, if the answers were meant to be in the same order as the table, and thus have 1-pentanol last.

 

[CM_Tutor]

Question 15:

pK_a(ethanoic acid) = 4.76 at 25 ^oC

So, pK_b(ethanoate ion) = pK_w - pK_a(ethanoic acid) = 14.00 - 4.76 = 9.24

Constructing the usual ICE table with being the [OH^-] at equilibrium, you should get

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The equilibrium clearly lies to the left with such a small equilibrium constant, so the assumption that and thus that is justified, and so:

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The answer is C.

 

[CM_Tutor]

Question 14:

The overall equation for the use of glucose requires combining reactions 2 and 3 to eliminate the intermediate (ethanol). When this is done, you see that the overall equation is reaction 1 backwards, so the CO₂ produced in reactions 2 and 3 exactly matches the CO₂ consumed in reaction 1. Hence, the answer is (C).

 

[hs17]

WOW thank youuu!!!

 

[jimmysmith560]

View attachment 32804@jazz519 bless me with your detailed answers

The answer is (A) - Add hydrochloric acid, then ammonia

A precipitate with HCl indicates either lead(II) or silver ions. Ammonia can then be added to produce different coloured solutions according to the cations present.

I hope this helps!

 

[jimmysmith560]

So why are the other answers wrong

As established in the question, the identity of an ion can be determined by testing whether it forms precipitates and if it does, by noting the colour of the precipitate.

If I understand this correctly, ammonia needs to be added for confirmation. The other elements don't allow for this in this particular case, therefore making the other options incorrect.

 

[CM_Tutor]

I think they mean A, but it is a badly written question.

Consider option B. If I had copper(II) carbonate, I could add HCl and see it dissolve to form a blue solution, and I would get a pale blue precipitate when enough sodium hydroxide was added. If I had a white solid that was barium carbonate or magnesium oxide, both would dissolve in HCl but magnesium hydroxide would form much more readily as sodium hydroxide was added than would have barium hydroxide.

I think it is based on a single method of cation analysis that is in that book, but the syllabus does not mandate a particular method and there are multiple approaches that are practical.