HSC 2012-2015 Chemistry Marathon (archive) (1 Viewer)

[BlueGas]

re: HSC Chemistry Marathon Archive

So I can't use barium nitrate to test for phosphates?

 

[BlueGas]

re: HSC Chemistry Marathon Archive

It says this on Easychem: "NH4 followed by Ba(NO3)2: White"

 

[leehuan]

re: HSC Chemistry Marathon Archive

As for the white, maybe I'm a bit wrong about my pale blue observation. Stick to the easychem colour.
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That there is my test. The reason I said add HNO3 was to remove any traces of carbonate ions which can always mess up anion tests. Note that I did use Ba(NO3)2 as my source of Ba(2+)

NH4 is not a compound. I think easychem had a typo. But I did use NH3 because the solution does have to be basified.

 

[BlueGas]

re: HSC Chemistry Marathon Archive

As for the white, maybe I'm a bit wrong about my pale blue observation. Stick to the easychem colour.
__________
That there is my test. The reason I said add HNO3 was to remove any traces of carbonate ions which can always mess up anion tests. Note that I did use Ba(NO3)2 as my source of Ba(2+)

NH4 is not a compound. I think easychem had a typo. But I did use NH3 because the solution does have to be basified.

So to test for phosphates:
- Add ammonia to basify the solution
- Add barium nitrate
- White precipitate forms?

What's the reaction?

 

[leehuan]

re: HSC Chemistry Marathon Archive

Net ionic equation

3 Ba(2+) + 2 PO4(3-) -> Ba3(PO4)2(s)
Ba3(PO4)2 is evidently white, yes.
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An example of a full equation COULD be...

3 Ba(NO3)2(aq) + 2 Na3PO4(aq) -> 6 NaNO3(aq) + Ba3(PO4)2

 

[BlueGas]

re: HSC Chemistry Marathon Archive

What are the answers for these two questions, and why?

 

[Mr_Kap]

re: HSC Chemistry Marathon Archive

idk.

Maybe A for the first one.

and

Maybe B for the second one.

 

[Drsoccerball]

re: HSC Chemistry Marathon Archive

First one is B
Second one is A

 

[BlueGas]

re: HSC Chemistry Marathon Archive

First one is B
Second one is A

Why?

 

[Drsoccerball]

re: HSC Chemistry Marathon Archive

Why?

Reduction is gain in electrons if calculated the oxidation state decreases for hydrogen therefore its being reduced.

It lies outside of the zone of stability and is also heavier than lead therefore its reactive and since the only option that has reactive in it is A its A.

 

[Mr_Kap]

re: HSC Chemistry Marathon Archive

Reduction is gain in electrons if calculated the oxidation state decreases for hydrogen therefore its being reduced.

It lies outside of the zone of stability and is also heavier than lead therefore its reactive and since the only option that has reactive in it is A its A.

How does the oxidation state change for hydrogen?

 

[Ekman]

re: HSC Chemistry Marathon Archive

What are the answers for these two questions, and why?

First one is A because the oxidation state for oxygen and hydrogen never changes, they will always remain as -2 and +1 respectively. Also carbon did oxidise if you look at the oxidation states

Second one is C because element 118 fits in the final group on the periodic table, hence its a noble gas which is inert.

 

[Mr_Kap]

re: HSC Chemistry Marathon Archive

First one is A because the oxidation state for oxygen and hydrogen never changes, they will always remain as -2 and +1 respectively. Also carbon did oxidise if you look at the oxidation states

Second one is C because element 118 fits in the final group on the periodic table, hence its a noble gas which is inert.

That's what i thought EXACTLY!!! for both of them.

For the second question though, i looked up element 118 on google and it said it was solid, so that's why it was either A or B, and i ended up choosing B as a random guess. So i don't know the answer to the second question.

 

[rand_althor]

re: HSC Chemistry Marathon Archive

That's what i thought EXACTLY!!! for both of them.

For the second question though, i looked up element 118 on google and it said it was solid, so that's why it was either A or B, and i ended up choosing B as a random guess. So i don't know the answer to the second question.

Wikipedia says its element category is "unknown, but probably a noble gas", and that its phase being solid is only a prediction. I'd still go with C.

 

[BlueGas]

re: HSC Chemistry Marathon Archive

First one is A because the oxidation state for oxygen and hydrogen never changes, they will always remain as -2 and +1 respectively. Also carbon did oxidise if you look at the oxidation states

Second one is C because element 118 fits in the final group on the periodic table, hence its a noble gas which is inert.

Excuse me if this is a stupid question but how would I know if carbon is oxidized?

As in how was the oxidation state for carbon changed?

 

[rand_althor]

re: HSC Chemistry Marathon Archive

Excuse me if this is a stupid question but how would I know if carbon is oxidized?

The oxidation state of C in CH4 is: x + 4 = 0 -> x = -4. The oxidation state of C in CO2 is: x + (-4) = 0 -> x = +4. The oxidation state of C increased from -4 to +4, so Carbon has been oxidised.

 

[Ekman]

re: HSC Chemistry Marathon Archive

That's what i thought EXACTLY!!! for both of them.

For the second question though, i looked up element 118 on google and it said it was solid, so that's why it was either A or B, and i ended up choosing B as a random guess. So i don't know the answer to the second question.

Under exam conditions I would of still went with c because element 118 is in group 8, and all elements in group 8 are noble gases, hence c.

 

[Mr_Kap]

re: HSC Chemistry Marathon Archive

Under exam conditions I would of still went with c because element 118 is in group 8, and all elements in group 8 are noble gases, hence c.

I would have gone with C as well (because I ain't got internet in the HSC), and I actually posted C before I looked it up and edited my post.

 

[BlueGas]

re: HSC Chemistry Marathon Archive

Can anyone identify the filtrates and the precipitates?

 

[Drsoccerball]

re: HSC Chemistry Marathon Archive

Can anyone identify the filtrates and the precipitates?

Precipitate 1 : lead chloride
and you dont need to know the filtrates