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March 29, 2024, 12:46:29 am

Author Topic: VCE Chemistry Question Thread  (Read 2313163 times)  Share 

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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #120 on: January 14, 2014, 01:53:03 pm »
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Thanks for the advice! Also, about the chemical formula of sodium acetate:

Which one is better: NaCH3COO or CH3COONa? Thanks!

The secondone is preferred because the sodium ion is a cation and the negative charge in the acetate ion is on tje oxygen. Thus the plus and minus charges are put together in the formula
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PsychoT

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Re: VCE Chemistry Question Thread
« Reply #121 on: January 14, 2014, 03:30:57 pm »
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Anyone got a good online resource to learn about the work covered in AoS 1 (Analytical Chemistry) and AoS2 2 (Organic Reaction Pathways)?

Don't get my books till the 23rd, want to get a little headstart.

Cheers.
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BLACKCATT

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Re: VCE Chemistry Question Thread
« Reply #122 on: January 14, 2014, 03:34:57 pm »
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How do you work out the reaction between potassium dichromate (k2cr2o7) and tin chloride(sncl2)?

eagles

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Re: VCE Chemistry Question Thread
« Reply #123 on: January 14, 2014, 03:48:14 pm »
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K2Cr2O7 (aq) + SnCl2 (aq) -> 2KCl (aq) + SnCr2O7 (s)

1. "Swap" the partners of each reactant with the other.
2. Balance equation.
3. Include states.

**I'm not sure about the solubility of tin (II) dichromate, if anyone can confirm it'll be great!
« Last Edit: January 14, 2014, 03:51:40 pm by eagles »

BLACKCATT

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Re: VCE Chemistry Question Thread
« Reply #124 on: January 14, 2014, 03:56:29 pm »
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Ohhh right thanks.
What if the potassium was omitted? Whats the reaction between the dichromate ion(cr2o7^-2, thats a negative charge) and sncl2?

Daenerys Targaryen

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Re: VCE Chemistry Question Thread
« Reply #125 on: January 14, 2014, 04:11:12 pm »
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You would have 2Cl- on the right hand side. (the K also omitted)
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eagles

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Re: VCE Chemistry Question Thread
« Reply #126 on: January 15, 2014, 08:10:16 am »
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Washing soda is added to hard water to allow soap to lather.
A certain brand of washing soda contains partially hydrated sodium carbonate solid.
A 0.300 g sample completely reacts with 20.0 mL of 0.250 M hydrochloride acid.

The answer is:

Na2CO3 (s) + 2HCl (aq) -> 2NaCl (aq) + H2O (l) + CO2 (g)

Why does the reaction not take into account the "partially hydrated" term, in that the chemical formula for sodium carbonate be Na2CO3.xH2O (ss) thanks!

Daenerys Targaryen

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Re: VCE Chemistry Question Thread
« Reply #127 on: January 15, 2014, 10:59:32 am »
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Washing soda is added to hard water to allow soap to lather.
A certain brand of washing soda contains partially hydrated sodium carbonate solid.
A 0.300 g sample completely reacts with 20.0 mL of 0.250 M hydrochloride acid.

The answer is:

Na2CO3 (s) + 2HCl (aq) -> 2NaCl (aq) + H2O (l) + CO2 (g)

Why does the reaction not take into account the "partially hydrated" term, in that the chemical formula for sodium carbonate be Na2CO3.xH2O (ss) thanks!

I think it is because the water on the carbonate wouldn't react with HCl and thus just appear as water on the right hand side. Thus not really needing to take it into account
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #128 on: January 15, 2014, 12:10:43 pm »
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Washing soda is added to hard water to allow soap to lather.
A certain brand of washing soda contains partially hydrated sodium carbonate solid.
A 0.300 g sample completely reacts with 20.0 mL of 0.250 M hydrochloride acid.

The answer is:

Na2CO3 (s) + 2HCl (aq) -> 2NaCl (aq) + H2O (l) + CO2 (g)

Why does the reaction not take into account the "partially hydrated" term, in that the chemical formula for sodium carbonate be Na2CO3.xH2O (ss) thanks!

I think it is because the water on the carbonate wouldn't react with HCl and thus just appear as water on the right hand side. Thus not really needing to take it into account

I was initially thinking that too, but then, having not done the calculations, I thought that perhaps the question wanted us to find the water of crystallisation (the x term in the formula given). Then I saw that the numbers were too nice for that to happen.

What's the question though?
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eagles

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Re: VCE Chemistry Question Thread
« Reply #129 on: January 15, 2014, 01:25:31 pm »
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Thanks for your responses!
The question is to write a balanced equation.

Edward Elric

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Re: VCE Chemistry Question Thread
« Reply #130 on: January 15, 2014, 01:55:55 pm »
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Hey Chemistry folks
Can someone help me out with the following Q
1) write the half equation(partial ionic equation) for the oxidation of chlorine gas (Cl2) to Hypochlorous acid (HOCL)

2) the oxidant and the reactant of : 2KMnO4(aq) + 5H2S(aq) + 6HCL(aq) -----> 2MnCl2(aq) + 5S(s) + 2KCl(aq) + 8H2O(l)

3)  write balanced equations for the following redox reactions: Under certain conditions, zinc can react with concentrated nitric acid (HNO3) to form the zinc ion (Zn^2+) the ammonium ion (NH4^+) and water.
Any solutions to the Q, will be much appreciated!

Edward21

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Re: VCE Chemistry Question Thread
« Reply #131 on: January 15, 2014, 07:36:24 pm »
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Hey Chemistry folks
Can someone help me out with the following Q
1) write the half equation(partial ionic equation) for the oxidation of chlorine gas (Cl2) to Hypochlorous acid (HOCL)

2) the oxidant and the reactant of : 2KMnO4(aq) + 5H2S(aq) + 6HCL(aq) -----> 2MnCl2(aq) + 5S(s) + 2KCl(aq) + 8H2O(l)

3)  write balanced equations for the following redox reactions: Under certain conditions, zinc can react with concentrated nitric acid (HNO3) to form the zinc ion (Zn^2+) the ammonium ion (NH4^+) and water.
Any solutions to the Q, will be much appreciated!

1. 2H2O(l) + Cl2(g) --> 2HOCl(aq) + 2H+(aq) + 2e-

2. oxidant and reductant? The oxidant gets reduced, and oxidises something else by taking electrons: [KMnO4] (+7 to +2) (Mn)
Reductant will give away electrons to reduce something and is oxidised: [H2S/S2-] (-2 to 0) (S) I've separated S2- from H2S as it's in the aqueous state..

3. (Zn(s) --> Zn2+(aq) + 2e-)*4
8e- + 9H+(aq) + HNO3(aq) --> NH4+(aq) +3H2O(l)

4Zn(s) + 9H+(aq) + HNO3(aq) --> 4Zn2+(aq) + NH4+(aq) + 3H2O(l)
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Edward Elric

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Re: VCE Chemistry Question Thread
« Reply #132 on: January 15, 2014, 09:26:06 pm »
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1. 2H2O(l) + Cl2(g) --> 2HOCl(aq) + 2H+(aq) + 2e-

2. oxidant and reductant? The oxidant gets reduced, and oxidises something else by taking electrons: [KMnO4] (+7 to +2) (Mn)
Reductant will give away electrons to reduce something and is oxidised: [H2S/S2-] (-2 to 0) (S) I've separated S2- from H2S as it's in the aqueous state..

3. (Zn(s) --> Zn2+(aq) + 2e-)*4
8e- + 9H+(aq) + HNO3(aq) --> NH4+(aq) +3H2O(l)

4Zn(s) + 9H+(aq) + HNO3(aq) --> 4Zn2+(aq) + NH4+(aq) + 3H2O(l)

Thank you so much! Legend! But i Think the oxidation no of 2KMnO4 to 2MnCl2 is from +12 to +2 instead of +7 to +2 :P but either way it doesn't make a difference.
« Last Edit: January 15, 2014, 10:42:36 pm by Edward Elric »

lzxnl

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Re: VCE Chemistry Question Thread
« Reply #133 on: January 16, 2014, 12:24:29 am »
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The manganese in permanganate has oxidation number +7, not +12. There is no substance in existence that has anything in such a high oxidation number. I can't imagine anything losing twelve electrons.
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Edward Elric

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Re: VCE Chemistry Question Thread
« Reply #134 on: January 16, 2014, 03:10:56 am »
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The manganese in permanganate has oxidation number +7, not +12. There is no substance in existence that has anything in such a high oxidation number. I can't imagine anything losing twelve electrons.

Ohh okay, sorry i must have done my oxidation nos incorrectly, cheers :) I think I used the coefficient to find the oxidation no instead of the subscript lol
« Last Edit: January 16, 2014, 03:29:54 am by Edward Elric »