chemistry multiple choice questions that I didn't get

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dasda
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so here are 6 multiple questions that I got wrong when doing a practice paperName:  number 1.jpg
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Size:  18.8 KBName:  number 2.jpg
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Size:  45.0 KBName:  number 6.jpg
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Last edited by dasda; 11 months ago
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dasda
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charco
Pigster
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Pigster
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Quite what do you want either of us to do?
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dasda
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(Original post by Pigster)
Quite what do you want either of us to do?
since you guys are the chemistry masters, I would like you to enlighten me on how to do these questions. for the first one, I know the answer can't be c and D because they are not stable. B is not in its simplest form. so I am left with A. but why is the answer A.
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gd99
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I can only see Qs 5 and 10 on my device. Q5 would be A, because A=an actual formula of a molecule, B is also correct, but is not empirical as you could divide through the formula by 2. C would be a radical so unstable, and likewise NH2 is not a stable molecule.

In Q10, D is probably right as the CN- ion would react at the C-Br bond on the ring, giving a product with CN instead of the Br. B is potentially plausible, but it is more likely that the ketone would be reduced than an addition-elimination would take place. A would not change the product, and C is not possible as there are no sites for elimination.
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dasda
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(Original post by gd99)
I can only see Qs 5 and 10 on my device. Q5 would be A, because A=an actual formula of a molecule, B is also correct, but is not empirical as you could divide through the formula by 2. C would be a radical so unstable, and likewise NH2 is not a stable molecule.

In Q10, D is probably right as the CN- ion would react at the C-Br bond on the ring, giving a product with CN instead of the Br. B is potentially plausible, but it is more likely that the ketone would be reduced than an addition-elimination would take place. A would not change the product, and C is not possible as there are no sites for elimination.
so A would be methanal. and its the empiracal formula since you cant simplify it any further
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dasda
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(Original post by Pigster)
Quite what do you want either of us to do?
How would you do number 29. I understand the anticlockwise rule means that Iodide ions would be oxidised and s2082- would be reduced. What does the e nought value have to do with the ions catalysing the reaction
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(Original post by dasda)
How would you do number 29. I understand the anticlockwise rule means that Iodide ions would be oxidised and s2082- would be reduced. What does the e nought value have to do with the ions catalysing the reaction
S2O82- doesn't like reacting with I- since they're both -ve ions and will repel each other.

The reaction can be catalysed by the addition of a +ve ion which can switch between two oxidation numbers.

You can add Co3+ OR Co2+. If you add 3+ then it can be reduced by I- (forming I2) the 2+ that is formed is then oxidised by the S2O82- (back to 3+ and forming SO42-). If you added 2+, the same two processes happen (oxidation by S2O82- and reduction of the 3+ formed back to 2+). It doesn't matter which one you added. Ditto Fe2+ and 3+. The reason either work is that the E value for the 2+/3+ reaction is between those for S2O82- and I-. The anticlockwise rule means that S2O82- can oxiside 2+ to 3+ and the I- can reduce the 3+ back to 2+.

Cr2+ is the correct answer since S2O82- can oxidise it to 3+ but the I- cannot reduce it back again to 2+ (so sayeth the anticlockwise rule).
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dasda
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(Original post by Pigster)
S2O82- doesn't like reacting with I- since they're both -ve ions and will repel each other.

The reaction can be catalysed by the addition of a +ve ion which can switch between two oxidation numbers.

You can add Co3+ OR Co2+. If you add 3+ then it can be reduced by I- (forming I2) the 2+ that is formed is then oxidised by the S2O82- (back to 3+ and forming SO42-). If you added 2+, the same two processes happen (oxidation by S2O82- and reduction of the 3+ formed back to 2+). It doesn't matter which one you added. Ditto Fe2+ and 3+. The reason either work is that the E value for the 2+/3+ reaction is between those for S2O82- and I-. The anticlockwise rule means that S2O82- can oxiside 2+ to 3+ and the I- can reduce the 3+ back to 2+.

Cr2+ is the correct answer since S2O82- can oxidise it to 3+ but the I- cannot reduce it back again to 2+ (so sayeth the anticlockwise rule).
This much thinking for 1 marker.
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dasda
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(Original post by Pigster)
S2O82- doesn't like reacting with I- since they're both -ve ions and will repel each other.

The reaction can be catalysed by the addition of a +ve ion which can switch between two oxidation numbers.

You can add Co3+ OR Co2+. If you add 3+ then it can be reduced by I- (forming I2) the 2+ that is formed is then oxidised by the S2O82- (back to 3+ and forming SO42-). If you added 2+, the same two processes happen (oxidation by S2O82- and reduction of the 3+ formed back to 2+). It doesn't matter which one you added. Ditto Fe2+ and 3+. The reason either work is that the E value for the 2+/3+ reaction is between those for S2O82- and I-. The anticlockwise rule means that S2O82- can oxiside 2+ to 3+ and the I- can reduce the 3+ back to 2+.

Cr2+ is the correct answer since S2O82- can oxidise it to 3+ but the I- cannot reduce it back again to 2+ (so sayeth the anticlockwise rule).
So let me try to understand this:
if we use the co3 one, I think the reaction would be 2 I- + 2co(3+) forming 2co(2+) + I2
and then this reacts with s2o4. why does this reaction occur. how you would you know it can be oxides by s208 since it contains Co and I2 which have difference e values
Last edited by dasda; 11 months ago
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(Original post by dasda)
So let me try to understand this:
if we use the co3 one, I think the reaction would be 2 I- + 2co(3+) forming 2co(2+) + I2
and then this reacts with s2o4. why does this reaction occur. how you would you know it can be oxides by s208 since it contains Co and I2 which have difference e values
The I2 that just formed would be oxidised by the S2O82- back to I-, which is where you started.

It is a 1 marker since you should know that redox systems with values in between an oxidant and reducing agent can act as a catalyst. I expect. I don't teach AQA (I think it is).
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(Original post by Pigster)
The I2 that just formed would be oxidised by the S2O82- back to I-, which is where you started.

It is a 1 marker since you should know that redox systems with values in between an oxidant and reducing agent can act as a catalyst. I expect. I don't teach AQA (I think it is).
Its my first time seeing a question of its type plus not so so great teachers.
Last edited by dasda; 11 months ago
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(Original post by Pigster)
The I2 that just formed would be oxidised by the S2O82- back to I-, which is where you started.

It is a 1 marker since you should know that redox systems with values in between an oxidant and reducing agent can act as a catalyst. I expect. I don't teach AQA (I think it is).
How would you do number 12. I have discounted d since it is a product. Am I supposed to solve the question algebraically or use an arbitrary value of 1 mol dm-3 to start off with
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(Original post by dasda)
How would you do number 12. I have discounted d since it is a product. Am I supposed to solve the question algebraically or use an arbitrary value of 1 mol dm-3 to start off with
Considering the provided rate equation in answer A, if you doubled [W] AND [X] AND [Z], what would happen to the rate?
Repeat that process for B and C.

(you're right it can't be D, since Z is the product)
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