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Chemistry: Post your doubts here!

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When making sparkler fireworks, a mixture of barium nitrate powder with aluminium powder, water and glue is coated onto wires and allowed to dry. At this stage, the following exothermic reaction may occur. 16Al + 3Ba(NO3)2 + 36H2O → 3Ba(OH)2 + 16Al(OH)3 + 6NH3 Which conditions would be best to reduce the rate of this reaction during the drying process, and would also keep the aluminium and barium nitrate unchanged? temperature/K pH A 298 7 B 298 14 C 398 7 D 398 14 https://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s14_qp_11.pdf plz help
to reduce the rate of reaction lower temperature will be used i.e 298 K and to keep the amount of barium nitrate and aluminum constant a lower or neutral pH will be used for if the pH is increased more hydroxides on the right side will be produced resulting in increased amount of aluminium reacting with the nitrates thus decreasing their amounts. to keep the reactants amount constant lower pH will be maintained.
 
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Hello guys..so i found out tht this page is so helpful n the people in here are so f(x)..so i wnt to share something..me n my friends back in my college have a whatsapp group called “Dr. thinker”...most of us in there are A level students..n there are a lot of genius people in there..imagine u ask a question in the group, get a respond within 4 minutes (average)..i literally send my the questions idk how to answer in the group..n i love to read the debate n discussions in the group...so u guys are welcome but bear in mind tht the group is strictly for education purposes, no advertising or whtsoevr..trust me the group will only ring ur phone for something tht will worth your time..n please speak in english..we have engineering n accounting students in there..so basically u can ask anything..be polite n serious..so here’s the link
https://chat.whatsapp.com/GUVCDHkNyvgDSV1uQljswk ..you’re welcome
 
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The two equations are:

CH4 + 2O2 --> CO2 + 2H2O
CH4 + 3/2O2 --> CO + 2H2O

multiply equation 1 by 99 and add to equation 2

99CH4 + 198O2 --> 99CO2 + 198H2O
CH4 + 3/2O2 --> CO + 2H2O
---------------------------------------------------------------------------------------- add
100CH4 + 199(1/2)O2 --> 99CO2 + CO + 200H2O

So, burning 100 moles needs 199(1/2) moles of oxygen


Therefore burning y moles needs 2y - y/200

or expressed as a decimal fraction:

2y - (0.01/2)y

i.e. answer A
Thanks again bro :)
 
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From part b we we know that:
0.04 mol of NaOH reacted with ethanoic acid
mol ratio of NaOH to ethanoic acid in the reaction is 1:1

therefore, there was 0.04 mol of ethanoic acid in the equilibrium state
before the equilibrium state there was 0.1 mol of ethanoic acid
so the other 0.06 mol of ethanoic acid underwent the forward reaction

mole ratio of acid to alcohol is 1:1
0.06 mol of alcohol reacted with the 0.06 mol of acid
this leaves 0.04 mol of alcohol

mole ratio of acid to both products is 1:1
0.06 mol of both products was formed from the 0.06 mol of acid

therefore, table is:
0.04 0.04 0.06 0.06
 
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how can we determine element b and c are in group 2 and 1,but not group 12 and 11? should we refer to data booklet for checking the standard 1st ionisation energies to get the elements and correct group?
 

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how can we determine element b and c are in group 2 and 1,but not group 12 and 11? should we refer to data booklet for checking the standard 1st ionisation energies to get the elements and correct group?

You're looking for a jump in successive ionization energies, which means that all the outermost electrons were removed and the next one is in an inner shell.

B has a jump between two and three, which means that it has two valence electrons, group 2.
C has a jump between one and two, one valence electron, group 1.
 
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You're looking for a jump in successive ionization energies, which means that all the outermost electrons were removed and the next one is in an inner shell.

B has a jump between two and three, which means that it has two valence electrons, group 2.
C has a jump between one and two, one valence electron, group 1.
why can't theybe group 11 and 12?
 
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Can someone help with this winter 2009 paper? Question 7, I got how to get carbon and hydrogen but how do you find the number Of oxygen and the overall structure of the compound ?
 

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Screenshot (12).png Screenshot (13).png
I'm confused with this one. The CH group is connected to an benzene and an oxygen, so both fit the 2.7 ppm and 4.0 pp,m option in the question.
and OH can be any of the two (because its range is anywhere between 0.5-6.0 ppm). How can I know whether that the CH group shows ppm of 2.7 because its an Alkyl next to an aromatic group OR whether it shows ppm of 4.0 because its next to an electronegative atom (Oxygen).

also if there are some good notes or examples on analytical techniques ( NMR especially) pls provide a link to me, I'm confused with it.

thank you!!!
 
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Those studying A2 this year.
How are you guys Keep revising this Reaction of Organic?
If i dont revise for 4-5 days,I will forget the Catalyst or reagent.
Cant Find any summary on Internet like we have of AS reactions.
Like 1 or 2 page summary reactions
 
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Those studying A2 this year.
How are you guys Keep revising this Reaction of Organic?
If i dont revise for 4-5 days,I will forget the Catalyst or reagent.
Cant Find any summary on Internet like we have of AS reactions.
Like 1 or 2 page summary reactions
Which reaction?
I hope this helps
I found this a while ago from somewhere in this forum, full credit goes to that guy. 0bb7821c (1).jpg
 
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Hey guys could you please explain why its not [Zr2(C2O4)4] the mark scheme is saying the answer should be [Zr(C2O4)4]^4-
 

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Hey guys could you please explain why its not [Zr2(C2O4)4] the mark scheme is saying the answer should be [Zr(C2O4)4]^4-
Because the complex you formed contains two Zn^4+ while it says in the question it should include only one.
Don't try to balance out the complex to make the net charge zero.
 
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Can somebody provide me with the link for February March 2019 CIE As level Chemistry, physics and maths question paper.
If somebody have it only downloaded please email me the file.
[email protected]

Thanks.
 
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