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Chemistry: Post your doubts here!

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For ethanoic acid in the -COOH structure the electron density is shifted towards the doubly bonded oxygen atom and away from the hydrogen atom due to it's high electronegativity. This results in the weakening of the -OH bond so that hydrogen can be lost more easily to the water molecules.
In ethanol there is no extra electron withdrawing atom as a result the -OH bond is much stronger and making H less available to the H2O molecules. So ethanoic acid is a much stronger acid.
is this question really in as level??
 
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is this question really in as level??

Well....Mid-territory perhaps, it's AS level if you consider it an application question since you do know how high electronegativity and bonding tends to effect a molecule but it's easier to answer if you have A2 knowledge, but that's true for a lot of cases.
 
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Well....Mid-territory perhaps, it's AS level if you consider it an application question since you do know how high electronegativity and bonding tends to effect a molecule but it's easier to answer if you have A2 knowledge, but that's true for a lot of cases.
how should i calculate maximun error in the volume run from burette recorded in any titration ?
like if i have maximum error of burette is +-0.05 cm^3
or Pipettes are calibrated to +-0.06cm^3
Please reply
Thanks In Advance:)
 
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Q: why is HCl a stronger acid in aqueous ammonia then in water?

its a MCQ, and the answer is that pH of ammonia is higher than water..but how does this relate?

Ammonia is a stronger base than water (the higher pH is a reflection) and tends to react more easily with HCl. This causes HCl to dissociate more in presence of NH3 than H2O. The strength of an acid is based on it's extent of dissociation and as HCl dissociates more in case of NH3 it is a stronger acid in NH3.
(I won't delve into bonding considering it's an AS question, but if you have questions, do ask.)
 
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how should i calculate maximun error in the volume run from burette recorded in any titration ?
like if i have maximum error of burette is +-0.05 cm^3
or Pipettes are calibrated to +-0.06cm^3
Please reply
Thanks In Advance:)

Any burette reading has an error of +/- .05, while taking readings during titration you have to take the initial and final reading, so you're taking the reading twice each with an error of +/- .05
So the max error is .05*2 = +/- .1 cm^3
for titration you usually consider the error in the burette reading.
 
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why you have multiplied 0.5 by 2?

Because you have to take the reading twice, initial and final. When ever you add or subtract the volumes the uncertainty is always added.
In this case the certainty is +/- .05 for each reading, so you have to add the two values for max. error.
.05 +.05 = .1 cm^3
Multiplying by 2 would be the same.
 
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Because you have to take the reading twice, initial and final. When ever you add or subtract the volumes the uncertainty is always added.
In this case the certainty is +/- .05 for each reading, so you have to add the two values for max. error.
.05 +.05 = .1 cm^3
Multiplying by 2 would be the same.
yeah i have understand that thats why i have delted that post?
Anyway thanks,and May Allah Bless You
 
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Well....Mid-territory perhaps, it's AS level if you consider it an application question since you do know how high electronegativity and bonding tends to effect a molecule but it's easier to answer if you have A2 knowledge, but that's true for a lot of cases.
correct answer is A tell me why
 
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N triple bond with N is 944 in my data booklet from sy 2014 are u sure that triple bond is 994 not 944???

My booklet is from sy 2012 and mine says 994. Since your booklet has a different value your answer won't match, maybe pick the closest value for the answer, which will be A. You'll get the mark anyway considering none of the answer match the actual.
 
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Question 1 2003 october

there are 3 x 10^-21 moles of amino acids. we need the molecules that this amount of amino acids contains.
1 mole = 6.02 x 10^23 particles (whether its atoms or molecules)
so 3 x 10^-21 moles would be 1806 molecules because we multiply 6.02 x 10^23 by 3 x 10^-21.
1806 is closest to the answer 1800.
do not get confused because they gave you Mr as 200, that's extra information designed to distract you and confuse you.

Question 3 2004 may june

2NaN3 ----> 3N2 + 2Na there are 50 grams of NaN3 - from this we can find the moles

moles = mass/Mr ------> 50/ 23 +(3 x 14) = 0.77 moles of NaN3
now, let's look at the mole ratio of NaN3 to N2 from the equation (big yellow numbers)------>

NaN3 : N2
2 : 3
0.77 : x ------------> find x = (0.77 x 3)/ 2 = 1.155 moles of N2
volume of N2, use the equation ----> moles (only for gases)= volume/24

so, volume = 1.155 moles x 24 =27.7 dm^-3
the answer is C.

Hope this helped, sorry I couldn't help with the other questions.
Please keep in mind, next time give links to the questions, makes it much easier.

Thankyou so much Oliveme for taking your time and replying to my post. Ahaa that's why no one was replying to my post :/ sorry for the extra trouble.
Can you please explain me this one?
Untitled.jpg
 
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10 cm meth reacts with 30 cm O2.. which produces 10 cm So2 and CO2.. so in the end u have 30 oxygen remaining + 20 cm from the extra products = 50 cm.. and H2O can be neglected since its liquid at room temperature..
Aha i got it ! thankyou so much for the help !! :)
 
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U guys..if you love Allah pleasee Help me :(( i cant solve pp1 Chem at all. Idk y and idk how to do it. I am so stressed idk wat to do.. i wasted alors of time just for solving one question. Most of the questions dont make any sense to me especially the ones with structural formulas. I am so worried I ended up crying cuz i cant even solve one paper :(
 
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U guys..if you love Allah pleasee Help me :(( i cant solve pp1 Chem at all. Idk y and idk how to do it. I am so stressed idk wat to do.. i wasted alors of time just for solving one question. Most of the questions dont make any sense to me especially the ones with structural formulas. I am so worried I ended up crying cuz i cant even solve one paper :(
Sister, there is no easy way other than solving more and more papers, u myt be slow now, but soon u will be faster...keep solving again and again..and make sure u read the book properly, keep the book beside you, when you are stuck read the book on that topic and make sure u understand it. May Allah help you succeed..! Don't waste time.
 
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