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Chemistry: Post your doubts here!

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Solve this question please
A student calculated the standard enthalpy change of formation of ethane, C2H6, using a method based on standard enthalpy changes of combustion.
He used correct values for the standard enthalpy change of combustion of ethane (–1560 kJ mol–1) and hydrogen (–286 kJ mol–1) but he used an incorrect value for the standard enthalpy change of combustion of carbon. He then performed his calculation correctly. His final answer was –158 kJ mol–1.
What did he use for the standard enthalpy change of combustion of carbon?
A –1432 kJ mol–1
B –860 kJ mol–1
C –430 kJ mol–1
D –272 kJ mol–1
Ans is C...i cant get the right answer :/
 
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B is ur answer bcz in A and D there are no poles bcz similar elements on both sides of the compound cancel the polarity of the compound. with those options eliminated, C has O and Cl at the two oppoite sides while B has Cl and H. the greater difference of electronegativity or polraity is between H and Cl so B is the sure answer.
 
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B is ur answer bcz in A and D there are no poles bcz similar elements on both sides of the compound cancel the polarity of the compound. with those options eliminated, C has O and Cl at the two oppoite sides while B has Cl and H. the greater difference of electronegativity or polraity is between H and Cl so B is the sure answer.
Hay there,

Thanks for helping out, but its still not solved :(
Option B has H and H on both sides.. (not H and Cl, if that would been, then it would be a sure answer, but its not like that)
Option C has Cl and Cl on both sides..


And I'll be grateful if you explain the question number 10 of the same paper. Why its answer is C, whats wrong in B ??\

Regards
 
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And I'll be grateful if you explain the question number 10 of the same paper. Why its answer is C, whats wrong in B ??
Regards


Because there is always some activation energy required. Option (C) has the smallest activation energy, which is the leap in the curve after the initial straight line. For option (B), the quick initial leap represents the activation energy. Hope this helps.
 
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oo my no ...thts a tough one -__- :/


Hot KMnO4 causes splitting of double bond (-C=C-), so the ring breaks where the double bond occurs. Each carbon involved in the double bond becomes attached to an oxygen by a double bond (-C=O and O=C-), making it an aldehyde or ketone. If it's an aldehyde group, it is further oxidised to a carboxylic acid.
Apply this to each of the structures.
1. You get two compounds, both have a ketone group and an acid group.
2. Only one double bond, so you don't actually get two organic products, so B is not possible.
3. You get one compound which is a di-oic acid and another compound which is a di-ketone. So 3 doesn't work either.
 
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guys please there is oct nov 2004 q10 please abt eqbm pressure how to solve this please i need detail much appreiiciated i try to figure but how plesse help
 
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guys please there is oct nov 2004 q10 please abt eqbm pressure how to solve this please i need detail much appreiiciated i try to figure but how plesse help

Draw an ICE chart, i.e. initial concentration, change in concentration and equilibrium concentration for all the reactants and products.
 
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Can someone please please help me with question 1 of this paper. Thanking you in advance
 

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  • 9701_w03_qp_1.pdf
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question no21 may june 2002
question no 2 may june 2002
question no 8 may june 20025
question no 3 ,8,9,10 oct nov 2002
question no 3,20,28 oct nov 2003
may june 2003 question no 8,20,32
question no 40 oct nov 2003
question no 5 may june 2004
quesyion no 5,8,18,28,35,39 may june 2004
oct nov 2005 queston no 2, 5,23,28,31
oct nov 2004 question 20
may june 2005 quetion 11,18
mayjune 06 q10,30
oct/nov questions 06 4,9,11,21
may june 07 questions 5,26,34,40
oct nov 07 questons 33
winter 08 2,8,30
winter 09 21,28,31
summer 10 qp 11 question no 4, 27
winter 10 questions 7 ,8 any one please answer thest ones
 
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DO NOT USE INAPPROPRIATE LANGUAGE.
just <content removed> help me in this moles u <content removed> cmon the eqbm please
 
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