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That question.....................................What was the answer for how's the percentage decomposition affected by increasing temperature[/QUOTE
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That question.....................................What was the answer for how's the percentage decomposition affected by increasing temperature[/QUOTE
this might be good enough for 2 marks. It clearly asked for le chatlier's principle as far as i can remember.I think these were the points
Increase in temp will shift the equilibirum towards the RHS
Because forward reaction and endo
Hence %of HBr decomposition increase
Sorry but no. When names are asked, they are the only option you have.I wrote structural formula of eyhanoic acid and methyl methanaote instead of.their names will it be acceptable?
the rate increasesWhat was the answer for how's the percentage decomposition affected by increasing temperature
Thats fine for 3 marksthis might be good enough for 2 marks. It clearly asked for le chatlier's principle as far as i can remember.
Unbranched hydrocarbons have a higher boiling point. In unbranched hydrocarbons, molecules are closer together compared to branched hydrocarbons. Unbranched hydrocarbons have stronger V.W. forces between them which are harder to overcome compared to branched hydrocarbons, so unbranched hydrocarbons have a higher boiling point. Just my answer. Not sure if good enough for 2 marks.What was the nucleon number for Fe? I forgot xD
Secondly what was the answer to high boiling point of unchained hydrocarbons?
I wrote that unbranched can overlap each other, increasing the molecular forces and bla bla.... lets seeUnbranched hydrocarbons have a higher boiling point. In unbranched hydrocarbons, molecules are closer together compared to branched hydrocarbons. Unbranched hydrocarbons have stronger V.W. forces between them which are harder to overcome compared to branched hydrocarbons, so unbranched hydrocarbons have a higher boiling point. Just my answer. Not sure if good enough for 2 marks.
These were the simple pointsUnbranched hydrocarbons have a higher boiling point. In unbranched hydrocarbons, molecules are closer together compared to branched hydrocarbons. Unbranched hydrocarbons have stronger V.W. forces between them which are harder to overcome compared to branched hydrocarbons, so unbranched hydrocarbons have a higher boiling point. Just my answer. Not sure if good enough for 2 marks.
Equal number of moles of gas present on both sides of equationWhat was the answer for that no need of total pressure ?
58 I guessNucleon number of Fe, anyone?? xD
what did you wrote?What was the answer for that no need of total pressure ?
Hbr bond strength greater than HI due to short bond length and greater overlap of orbitals hence electrons are closer to nucleusWhat did you guys wrote in that Bond strength effecting the decomposition percentages?
yeah i wrote the same thing that HBr has greater bond energy than HI because HI has a greater bond length which reduces the attraction of nucleus with the bonded electrons and so with the same amount of energy the HI bonds will break in more amount as compare to HBr due to lower requirement of energy for HI bonds to break.Hbr bond strength greater than HI due to short bond length and greater overlap of orbitals hence electrons are closer to nucleus
I dont know exactly but the answer was around 10.9940 something this isnt an accurate value its just my rough estimate of what i remember.Expecting around 54-55,i screwed up the easiest part to calculate isptope mass where answer was required to 6 sig figs.What was its answer?
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