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Edexcel IAL Chemistry Unit 1 (WCH01 )/ May 27 June 2016

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Original post by OnTheHorizon
I wrote that is light and also very small so can pass through the smallest of gaps.
For shells I included the spins as it asked for orbital.The assumption was for the RAM which was that Carbon has an RAM of 12


Oh yeah! That's what I wrote too. The RAM of C is assumed to be 12. :smile:
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Original post by tejsanghavi
What was the helium question?
what was the answer to the question where u had to state an assumption to the enthalpy change for the hess law?
what was the answer to the shell question?


light
-2786.. something -2000 (not sure about the answer)
Al-1s2 2s2 2p6 3s2 3p1 so p orbital, s orbital, s orbital, p orbital, p orbital...
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Original post by Sandy_Vega30
Oh yeah! That's what I wrote too. The RAM of C is assumed to be 12. :smile:


Said that too!!
What did you write about that catalyst question and the one about the enthalpy of H20(g) being less than that of H20(l)?
I wrote that once the catalyst is heated it starts/Speeds up the combustion reaction which also produces heat(So the tool or whatever is no longer needed,Didn't write this bit but I implied it).
For Enthalpy,I wrote that the enthalpy change of formation without changing state is less than that of changing state or something along those lines.
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Original post by waleedlat
What did you write for an extra information other than vaporization of water


formation of CO2
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Original post by OnTheHorizon
What did you write about that catalyst question and the one about the enthalpy of H20(g) being less than that of H20(l)?
I wrote that once the catalyst is heated it starts/Speeds up the combustion reaction which also produces heat(So the tool or whatever is no longer needed,Didn't write this bit but I implied it).
For Enthalpy,I wrote that the enthalpy change of formation without changing state is less than that of changing state or something along those lines.


That would be right. However since gaseous reactants were involved, the adsorption and desorption would also increase heat energy.
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Original post by waleedlat
Bond entrapped is wrong ?


I would not be so sure. the arrow pointed from co2 to the bottom of the cycle hence reverse would give you the enthalpy of formation of co2
Original post by OnTheHorizon
What did you write about that catalyst question and the one about the enthalpy of H20(g) being less than that of H20(l)?
I wrote that once the catalyst is heated it starts/Speeds up the combustion reaction which also produces heat(So the tool or whatever is no longer needed,Didn't write this bit but I implied it).
For Enthalpy,I wrote that the enthalpy change of formation without changing state is less than that of changing state or something along those lines.


I said that forming H2O(g) requires less energy because of weak bonds. In h2O(l), more energy is required due to the formation of Hydrogen bonding.

I said that reaction between C4H10 and O2 is exothermic and releases enough energy hence does no require continued heating. I don't think that worth 2 marks. Maybe 1. :redface:
What was the catalyst assumption I wrote because now catalyst carry out reaction itself I wrote RAM 12 I wrote 150 on gas volume
Original post by Kk1999
the mcq with 50cm3 and 25 cm3??
I chose 75cm3 yay:banana2::banana2::banana2::banana2::banana2::banana2:!!!


it wasn't 75! it was 100, you had to find the moles of the H2 by dividing the volume by 24 and then using the ratio of moles and the 24 dm3 to find the volume of the HCL
Original post by anujsr
formation of CO2


was it not enthalpy change of atomisation? since some of the molecules were turned into a gaseous state?
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Original post by Yara El-Bardisi
was it not enthalpy change of atomisation? since some of the molecules were turned into a gaseous state?


Not really. if you form co2, co2 is a gas and it is formed from solid carbon and oxygen. Both were part of the cycle. There could also be many other alternatives.
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Original post by Yara El-Bardisi
it wasn't 75! it was 100, you had to find the moles of the H2 by dividing the volume by 24 and then using the ratio of moles and the 24 dm3 to find the volume of the HCL


Volume of reacting gasses. 25 cm3 of h2 would react with 25 cm3 of hcl to form 5o cm3 of whatever the gas was. I remember mole ration was 1 is to 1 is to 2 hence out of the 50cm3 only 25 reacted and 25 remained. 50 formed hence answer= 50+25 = 75
Guys!! What were the answers to the enthalphy change questions ??😱😱
What was the answer to the question where it was asked to suggest why helium is suitable to use in the mass spectrometer?

And also, the answer to the catalyst question...why isn't a battery needed once the catalyst is heated up?
For the increasing atomic number from mg to Al what was the effect on boiling point caused by?
I thought mean bond enthalpy of C4H10
Original post by Hiyoriiki
For the increasing atomic number from mg to Al what was the effect on boiling point caused by?


Mg has fewer protons than Al hence Mg has weaker metallic bonds than Al. This means Mg requires less energy to break these points hence has a lower melting point.
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Original post by Hiyoriiki
For the increasing atomic number from mg to Al what was the effect on boiling point caused by?

Al contributes 3 electrons to the sea of electrons whereas 2 electrons are contributed to the sea of electron of MG. Hence stronger metallic bonds in AL as compared to mg
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Original post by Rahatara Sadique
What was the answer to the question where it was asked to suggest why helium is suitable to use in the mass spectrometer?

And also, the answer to the catalyst question...why isn't a battery needed once the catalyst is heated up?


adsorption and desorption of the gases also leads to more heat energy due to the catalyst surface

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