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

Hey guys! This is the official thread for the Chem Unit 1 exam on May 27, Who else is giving it? How is your revision?

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Reply 1
Original post by Sandy_Vega30
Hey guys! This is the official thread for the Chem Unit 1 exam on May 27, Who else is giving it? How is your revision?

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hey
yeah i'm doing it on the same date as you are
lets just hope its easy
on a completely random note can anyone tell me why the answer is C and not A
don't all ionic cpds have little covalent character?
(edited 7 years ago)
Original post by Sandy_Vega30
Hey guys! This is the official thread for the Chem Unit 1 exam on May 27, Who else is giving it? How is your revision?


Hey I'm also doing Chem unit 1
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Original post by Bliss_
Capture.PNG
hey
yeah i'm doing it on the same date as you are
lets just hope its easy
on a completely random note can anyone tell me why the answer is C and not A
don't all ionic cpds have little covalent character?


I think its because Sodium has a very low charge density and fluorine is the smallest anion of the halogens. Hence the polarizing power of Sodium and the measure of how polarizable the Fluorine atom is low. It is hence an Ionic Bond.
Reply 4
Original post by Sandy_Vega30
I think its because Sodium has a very low charge density and fluorine is the smallest anion of the halogens. Hence the polarizing power of Sodium and the measure of how polarizable the Fluorine atom is low. It is hence an Ionic Bond.


oh alright thanks


To measure enthalpy change, there must be a temperature change hence temperature cannot be constant. And any experiment related to gases, Pressure must be controlled because it is the main factor that affects gases. :smile:
Hey guys please could you help me out 😃
Q- 5 (c) is C
Q- 7 is A
Q- 8 is D
Q- 9 is C
Q- 10 is D
Q- 16 is D
Please tell what's the reason for these answers because I don't get them
Btw: its the IAL Jan 2014 paper

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Reply 8
Original post by dinithisara
Hey guys please could you help me out 😃
Q- 5 (c) is C
Q- 7 is A
Q- 8 is D
Q- 9 is C
Q- 10 is D
Q- 16 is D
Please tell what's the reason for these answers because I don't get them
Btw: its the IAL Jan 2014 paper

Posted from TSR Mobile


see, for question 5(c) it asks for trend in first ionisation energies from group 6 onwards. from group 6 to group 8 the ie increases as we are moving across a period. then we will move back to group 1 and to the next period so shielding will increase and ie will decrease . With this conclusion you should come up with an answer that has first three ie high followed by last two to be low due to increased shielding so the answer is (c)

for question 7 the answer is not b or d because CCl4 and Br have covalent bonds and no holes to carry the charge and conduuct electricity. the answer is not molten NaCl because it doesnt conduct in solid state due to its ionic nature

for question 8 H2O2 has four atoms in one molecule so you'll multiple the answer by 4 like so 1(moles) * avogadros constant * 4 and you'll get D

for question 16 you should figure out the legnth of the longest possible carbon chain and use it in this case the longest possible carbon chain has 6 carbons. Next you should choose the order of numbering the carbons so that your groups such as methyl get lowest possible numbers.
I hope i made sense. good luck btw hope it helped :smile:
(edited 7 years ago)
Original post by dinithisara
Hey guys please could you help me out 😃
Q- 5 (c) is C
Q- 7 is A
Q- 8 is D
Q- 9 is C
Q- 10 is D
Q- 16 is D
Please tell what's the reason for these answers because I don't get them
Btw: its the IAL Jan 2014 paper

Posted from TSR Mobile


Q5) - They want the ionization energies of elements whose atomic no. increases starting from group 6. Take the following example:

Element : O F Ne Na Mg
Atomic no: 8 9 10 11 12
Configuration: 2:6 2:7 2:8 2:8:1 2:8:2


The trend shows that ionization energies increase upto Ne because atomic no. increases. This means that there are more electrons and hence there is a stronger nuclear attraction in the atoms. But after Ne, Ionization energies decreases because there is a 3rd subshell. This increases the distance between the 3rd valence electrons and the nucleus hence less nuclear attraction and hence less energy required to remove an electron from Na. Mg has more electrons than Na, so it must have a greater ionization energy. The pattern showing increasing ionization energies for the first 3 elements and less increasing ionization energies for the 4th and 5th element is C.


Q7) - I am not sure how to explain this. But I think its coz Mercury has a very low melting temperature and is a metal hence can conduct electricity in both liquid and solid state.

Q8) - The no. of atoms in H2O2 is 4, coz there are 2 hydrogen atoms and 2 oxygen atoms. Therefore, all you need to do is:

1 atom ----> 6.0x10^23 particles
4 atoms -----> ? = 4 X 6.0x10^23 = 2.4x10^24


Q9) - I am also not sure of this but this is how i think C is the answer.

Balanced equation if Mg is the metal:
Mg + Cl2 ----> MgCl2

According to mole ratio, Cl2 also is of 0.1 moles. Hence find the mass of Cl2 used ( 0.1x 35.5 x 2 = 7.1g).

Q10) - The gases in this equation are both NO2 and O2. Hence mole ratio respectively is 0.2 and 0.05. Total moles of gases is 0.25.
Therefore, Volume = 0.25x24 = 6.0dm3

Q16) - Try converting the skeletal formula to displayed formula:

Start counting from the smaller side. i.e if you start from the left the methyl group will be on Carbon 4 and 5, but from right side, they will be on 3 and 4. Always aim on counting from the smaller side. The parent chain has 7 carbons hence its a "Heptane". The name is 3,4- dimethylheptane. The 'di' is because there are 2 methyl groups in this molecule.

Sorry for the late reply. Hope that helped. :smile:
(edited 7 years ago)
Here is the picture of the displayed formula.
Original post by Bliss_
see, for question 5(c) it asks for trend in first ionisation energies from group 6 onwards. from group 6 to group 8 the ie increases as we are moving across a period. then we will move back to group 1 and to the next period so shielding will increase and ie will decrease . With this conclusion you should come up with an answer that has first three ie high followed by last two to be low due to increased shielding so the answer is (c)

for question 7 the answer is not b or d because CCl4 and Br have covalent bonds and no holes to carry the charge and conduuct electricity. the answer is not molten NaCl because it doesnt conduct in solid state due to its ionic nature

for question 8 H2O2 has four atoms in one molecule so you'll multiple the answer by 4 like so 1(moles) * avogadros constant * 4 and you'll get D

for question 16 you should figure out the legnth of the longest possible carbon chain and use it in this case the longest possible carbon chain has 6 carbons. Next you should choose the order of numbering the carbons so that your groups such as methyl get lowest possible numbers.
I hope i made sense. good luck btw hope it helped :smile:


Thank you so much 😃
Original post by Sandy_Vega30
Q5) - They want the ionization energies of elements whose atomic no. increases starting from group 6. Take the following example:

Element : O F Ne Na Mg
Atomic no: 8 9 10 11 12
Configuration: 2:6 2:7 2:8 2:8:1 2:8:2


The trend shows that ionization energies increase upto Ne because atomic no. increases. This means that there are more electrons and hence there is a stronger nuclear attraction in the atoms. But after Ne, Ionization energies decreases because there is a 3rd subshell. This increases the distance between the 3rd valence electrons and the nucleus hence less nuclear attraction and hence less energy required to remove an electron from Na. Mg has more electrons than Na, so it must have a greater ionization energy. The pattern showing increasing ionization energies for the first 3 elements and less increasing ionization energies for the 4th and 5th element is C.


Q7) - I am not sure how to explain this. But I think its coz Mercury has a very low melting temperature and is a metal hence can conduct electricity in both liquid and solid state.

Q8) - The no. of atoms in H2O2 is 4, coz there are 2 hydrogen atoms and 2 oxygen atoms. Therefore, all you need to do is:

1 atom ----> 6.0x10^23 particles
4 atoms -----> ? = 4 X 6.0x10^23 = 2.4x10^24


Q9) - I am also not sure of this but this is how i think C is the answer.

Balanced equation if Mg is the metal:
Mg + Cl2 ----> MgCl2

According to mole ratio, Cl2 also is of 0.1 moles. Hence find the mass of Cl2 used ( 0.1x 35.5 x 2 = 7.1g).

Q10) - The gases in this equation are both NO2 and O2. Hence mole ratio respectively is 0.2 and 0.05. Total moles of gases is 0.25.
Therefore, Volume = 0.25x24 = 6.0dm3

Q16) - Try converting the skeletal formula to displayed formula:

Start counting from the smaller side. i.e if you start from the left the methyl group will be on Carbon 4 and 5, but from right side, they will be on 3 and 4. Always aim on counting from the smaller side. The parent chain has 7 carbons hence its a "Heptane". The name is 3,4- dimethylheptane. The 'di' is because there are 2 methyl groups in this molecule.

Sorry for the late reply. Hope that helped. :smile:


Thank you 😃😃😃 yes it did help a lot
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Could someone explain that to me? The mark scheme says the answer is B.
hey guys have anyone written any notes for each specification point?
Original post by daddydevil
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Could someone explain that to me? The mark scheme says the answer is B.


You have to rearrange the equations in such a way to produce the required equation. The first equation does not need to be flipped because both C and O2 are on the reactant side. The second equation needs to be flipped a half mole of O2 can be cancelled with one mole of O2 in the first equation to give a half mole of O2 as required by the main equation. The third equation also needs to be flipped do that CO2, PbO, and Pb can be cancelled out. Note that when an equation is flipped the sign of the enthalpy changes. Write the overall equation and add the enthalpies. The answer is -111.
Which of the following quantities, used in the calculation of the lattice energy oflithium oxide, Li2O, has a negative value?

1.

A The enthalpy change of atomization of lithium.

2.

B The first ionization energy of lithium.

3.

C The first electron affinity of oxygen.

4.

D The second electron affinity of oxygen

the answer is C can anyone explain to me why?
Original post by zahragoli97
Which of the following quantities, used in the calculation of the lattice energy oflithium oxide, Li2O, has a negative value?

1.

A The enthalpy change of atomization of lithium.

2.

B The first ionization energy of lithium.

3.

C The first electron affinity of oxygen.

4.

D The second electron affinity of oxygen

the answer is C can anyone explain to me why?


1st electron affinity is exothermic because electrons have to lose energy to occupy a lower energy orbital. 2nd electron affinity is always endothermic (positive) because the electron is added to an ion which is already negative therefore it must overcome the repulsion. The ionization and atomization energies are always positive 9Refer to the Born - Haber Cycle).
(edited 7 years ago)
https://110d0b1a7b584219c71ba4cfe896ad1711fa3ef7.googledrive.com/host/0B1ZiqBksUHNYR00tYjdZN052azQ/January%202013%20QP%20-%20Unit%201%20Edexcel%20Chemistry.pdf
Q9, answers A but why is the temp change the same? if e=mct then shouldnt the heat evolved be more bc the "mass" (volume) has increased?
Original post by Sandy_Vega30
To measure enthalpy change, there must be a temperature change hence temperature cannot be constant. And any experiment related to gases, Pressure must be controlled because it is the main factor that affects gases. :smile:


i now realise how dumb the question was - thankyou!

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