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# AQA AS Chemistry, unit 1 watch

1. (Original post by Tabby.1.2)
please can someone explain this to me, i think i'm having a blonde moment ?!
i was doing a past paper and i cannot work out this 1 mark question hahah :

Norgessaltpeter was the first nitrogen fertiliser to be manufactured in Norway.
It has the formula Ca(NO3)2
Norgessaltpeter can be made by the reaction of calcium carbonate with
dilute nitric acid as shown by the following equation.

CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + CO2(g) + H2O(I)

In an experiment, an excess of powdered calcium carbonate was added to 36.2 cm3
of 0.586 mol dm–3 nitric acid.

Calculate the amount, in moles, of HNO3 in 36.2 cm3 of 0.586 mol dm–3 nitric acid.
Give your answer to 3 significant figures.
moles = conc x vol
moles = 0.586 x (36.2/1000) and you should get your answer
2. (Original post by maryamnc)
moles = conc x vol
moles = 0.586 x (36.2/1000) and you should get your answer
hahah thankyou, i was definitely having a blonde moment :')
3. (Original post by Tabby.1.2)
please can someone explain this to me, i think i'm having a blonde moment ?!
i was doing a past paper and i cannot work out this 1 mark question hahah :

Norgessaltpeter was the first nitrogen fertiliser to be manufactured in Norway.
It has the formula Ca(NO3)2
Norgessaltpeter can be made by the reaction of calcium carbonate with
dilute nitric acid as shown by the following equation.

CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + CO2(g) + H2O(I)

In an experiment, an excess of powdered calcium carbonate was added to 36.2 cm3
of 0.586 mol dm–3 nitric acid.

Calculate the amount, in moles, of HNO3 in 36.2 cm3 of 0.586 mol dm–3 nitric acid.
Give your answer to 3 significant figures.
The equation is:
n = m x v / 1000
m = concentration
v = volume
n = moles

0.586 x 36.2 / 1000 = 0.0212132
= 0.021mol

I think that's how you do it, I would do it like that anyways :P
4. (Original post by IWantSomeMushu)
Can anybody explain the effect surface area has on the boiling points of hydrocarbons?

As surface area (number of atoms within the hydrocarbon molecule) increases, so does the boiling point.
This is because there are stronger/more van der waals forces between the molecules (As larger surface area) therefore more energy is needed to overcome these forces (higher BP)

that should be right, anyone correct me if am wrong please!!
5. (Original post by IWantSomeMushu)
Can anybody explain the effect surface area has on the boiling points of hydrocarbons?
Larger SA=More Electrons and more electrons means the Van der Waals forces are stronger and require more energy to overcome

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6. (Original post by Tabby.1.2)
please can someone explain this to me, i think i'm having a blonde moment ?!
i was doing a past paper and i cannot work out this 1 mark question hahah :

Norgessaltpeter was the first nitrogen fertiliser to be manufactured in Norway.
It has the formula Ca(NO3)2
Norgessaltpeter can be made by the reaction of calcium carbonate with
dilute nitric acid as shown by the following equation.

CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + CO2(g) + H2O(I)

In an experiment, an excess of powdered calcium carbonate was added to 36.2 cm3
of 0.586 mol dm–3 nitric acid.

Calculate the amount, in moles, of HNO3 in 36.2 cm3 of 0.586 mol dm–3 nitric acid.
Give your answer to 3 significant figures.
see attatched its way easier than you think
Attached Images

7. (Original post by a.jiwa)
Yeahhh i dont remember seeing it anywhere and hopefully they dont thorw it in! i think ill probably have memorized the explanation just to be safe however if they ask me to apply it i have no idea what the explanation actually means either

aaah well, im doing the shapes now and its working well thanks to your earlier suggestion, best of luck for tomorrow and thanks for your help!
Not on the spec for that specifically so hopefully it's not in the exam tomorrow
No problem, glad it's working You too, best of luck!

(Original post by IWantSomeMushu)
Can anybody explain the effect surface area has on the boiling points of hydrocarbons?
Boiling points decreases when there is branching in hydrocarbons because of less effective van der Waals forces. Smaller SA = more branching = less vdw?? That kinda sounds logical to me lol xD Correct me if it's totally wrong :P
8. (Original post by NabRoh)
Predict the bond angle in Na2S?

How do you guys approach this and get the right answer? Cost me my 90 on a past paper. :/

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9. (Original post by Paulineuh)
Not on the spec for that specifically so hopefully it's not in the exam tomorrow
No problem, glad it's working You too, best of luck!

Boiling points decreases when there is branching in hydrocarbons because of less effective van der Waals forces. Smaller SA = more branching = less vdw?? That kinda sounds logical to me lol xD Correct me if it's totally wrong :P

I remember it as:

Branching means less/lower Surface area
so weaker VDW

Hence lower BP
10. (Original post by a.jiwa)
people,
for AS AQA do we need to know about the anomalous electron configurations of chromium and copper?

I just had a read through the old text book and it does mention it there but im not sure if its on specification or if aqa will decide to be nasty and throw it in the exam?

The text book says :
Copper and chromium electron configurations do not follow the pattern and are anomalous:

- Chromium is 1s2 2s2 2p6 3s2 3p6 4s1 3d5
NOT 1s2 2s2 2p6 3s2 3p6 4s2 3d4

- copper is 1s2 2s2 2p6 3s2 3p6 4s1 3d10
NOT 1s2 2s2 2p6 3s2 3p6 4s2 3d9

The explanation of this which i didnt fully understand according to ye olde text book is:
The reason for these anomoules is that both cases the outer sub levels are either full or half full and the structures shown are of lower energy and represent more stable arrangements..

has ANYONE seen this come up in a past paper?

I've never seen it.

Thanks for posting the explanations and the configurations though. I'll make a note of them so I don't get caught out!
11. (Original post by NabRoh)
Predict the bond angle in Na2S?

How do you guys approach this and get the right answer? Cost me my 90 on a past paper. :/
Well, I'd draw it out.

Find the central atom which in this case is S.

Then draw out the e- on it's outer shell e.g. 6 here (S, Group 6)

Find how many bonding atoms are bonded to the central atom - 2 here (Na2)

Then I'd count how many pairs of e- are surrounding that central atom, in this case, 4. (2 lone pairs, 2 bonded pairs)

Name it then remember the bond angle for that name so tetrahedral here which is 109.5.

But then take any lone pairs into account and taking away 2.5 from the "original" angle, e.g. here there's 2

Thus concluding with 109.5-2.5(2) = 104.5 which is the same as h20 = v-shaped/bent

(Original post by gooner1886)
I remember it as:

Branching means less/lower Surface area
so weaker VDW

Hence lower BP
Thanks, got it!
12. (Original post by a.jiwa)
As surface area (number of atoms within the hydrocarbon molecule) increases, so does the boiling point.
This is because there are stronger/more van der waals forces between the molecules (As larger surface area) therefore more energy is needed to overcome these forces (higher BP)

that should be right, anyone correct me if am wrong please!!

(Original post by homefind)
Larger SA=More Electrons and more electrons means the Van der Waals forces are stronger and require more energy to overcome

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(Original post by Paulineuh)
Not on the spec for that specifically so hopefully it's not in the exam tomorrow
No problem, glad it's working You too, best of luck!

Boiling points decreases when there is branching in hydrocarbons because of less effective van der Waals forces. Smaller SA = more branching = less vdw?? That kinda sounds logical to me lol xD Correct me if it's totally wrong :P
Cheers
13. (Original post by a.jiwa)
see attatched its way easier than you think
(Original post by Paulineuh)
The equation is:
n = m x v / 1000
m = concentration
v = volume
n = moles

0.586 x 36.2 / 1000 = 0.0212132
= 0.021mol

I think that's how you do it, I would do it like that anyways :P
thankyou
14. Predict the type of bonding in Na2S?

How do you guys approach this and get the right answer? Cost me my 90 on a past paper. :/

Reposted because old question was wrong. Want to know if its covalent, dative, ionic, etc.
15. (Original post by Paulineuh)
Well, I'd draw it out.

Find the central atom which in this case is S.

Then draw out the e- on it's outer shell e.g. 6 here (S, Group 6)

Find how many bonding atoms are bonded to the central atom - 2 here (Na2)

Then I'd count how many pairs of e- are surrounding that central atom, in this case, 4. (2 lone pairs, 2 bonded pairs)

Name it then remember the bond angle for that name so tetrahedral here which is 109.5.

But then take any lone pairs into account and taking away 2.5 from the "original" angle, e.g. here there's 2

Thus concluding with 109.5-2.5(2) = 104.5 which is the same as h20 = v-shaped/bent

Thanks, got it!
There is a faster way to do it...

Na is in group 1 so therefore it has 1 outer electron, there are two Na so that is 2 electrons
S is in group 6, therefore it has 6 outer electrons
6+2=8 then 8/2=4
Therefore there are 4 electron pairs, only 2 are involved in bonding therefore there are two lone pairs.
So it will be bent with a bond angle of 104.5

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16. (Original post by NabRoh)
Predict the type of bonding in Na2S?

How do you guys approach this and get the right answer? Cost me my 90 on a past paper. :/

Reposted because old question was wrong. Want to know if its covalent, dative, ionic, etc.
It is ionic - one non-metal and one metal atom

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17. (Original post by homefind)
There is a faster way to do it...

Na is in group 1 so therefore it has 1 outer electron, there are two Na so that is 2 electrons
S is in group 6, therefore it has 6 outer electrons
6+2=8 then 8/2=4
Therefore there are 4 electron pairs, only 2 are involved in bonding therefore there are two lone pairs.
So it will be bent with a bond angle of 104.5

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I never knew you could do it like that! Thank you Will save me a lot of time now, lol XD
18. (Original post by Paulineuh)
I never knew you could do it like that! Thank you Will save me a lot of time now, lol XD
Haha, sure will! no problem

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19. (Original post by NabRoh)

According to the mark scheme, no f orbitals...

When we fill orbitals we fill the one with a lower energy level first.... so for example 4s gets filled before 3d. So generally there is f orbital present in the 4th subshell but its not relevant here because the 5p orbital is of a lower energy level than the 4f orbital hence it gets filled first

well thats the way ive been taught it.... but generally f orbitals aren't relevant till A2
20. Did anyone resit bio1? I did and found it went good, it was a decent paper so hopefully chemistry will be the same

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