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# AS Chemistry- helping each other out! watch

1. Guys I think I'm having a stupid moment but doing this empa past paper from june 2010 and can't seem to get the answer right for this question! The MS says the answer's 17.6, could anyone explain to me how they got that?

"A saturated solution of ammonia contains 300 g of ammonia in 1.00dm3 of solution.

Calculate the concentration, in mol dm–3, of ammonia in this solution."

2. Posted from TSR Mobile

(Original post by sophiegashman)
Guys I think I'm having a stupid moment but doing this empa past paper from june 2010 and can't seem to get the answer right for this question! The MS says the answer's 17.6, could anyone explain to me how they got that?

"A saturated solution of ammonia contains 300 g of ammonia in 1.00dm3 of solution.

Calculate the concentration, in mol dm–3, of ammonia in this solution."

Pretty straightforward.

First find the moles of NH3 using N = Mass/Mr ---------- 300g/17 gmol-1 = 17.647 mol

Then use the equation C = N/V so 17.647 mol/1 dm3 volume (no convertion of units required = 17.6 mol dm-3 to 3 sig figs.

Make sure you understand that I got the Mr from adding the Ar (or mass number) of Nitrogen which is 14 on the OCR table. With 3 x 1.0 (1.0 is the mass number of hydrogen, since it's Ammonia NH3. We multiply by 3). 14 + 3 = 17 g mol-1

Remember the equations N = Mass/Mr for moles. N = C x V for solutions and N = V/Vm for gas volumes.

Practice loads and you'll get better
3. Welp incoming mod to delete these posts, you guys know talking about the EMPA is against the rules so stop plserino thankerino.
4. (Original post by ozzie2)
Welp incoming mod to delete these posts, you guys know talking about the EMPA is against the rules so stop plserino thankerino.
Actually, it is only against the rules if you talk about EMPA/ISA specifics. That is, the questions that are on the paper. Talking about TOPICS is not against the rules unless explicitly stated that they were on the paper AND that significant help was administered to warrant a correct answer on that question.
5. This is a question from 2014 paper unit 1 I'm confused.

The equation is 4Fe(NO3)3 ---> 2FeO3 + 12NO2 + 3O2

4Fe(NO3)3 mass is 2.16g and Mr is 241.8

it it said to work out moles of iron nitrate I got it as 0.00893n which was right

then it said work out the moles of oxygen gas produced?
I got the mole ratio 4:3 but how do you work out the moles in 3O2 so confused
6. (Original post by samwillettsxxx)
This is a question from 2014 paper unit 1 I'm confused.

The equation is 4Fe(NO3)3 ---> 2FeO3 + 12NO2 + 3O2

4Fe(NO3)3 mass is 2.16g and Mr is 241.8

it it said to work out moles of iron nitrate I got it as 0.00893n which was right

then it said work out the moles of oxygen gas produced?
I got the mole ratio 4:3 but how do you work out the moles in 3O2 so confused
Looking at the mole ratio - for every 4 moles of iron(III) nitrate, there are 3 moles of oxygen (or if it's easier, you will have 1 mole of iron(III) nitrate for 3/4 mole of oxygen). That means you need to multiply the number of moles of iron(III) nitrate by 3/4 to get the number of moles of oxygen.
7. (Original post by usycool1)
Looking at the mole ratio - for every 4 moles of iron(III) nitrate, there are 3 moles of oxygen (or if it's easier, you will have 1 mole of iron(III) nitrate for 3/4 mole of oxygen). That means you need to multiply the number of moles of iron(III) nitrate by 3/4 to get the number of moles of oxygen.
Oh I see thank you that makes sense now! I was dividing 4/3 instead of 3/4 lmao
8. (Original post by samwillettsxxx)
Oh I see thank you that makes sense now! I was dividing 4/3 instead of 3/4 lmao
No worries.
9. (Original post by usycool1)
No worries.
So for the nitrogen dioxide 4/12 = 3
3x 0.00893 is moles in nitrogen dioxide
10. (Original post by samwillettsxxx)
So for the nitrogen dioxide 4/12 = 3
3x 0.00893 is moles in nitrogen dioxide
Yup
11. (Original post by usycool1)
Yup
Thank you so much, very helpful!
12. (Original post by C0balt)
The carbon which has hydroxyl group attached to it has three carbon atoms so there is no H atom on its own attached to the said carbon. Draw a tertiary alcohol and you will see. But I don't think the explanation is on any syllabus.

The chromium (VI) in dichromate has orange colour and chromium (III) is green. It is reduced in oxidation of alcohol this is where the colour change comes from.

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Stuck on a physics question .
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Stuck on a physics question .
Which?

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15. How do you know how many lone pair of electrons an element when it's connected to a compound, eg. the Oxygen in Phenol?
without drawing a dot and cross diagram?
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I just responded

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17. (Original post by Georgiam247)

No I understand but it says the method for the reaction is to bubble hydrogen halides (gases @ rt) into liquid alkANE, but if this happened no reaction would occur. The hydrogen halide gas should be bubbled into a liquid alkENE....right?
I can see where you're coming from, but I'm presuming it means bubbled and turned into liquid alkane product?! But yes, they are reacted with alkenes and made into alkanes, so not very well written...

Chemguide.co.uk says:
Conditions
The alkenes react with gaseous hydrogen halides at room temperature. If the alkene is also a gas, you can simply mix the gases. If the alkene is a liquid, you can bubble the hydrogen halide through the liquid.
18. (Original post by _NMcC_)
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Pretty straightforward.

First find the moles of NH3 using N = Mass/Mr ---------- 300g/17 gmol-1 = 17.647 mol

Then use the equation C = N/V so 17.647 mol/1 dm3 volume (no convertion of units required = 17.6 mol dm-3 to 3 sig figs.

Make sure you understand that I got the Mr from adding the Ar (or mass number) of Nitrogen which is 14 on the OCR table. With 3 x 1.0 (1.0 is the mass number of hydrogen, since it's Ammonia NH3. We multiply by 3). 14 + 3 = 17 g mol-1

Remember the equations N = Mass/Mr for moles. N = C x V for solutions and N = V/Vm for gas volumes.

Practice loads and you'll get better

Sorry I was having a totally stupid moment and using the Mr of Ammonium instead of Ammonia! Thanks for the help though A-levels are most definitely frazzling my brain!
19. Can anyone help me out with percentage abundance questions? For isotopes, I am awful with maths.
So for example 10B and 11B are 2 isotopes with an relative atomic mass of 10.8

i I only got ratio 10:11 I don't know what to do next
20. (Original post by frozo123)
How do you know how many lone pair of electrons an element when it's connected to a compound, eg. the Oxygen in Phenol?
without drawing a dot and cross diagram?
What I do is find the amount of electrons on the outer shell of the bonding atom.
Then I work out the electrons that the bonding atom wants - eg chlorine wants 1 (as it will be sharing the same as the electron it wants in the bond if you get me)
Then I add these both together to get the total electrons and divide by 2 as there are 2 electrons in a pair. You can work out how many lone pairs after taking away the shared electrons in the compound

Eg NH3 : So nitrogen has 5 outer electrons. Hydrogen wants a total of 3 electrons.
Now add these both together and there are 8 electrons so a total of 4 bonds. As there are 3 hydrogens there are 3 bonds with atoms so lose 6 electrons (2 electrons per pair) and you are left with 2 electrons, which are a lone pair as it doesn't bond with any atoms.

Hope this helps, dunno if I even answered the question properly haha

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