OCR chemistry unit 1 Friday 21st May 2010 Watch

FantAbyLus
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#81
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hopefully its right
i got 9.84% but i thot dat cant be right so i went back and realised i had to scale down from 250 to 25
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lsahota
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#82
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#82
what did people get for the empirical formula on 1st page
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Mr van der WAALs
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#83
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(Original post by lsahota)
what did people get for the empirical formula on 1st page
Sn02? It was a ratio of 1:2; the first question was a really nice easy 9 marks
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Jsmalley1
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#84
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i think it went okkkk how do you work out the % composition of NaOH? my mind went completely blank! x
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StephanRamone
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(Original post by Jsmalley1)
i think it went okkkk how do you work out the % composition of NaOH? my mind went completely blank! x
you just double the moles of the acid 2.46x10 to the -3 times 2. so 4.96x10 to the -3 is the moles of NaOH in the equation.
then, you use the fact there were 2grams of impure NaOH in 250cm3 and divide by 10 to get it into 25cm3, so you times 0.2grams by 25/1000 and get 5x10 to the -3.
then take the percentage

(4.92/5)x100 equals 98.4%
not that, that cleared things up at all :p:
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Sean5861
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#86
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Ouch, I got somewhere close to 17%. I should have really got 9% or so because I used the answer to (a) rather than (b).

I did not like that I'm afraid I probably did a bit better than my Jan attempt though. I probably wasn't very clear in my answer about the orbital.

Working mark pleeeaaaase :surprise:
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ziggles
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#87
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hey, the last question about atomic radius accross the period worth four marks, would i get it if i wrote:

As you go accross the periodic table, atoms gain a extra proton which increases the nuclear charge, this attracts electrons closer towards it, decreasing the atomic radius. also as you go across electrons are added to the same shell which means shielding is simular.

also i got 95% for the % compositon, what i did was take away the moles from H2SO4 from the moles of NaOH, this gave me the moles that did not react, and multiplied this by the Mr of Naoh which was 40, this gave me the mass which was something like 1.9, i then divided this by 2 which was the mass of naoh and multiplied a 100, which gave me 95%. also was the moles of Naoh 0.05? i think my % one is wrong??

finally, lols, the question about inoisation energys, where it gave u a table of I.E's and it sed how many electrons in the S shells, it was for Al, i got 2nd,3rd and the 10th and 11th? any ideas of what the answer is?
thanks
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niaaa
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#88
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(Original post by ziggles)
hey, the last question about atomic radius accross the period worth four marks, would i get it if i wrote:

As you go accross the periodic table, atoms gain a extra proton which increases the nuclear charge, this attracts electrons closer towards it, decreasing the atomic radius. also as you go across electrons are added to the same shell which means shielding is simular.

also i got 95% for the % compositon, what i did was take away the moles from H2SO4 from the moles of NaOH, this gave me the moles that did not react, and multiplied this by the Mr of Naoh which was 40, this gave me the mass which was something like 1.9, i then divided this by 2 which was the mass of naoh and multiplied a 100, which gave me 95%. also was the moles of Naoh 0.05? i think my % one is wrong??

finally, lols, the question about inoisation energys, where it gave u a table of I.E's and it sed how many electrons in the S shells, it was for Al, i got 2nd,3rd and the 10th and 11th? any ideas of what the answer is?
thanks

well your explanation about the trend across the periodic table is correct. so is your ionisation energies
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justDannyboy
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#89
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With the percentage mass question, you should have worked out the MR of NaOH and then divide it by the MR of the impure caustic acid.

This is percentage mass, youve all worked out a molar ratio between the two. The answer comes out to 98.3%.

This comes out to the same answer, with rounding, but you will miss a mark by not converting it to molar mass.
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TwilightKnight
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#90
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(Original post by justDannyboy)
With the percentage mass question, you should have worked out the MR of NaOH and then divide it by the MR of the impure caustic acid.

This is percentage mass, youve all worked out a molar ratio between the two. The answer comes out to 98.3%.

This comes out to the same answer, with rounding, but you will miss a mark by not converting it to molar mass.
Actually, we've worked out the percentage purity, which is 98.4%. Im absolutely, 1 bajillion percent sure of that.

You also didn't know the Mr of the impure caustic soda, because it's IMPURE. You weren't given the formula of any impurities / WoC or the like.
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TwilightKnight
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#91
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(Original post by ziggles)
hey, the last question about atomic radius accross the period worth four marks, would i get it if i wrote:

As you go accross the periodic table, atoms gain a extra proton which increases the nuclear charge, this attracts electrons closer towards it, decreasing the atomic radius. also as you go across electrons are added to the same shell which means shielding is simular.
Generally, in a mark scheme, they want you to say that electrons AND protons are added as you go across a period, and that because the amount of shielding remains the same or very similiar, the increased nuclear attraction between oppositely charged positive protons and negative electrons pulls the outer shell/ electrons closer to the nucleus.

You might get away with it by what you said at the end though.
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ziggles
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#92
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the question regaurding halide colours in an organic solvent, it gave us the colour of bromine whaich was orange was the order purple for idoine, orange and orange?
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user1111
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#93
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One of the questions was about the Orbitals... something like 3d and 4s what was the answer???
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StephanRamone
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#94
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(Original post by ziggles)
the question about inoisation energys, where it gave u a table of I.E's and it sed how many electrons in the S shells, it was for Al, i got 2nd,3rd and the 10th and 11th? any ideas of what the answer is?
thanks
i sed 2nd 3rd 10th and 11th like you, but i also put 12th and 13th as the question sed all of the s sub-shell ionisations not just the ones in the table
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user1111
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(Original post by StephanRamone)
i sed 2nd 3rd 10th and 11th like you, but i put 12th and 13th as the question sed all of the s sub-shell ionisations not just the ones in the table
What did you get for the question where you had to fill the gaps with 3d and 4s Was 4s first or 3d
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Mr van der WAALs
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#96
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(Original post by user1111)
What did you get for the question where you had to fill the gaps with 3d and 4s Was 4s first or 3d
It was 4s first then 3d. Two electrons fill the 4s orbital before any electrons begin to fill the 3d orbitals
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RickRoll
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#97
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post the question paper or mark scheme please
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user1111
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#98
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(Original post by Mr van der WAALs)
It was 4s first then 3d. Two electrons fill the 4s orbital before any electrons begin to fill the 3d orbitals
few!! lol got that 2
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justDannyboy
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#99
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(Original post by TwilightKnight)
Actually, we've worked out the percentage purity, which is 98.4%. Im absolutely, 1 bajillion percent sure of that.

You also didn't know the Mr of the impure caustic soda, because it's IMPURE. You weren't given the formula of any impurities / WoC or the like.
It gave you the mass of the impure caustic soda, and you worked out the mol. You can then work out the mr, and as you know pure caustic soda is NaOH(S), which has a MR of 40, and the other one was like 40.7,You can find the percentage purity by mass, by dividing the two, and mulitplying by 100.

It said "compostition BY MASS".

Also i believe the answer to the third halide test was pale green, as you were reacting BR2 with CL Ions. Thus no colour change.
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steph_v
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#100
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(Original post by justDannyboy)
It gave you the mass of the impure caustic soda, and you worked out the mol. You can then work out the mr, and as you know pure caustic soda is NaOH(S), which has a MR of 40, and the other one was like 40.7,You can find the percentage purity by mass, by dividing the two, and mulitplying by 100.

It said "compostition BY MASS".

Also i believe the answer to the third halide test was pale green, as you were reacting BR2 with CL Ions. Thus no colour change.
No, it wouldn't be pale green. Br2 is orange in water and organic solvents. CL2 would be pale green.

Edit: At least, I think it is. I'm doubting myself now!
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