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# Cambridge Chemistry Challenge watch

1. (Original post by HerrDudelmann)
Anyone got any hints for Gold about the actual specifications of the coin, e.g. mass and material?
U do realise if u type 1 yen into google it tells u...... Not very complicated!!!!!

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2. stuck on Rg ...

i have worked out how many triangles are in the hexagon, and worked out the fraction that are blue...

then I worked out the total surface area of the hexagon and then multiplied that by its depth (I) to get the volume.

then i multiplied that by the fraction of the hexagon which is shaded blue...

then I worked out the weight of that volume in water (x0.1) to get grams

then i used n=m/Mr to find the moles of water..

and then multiplied by avogrado's constant (6.022x10^23) to get the number of molecules

Am i along the right lines or am i going off on a tangent? any help much appriciated...
3. stuck on Rg ...

i have worked out how many triangles are in the hexagon, and worked out the fraction that are blue...

then I worked out the total surface area of the hexagon and then multiplied that by its depth (I) to get the volume.

then i multiplied that by the fraction of the hexagon which is shaded blue...

then I worked out the weight of that volume in water (x0.1) to get grams

then i used n=m/Mr to find the moles of water..

and then multiplied by avogrado's constant (6.022x10^23) to get the number of molecules

Am i along the right lines or am i going off on a tangent? any help much appriciated...
4. stuck on Rg ...

i have worked out how many triangles are in the hexagon, and worked out the fraction that are blue...

then I worked out the total surface area of the hexagon and then multiplied that by its depth (I) to get the volume.

then i multiplied that by the fraction of the hexagon which is shaded blue...

then I worked out the weight of that volume in water (x0.1) to get grams

then i used n=m/Mr to find the moles of water..

and then multiplied by avogrado's constant (6.022x10^23) to get the number of molecules

Am i along the right lines or am i going off on a tangent? any help much appriciated...
5. (Original post by Medicine1234)
stuck on Rg ...

i have worked out how many triangles are in the hexagon, and worked out the fraction that are blue...

then I worked out the total surface area of the hexagon and then multiplied that by its depth (I) to get the volume.

then i multiplied that by the fraction of the hexagon which is shaded blue...

then I worked out the weight of that volume in water (x0.1) to get grams

then i used n=m/Mr to find the moles of water..

and then multiplied by avogrado's constant (6.022x10^23) to get the number of molecules

Am i along the right lines or am i going off on a tangent? any help much appriciated...
Look in the earlier page some person tells u how to do it/ he is asking if he is right and he is

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6. I can't see what I'm doing wrong.. For rg
This is my calculation
Volume-3route3/2 x 0.035^2 x0.02
Times volume by 34/49 (area of shaded) x density (0.9167) /18 x 6.0221412927x23 (avagrdos number)
Any ideas ???. Much appreciated
7. (Original post by ltigers)
I can't see what I'm doing wrong.. For rg
This is my calculation
Volume-3route3/2 x 0.035^2 x0.02
Times volume by 34/49 (area of shaded) x density (0.9167) /18 x 6.0221412927x23 (avagrdos number)
Any ideas ???. Much appreciated
You may have your units wrong.

When calculating the moles of water, the mass has to be in grams. Also, 0.9167 is g/cm^3, not m^3. So you may need to take some of these into account. You could try adding or removing some zeros to see if that works.
8. (Original post by brittanna)
You may have your units wrong.

When calculating the moles of water, the mass has to be in grams. Also, 0.9167 is g/cm^3, not m^3. So you may need to take some of these into account. You could try adding or removing some zeros to see if that works.
What units do the volume need to be in ..mm/cm or m3?
9. (Original post by ltigers)
What units do the volume need to be in ..mm/cm or m3?
I would probably use cm^3 as all the values you need involve cm^3, but you can use any units, you will just need to adjust the density.
10. It just doesn't seem to work...I can't see how i go it wrong
11. (Original post by brittanna)
I would probably use cm^3 as all the values you need involve cm^3, but you can use any units, you will just need to adjust the density.
Did u manage to do it?
12. http://www.c3l6.org/challenges/play/1/23
What's the answer for the entry level question?
13. (Original post by GapYearMedic)
http://www.c3l6.org/challenges/play/1/23
What's the answer for the entry level question?
Malaria
14. (Original post by mu142857)
Malaria
Ah cool, it's quinine! Cheers! I'm not even doing the challenge, I just came across that question randomly and it really annoyed me lol
15. (Original post by ltigers)
Did u manage to do it?
Yeah, I finished it yesterday.
16. (Original post by brittanna)
Yeah, I finished it yesterday.
Are the numbers not in standard from...1.354585663?
17. (Original post by ltigers)
Are the numbers not in standard from...1.354585663?
I wrote mine out in full: e.g. 565000000000000000000000
18. (Original post by brittanna)
I wrote mine out in full: e.g. 565000000000000000000000
were mine the numbers not without the 0s.? im not sure whether ive made a caluclation error or a typing error.
19. (Original post by ltigers)
were mine the numbers not without the 0s.? im not sure whether ive made a caluclation error or a typing error.
Have u used 3 sig fig

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20. (Original post by little_miss_smiley)
Have u used 3 sig fig

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No, I've used the actual values. I can't see how I got it wrong.

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