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#1
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Last edited by English143; 2 months ago
0
2 months ago
#2
Could you show your attempt so far so we know where you’re stuck? Cheers x
1
#3
(Original post by Chi chi5)
Could you show your attempt so far so we know where you’re stuck? Cheers x
Hi, I honestly have no clue. I am stuck with turning the ratio into a percentage or fraction. I basically just need the first step. Thanks
0
2 months ago
#4

I called the abundance of Ti 46 just x

so

using the abundances given:

(46 x2) + (47 x2) + (48 x) + 49 x 1) / 5 + x = 47.8 5+ x is the sum of all the abundances

= 235 + 48x / 5 + x = 47.8

= 235 + 48x = 239 + 47.8x
x = 20 but this seems high when comapred to the other abundances, but then again the average mass is very close to 48

maybe I have made an arithmetic error!

so total abundances are just 2 = 2+ 20 + 1 = 25.

so % for 46 is just 2/20 x 100% = 10%
0
#5
(Original post by scimus63)

I called the abundance of Ti 46 just x

so

using the abundances given:

(46 x2) + (47 x2) + (48 x) + 49 x 1) / 5 + x = 47.8 5+ x is the sum of all the abundances

= 235 + 48x / 5 + x = 47.8

= 235 + 48x = 239 + 47.8x
x = 20 but this seems high when comapred to the other abundances, but then again the average mass is very close to 48

maybe I have made an arithmetic error!

so total abundances are just 2 = 2+ 20 + 1 = 25.

so % for 46 is just 2/20 x 100% = 10%
Scimus I honestly dont know what to say but you are honestly a legend thank you so much.

Edit: Sorry my account is being dumb and wont let me rep you
Last edited by English143; 2 months ago
0
2 months ago
#6
(Original post by English143)
Hi, I honestly have no clue. I am stuck with turning the ratio into a percentage or fraction. I basically just need the first step. Thanks
Think of it in terms of maths only
You have four variables where two are the same ratio and one is half of those two. This can be seen as 2x+2x+x+y=1
Also 46(2x)+47(2x)+48y+49x=47.8

Solve by substitution
0
2 months ago
#7
(Original post by scimus63)

I called the abundance of Ti 46 just x

so

using the abundances given:

(46 x2) + (47 x2) + (48 x) + 49 x 1) / 5 + x = 47.8 5+ x is the sum of all the abundances

= 235 + 48x / 5 + x = 47.8

= 235 + 48x = 239 + 47.8x
x = 20 but this seems high when comapred to the other abundances, but then again the average mass is very close to 48

maybe I have made an arithmetic error!

so total abundances are just 2 = 2+ 20 + 1 = 25.

so % for 46 is just 2/20 x 100% = 10%
You did it right
0
2 months ago
#8
(Original post by scimus63)

I called the abundance of Ti 46 just x

so

using the abundances given:

(46 x2) + (47 x2) + (48 x) + 49 x 1) / 5 + x = 47.8 5+ x is the sum of all the abundances

= 235 + 48x / 5 + x = 47.8

= 235 + 48x = 239 + 47.8x
x = 20 but this seems high when comapred to the other abundances, but then again the average mass is very close to 48

maybe I have made an arithmetic error!

so total abundances are just 2 = 2+ 20 + 1 = 25.

so % for 46 is just 2/20 x 100% = 10%
For your last line shouldn’t it be 2/25 x100 as you’re looking at the abundance relative to the complete sample
0
#9
(Original post by Chi chi5)
Think of it in terms of maths only
You have four variables where two are the same ratio and one is half of those two. This can be seen as 2x+2x+x+y=1
Also 46(2x)+47(2x)+48y+49x=47.8

Solve by substitution
Is the abundance of 46Ti 8%, as thats what I got
0
#10
(Original post by ashvinsingh)
You did it right
I think it may be 8%, as you add them up to 25 and then change the ratio to out of 100. You then should get 80 8:4.
0
2 months ago
#11
(Original post by English143)
Is the abundance of 46Ti 8%, as thats what I got
Well done 0
2 months ago
#12
(Original post by The A.G)
For your last line shouldn’t it be 2/25 x100 as you’re looking at the abundance relative to the complete sample
The mean of 46 47 and 49 Ti is 47. So 2Ti47: TI48 forms TI 47.8 so must be in ratio 4:1 so TI48 must be 0.5. 0.5/5 = 1/10 = 10%
0
2 months ago
#13
(Original post by scimus63)

I called the abundance of Ti 46 just x

so

using the abundances given:

(46 x2) + (47 x2) + (48 x) + 49 x 1) / 5 + x = 47.8 5+ x is the sum of all the abundances

= 235 + 48x / 5 + x = 47.8

= 235 + 48x = 239 + 47.8x
x = 20 but this seems high when comapred to the other abundances, but then again the average mass is very close to 48

maybe I have made an arithmetic error!

so total abundances are just 2 = 2+ 20 + 1 = 25.

so % for 46 is just 2/20 x 100% = 10%
Should be 49x not 49 y, and 48y
0
#14
(Original post by Chi chi5)
Should be 49x not 49 y, and 48y
Did you get 8%?
0
2 months ago
#15
Yes
0
#16
(Original post by Chi chi5)
Yes
Nice
0
2 months ago
#17
(Original post by ashvinsingh)
The mean of 46 47 and 49 Ti is 47. So 2Ti47: TI48 forms TI 47.8 so must be in ratio 4:1 so TI48 must be 0.5. 0.5/5 = 1/10 = 10%
Everyone’s gotten 8% and not quite following your method as you the relationship you created between Ti 47 and Ti 48 would be independent of the overall composition of the sample
0
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