chem q colourimetry (1 Viewer)

[Masaken]

i might be a bit stupid but i'm not sure where you start. like i know you have to start by using the formula for A they give above part (a) and i know where to go from there, but i don't know what values you're meant to sub in because the question doesn't give any other information about the light intensity or the absorbance or any concentration, unless i'm meant to deduce the intensity of the light from the graph?? (if so it might have slipped from my mind bc i have no idea what to do). could someone help with how to find the intensity of the light from the info of the q ?? sorry if it's so obvious i just have no idea, thank you in advance

 

[SadCeliac]

View attachment 40028
i might be a bit stupid but i'm not sure where you start. like i know you have to start by using the formula for A they give above part (a) and i know where to go from there, but i don't know what values you're meant to sub in because the question doesn't give any other information about the light intensity or the absorbance or any concentration, unless i'm meant to deduce the intensity of the light from the graph?? (if so it might have slipped from my mind bc i have no idea what to do). could someone help with how to find the intensity of the light from the info of the q ?? sorry if it's so obvious i just have no idea, thank you in advance

there's a formula A = elc
something like that??? not too sure though lol sorry

 

[Masaken]

there's a formula A = elc
something like that??? not too sure though lol sorry

yeah A = elc = log(I0/I) but i've no idea how that'll help cos the q doesn't provide any specific information for either of the formulas unless it's on the graph but idk what to do...
that's ok thanks tho

 

[carrotsss]

I ran into that a while back, the question in this copy is missing info for some reason iirc ill try find the version thats not missing it

 

[carrotsss]

here's the missing info

 

[SadCeliac]

View attachment 40048
here's the missing info

OH I REMEMBER THIS QUESTION we went through it in class lol
it's not too bad if I remember correctly? @carrotsss

 

[carrotsss]

OH I REMEMBER THIS QUESTION we went through it in class lol
it's not too bad if I remember correctly? @carrotsss

yeah its easy once u actually have the info

 

[Unovan]

yeah its easy once u actually have the info

For all colourimetrey questions do u assume that light before entering sample was 100% intensity

that's how ive been doing them but idk if flawed or if you need to state assumption or if its just complete wrong

 

[carrotsss]

For all colourimetrey questions do u assume that light before entering sample was 100% intensity

that's how ive been doing them but idk if flawed or if you need to state assumption or if its just complete wrong

yeah the final it’s like a percentage of the original

 

[Luukas.2]

For all colourimetrey questions do u assume that light before entering sample was 100% intensity

that's how ive been doing them but idk if flawed or if you need to state assumption or if its just complete wrong

Sort of...

The intensity before passing through the solution is defined to be I₀.

After passing through, the intensity is 42.6% of what it was before. That is, it has become I = 42.6% x I₀ = 0.426I₀.

You then know the value of the ratio I₀ / I = I₀ / 0.426I₀ = 1 / 0.426, on cancelling.

This is the same value that you will get if you assume that I₀ = 1 and thus take 42.6% as meaning I = 0.426.

The former approach is more mathematically sound, but the latter is probably fine given the Chemistry HSC attitudes towards mathematical rigour.

In either case, what you have is a number you can put into the calculator in order to find A.

 

[Masaken]

View attachment 40048
here's the missing info

ik how to do the q now i have that bit of info but i was genuinely so confused when i saw it cos it was nowhere to be found on that paper
ur a legend, thanks so much