alwaysneedadvice
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I don't know why I'm struggling with this one so much since it's just a converting units question (and I've done an unnecessary amount of these questions because our Chemistry teacher wants to torture us) but I keep getting wrong answers. The question is:

Amount= 2.0μmoldm^−3 × 75μm^3
I need to type in the units (standard SI units) and the value.
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lordaxil
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Amount= 2.0μmoldm^−3 × 75μm^3
I need to type in the units (standard SI units) and the value.
With these kind of questions, it always helps to break them down into chunks, for which typesetting helps. I assume that what you mean is 2.0~\mu \rm{mol}~\rm{dm}^{-3} \times 75~\mu m^{3}. Correct spacing is crucial to avoid confusion.

From the above, I can see immediately that you have a concentration (in \mu \rm{mol}~\rm{dm}^{-3}) multiplied by a volume (in \mu \rm{m}^{3}), which will give an amount of substance (in something proportional to \rm{mol}). That means your equation is dimensionally consistent, which is a good start! (many times they are not)

Next step after verifying dimensional correctness is to get rid of all the SI prefixes, like milli-, micro-, deci-, kilo-, etc. The key thing here is to be careful of the sign of the exponent for each unit. For example, \mu \rm{mol}=10^{-6}~\rm{mol} but \rm{dm}^{-3} = 10^{3}~\rm{m}^{-3} because the exponent is negative. This is very easy to get wrong, so double, triple check that step.

So, looking again at your expression above, I can see there is a \mu \rm{mol} and a product of three \mu \rm{m}. That means 4 lots of 10^{-6} = 10^{-24} to get rid of all the micros. To get rid of the \rm{dm}^{-3} we need a 10^{3} so in total we have 10^{-24} \times 10^{3} = 10^{-21} to get rid of all the prefixes.

So, the final answer in SI units is 2 \times 75 \times 10^{-21}= 1.5 \times 10^{-19}~\rm{mol}.
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alwaysneedadvice
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(Original post by lordaxil)
With these kind of questions, it always helps to break them down into chunks, for which typesetting helps. I assume that what you mean is 2.0~\mu \rm{mol}~\rm{dm}^{-3} \times 75~\mu m^{3}. Correct spacing is crucial to avoid confusion.

From the above, I can see immediately that you have a concentration (in \mu \rm{mol}~\rm{dm}^{-3}) multiplied by a volume (in \mu \rm{m}^{3}), which will give an amount of substance (in something proportional to \rm{mol}). That means your equation is dimensionally consistent, which is a good start! (many times they are not)

Next step after verifying dimensional correctness is to get rid of all the SI prefixes, like milli-, micro-, deci-, kilo-, etc. The key thing here is to be careful of the sign of the exponent for each unit. For example, \mu \rm{mol}=10^{-6}~\rm{mol} but \rm{dm}^{-3} = 10^{3}~\rm{m}^{-3} because the exponent is negative. This is very easy to get wrong, so double, triple check that step.

So, looking again at your expression above, I can see there is a \mu \rm{mol} and a product of three \mu \rm{m}. That means 4 lots of 10^{-6} = 10^{-24} to get rid of all the micros. To get rid of the \rm{dm}^{-3} we need a 10^{3} so in total we have 10^{-24} \times 10^{3} = 10^{-21} to get rid of all the prefixes.

So, the final answer in SI units is 2 \times 75 \times 10^{-21}= 1.5 \times 10^{-19}~\rm{mol}.
Thank you, I really appreciate it. I'm sorry I didn't format it well, I was kinda getting frustrated because I had done so many of these questions that were so similar but it kept saying I got this one wrong.
You're answer is right obviously, (so thank you again) but I'm still kinda confused at the you last paragraph which is how you finally get to the answer.

"I can see there is a \mu \rm{mol} and a product of three \mu \rm{m}."
Why is this relevant?

"That means 4 lots of 10^{-6} = 10^{-24} to get rid of all the micros."
Where does the 4 lots come from?

"To get rid of the \rm{dm}^{-3} we need a 10^{3}"
Why would you remove the \rm{dm}^{-3} wouldn't you have mol x cubic meter?

Maybe it's easier if I show you how I did it (so you can correct me) because I believe our methods are different. I started by converting the units. So micromol/dm^3 to mol/dm^3 would be 2x10^-6. Then cubic micrometer to cubic meter would be 75x10^-18 which turns into 7.5x10-17. I multiplied them together to get 1.5x10^-22.

Again, thank you!
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lordaxil
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(Original post by alwaysneedadvice)
Thank you, I really appreciate it. I'm sorry I didn't format it well, I was kinda getting frustrated because I had done so many of these questions that were so similar but it kept saying I got this one wrong.
No problem. Doing the formatting properly is time-consuming, especially if you don't know \LaTeX.

(Original post by alwaysneedadvice)
"I can see there is a \mu \rm{mol} and a product of three \mu \rm{m}."
Why is this relevant?
See below

(Original post by alwaysneedadvice)
"That means 4 lots of 10^{-6} = 10^{-24} to get rid of all the micros."
Where does the 4 lots come from?
1 lot comes from the micromol, and 3 lots from the micrometres (because they are multiplied together to get cubic micrometres)

(Original post by alwaysneedadvice)
"To get rid of the \rm{dm}^{-3} we need a 10^{3}"
Why would you remove the \rm{dm}^{-3} wouldn't you have mol x cubic meter?
Putting in the 10^{3} only removes the deci- prefix, it leaves behind the \rm{m}^{-3}. I dealt with the mol unit above.

(Original post by alwaysneedadvice)
Maybe it's easier if I show you how I did it (so you can correct me) because I believe our methods are different. I started by converting the units. So micromol/dm^3 to mol/dm^3 would be 2x10^-6. Then cubic micrometer to cubic meter would be 75x10^-18 which turns into 7.5x10-17. I multiplied them together to get 1.5x10^-22.
Ah - I see where you went wrong. You forgot to remove the deci- prefix, because when you multiply \rm{dm}^{-3} by \rm{m}^{3} you get an extra 10^{3} which makes your answer same as mine.
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alwaysneedadvice
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o

(Original post by lordaxil)
No problem. Doing the formatting properly is time-consuming, especially if you don't know \LaTeX.


See below


1 lot comes from the micromol, and 3 lots from the micrometres (because they are multiplied together to get cubic micrometres)


Putting in the 10^{3} only removes the deci- prefix, it leaves behind the \rm{m}^{-3}. I dealt with the mol unit above.



Ah - I see where you went wrong. You forgot to remove the deci- prefix, because when you multiply \rm{dm}^{-3} by \rm{m}^{3} you get an extra 10^{3} which makes your answer same as mine.
Sorry for the late reply. Yes, thank you! I understand now, I really appreciate your help.
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