anonymous
  • anonymous
what are the errors in the measured values g=9.910 m/s^2 amd 9.805? true value 9.797 for 9.910 i got 0.113 and for 9.805 i got 0.008. i know i need for significant figures but i got three. so what do i do
Chemistry
schrodinger
  • schrodinger
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Xishem
  • Xishem
Why do you feel like you "need" four significant figures, exactly? You calculated the answers with the correct number of significant figures. I see nothing wrong here so far.
anonymous
  • anonymous
9.910-9.797=0.113 ( since I started with 4 sigs figs , dont i need to retain it in my answer)
Xishem
  • Xishem
Well, no. You keep as many sigfigs as possible throughout your calculation. In this case, you were only able to keep 3 sigfigs through the above calculation, so you are only justified in keeping 3 sigfigs in your final answer.

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anonymous
  • anonymous
so i dont need to to write zeros after the decimal place to give it sig figs
Xishem
  • Xishem
You don't know the value to 4 significant figures, so no. However, keep in mind that you should NOT round intermediate calculations. Keep as many sigfigs as possible for any intermediate values. Do not round until the end. A good way to keep track of how many significant figures you can round to is to do something like this. For example, if you have 9.683457 * 0.12, you are only justified in keeping 2 significant figures, but you don't want to completely truncate the rest. You want to keep them for calculation, but you need to remember how many sigfigs you can round to in your final result. So you would write it like this...\[9.683457*0.12=1.1_{6201484}\]The regular numbers represent sigfigs, while the subscript numbers are insignificant digits. This way, you don't round too early, but you can keep track of how many significant figure EACH value holds. Then, once you are ready to round your final answer, look through your work, and see how many significant figures the least significant value has.
anonymous
  • anonymous
okay thank you so much
Xishem
  • Xishem
No problem. Feel free to ask any more questions you have, and I'll try to clear things up (:.
anonymous
  • anonymous
thanks actually i do have another question. by any chance do you know what synthetic intermediate means
Xishem
  • Xishem
Have you by chance heard the same term be called "reactive intermediate" or just "intermediate?" I've never heard the term synthetic intermediate.
anonymous
  • anonymous
no but could please explain
Xishem
  • Xishem
I believe that they all mean the same thing. Basically what it is... When a given reaction occurs, the reactants generally don't DIRECTLY form products. For example, in the following reaction...\[2NO_2 + F_2 = 2 NO_2F\]You might assume that all of the product formed is going to be formed by the NO2 and the F. However, more likely, it follows some specific reaction mechanism. An example of a reaction mechanism for this reaction would be...\[NO_2+F_2 \rightarrow NO_2F + F\]\[NO_2+F \rightarrow NO_2F\]As you can see, the reaction takes place in two steps. Now, an intermediate (or "synthetic intermediate" or "reactive intermediate") is a species which occurs on the products side of one of the elementary reactions, and on the reactants side of another. In other words, it's a species which shows up in the overall reaction mechanism, but doesn't actually show up in the final equation, which in this case is...\[2NO_2 + F_2 = 2 NO_2F\]As you can see, the fluoride anion doesn't show up in the overall reaction -- this is because it's an intermediate. If you were to add the two above elementary reactions together, cancelling things that occur on both sides, you would end up with the overall reaction. Make sense?
anonymous
  • anonymous
yah it does.. so if a textbook says that an ethers are not good synthetic intermediates what does that mean
anonymous
  • anonymous
i understood ur explnation.forget about what i asked

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