Chemistry

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Unfortunately I was hit by the sweep of illnesses going around due to changing whether and missed school today. Thus I missed the notes in Chemistry leaving me completely confuzzled about tonights homework on Dimmensional analysis or Unit Factoring. I called a friend but she isn't the note-taking type of person and was only able to tell me to do the problems in a table (??) so if someone could do an example problem or two for me I think I can handle the rest.

It says to perform the following metric conversion using dimnesional analysis and one conversion factor.

3.50 nm - ? m

Another said to do the same but with two conversion factors.

23 dL - ? cL

It would really help me out if someone could work these out for me. I'm not trying to ask someone to do my homework, just show me the right way to complete it.

Thanks, Tammy

  • Chemistry -

    Dimensional analysis is the art (or act) of multiplying some number by a factor to convert to another unit. And the units you don't want to keep cancel but the unit you want to keep stays. Here is the first one.
    The factor for converting nm (nanometers) and meters is 1 nm = 10^-9 m; therefore, this factor can be written one of two ways.

    1 nm/10^-9 m OR
    10^-9 m/1 nm AND

    the conversion then is
    3.50 nm x factor = ?? meters. Note that the only thing you need to do is to know the factors (which you should have done with the metric system) AND use the correct one of the two factors. BUT you always know which factor is correct because the units you don't want will cancel.
    3.50 nm x (10^-9 m/1 nm) = ?? m
    Note that the unit of nm cancels and leaves the unit of m so the unit we don't want to keep cancels and the unit we want to keep (meters) stays.

    IF you use the wrong factor, the units will NOT cancel, like so:
    3.50 nm x (1 nm/10^-9 m) = ?? nm^2/m and that certainly is not the unit we want in the answer. So we know the first one is correct and the second one is not. I hope this gets you started.

    For the other one, I will leave most of the discussion out.
    If they want two conversion factors, I suggest we convert dL to L, then L to cL.
    We know 10 dL = 1L and we know 100 cL = L so we do the following:

    23 dL x (1 L/10 dL) x (100 cL/1 L) = ?? cL. Note that dL in the numerator of the first term cancels with the dL in the denominator of the second term and L in the numerator of the seccond term cancels with L in the denominator of the third term to leave what we want, which is cL.
    That gives us 230 cL.
    Check my work.

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