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# Giancoli Solutions on Video

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## Giancoli 6th Edition, Chapter 8, Problem 43

Isn't it 1.4x10^3?

## Giancoli 7th Edition, Chapter 13, Problem 1

Hi Greg, thank you for your question. I think the difference between what I've done here and what is common in chemistry is that here I've used "atomic mass", whereas in chemistry "molar mass" is more common. When looking at the periodic table of elements, the mass given for each element can actually be interpreted either way. For gold, you can say it has an atomic mass (technically, this is a "relative atomic mass" which takes an average of all naturally occurring isotopes weighted by how common each isotope is) of $196.966569 \textrm{ u/atom}$ or you can read it as $196.966569 \textrm{ g/mol}$, depending on your preference. Since I went with u/atom, I needed a conversion factor to turn the atomic mass unit (that's the "u") into kilograms, so that the "kg" in the numerator would cancel with "kg" in the denominator. The is the number you asked about: $1.66 \times 10^{-27} \textrm{ kg/u}$, which is just a constant you can look up in the front cover of the textbook under the section "Fundamental Constants" and it tells you how many kilograms are in one atomic mass unit.

I can't really say why your approach didn't work since I don't understand what you mean by "I took it through the mole". In any case, the chemistry approach definitely works. If the video took that approach, the atomic mass would be replaced with "molar mass", which is the same number but just different units, and then divide it by Avogadro's number.

Probably the confusing thing about this question is having fractions within fractions. Perhaps I should have written it on one line by multiplying by the reciprocal of the denominator. When checking your work make sure you're putting things in the right place within the fractions.

Hope this helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 13, Problem 1

Hi, The process on this is a little confusing to me. In chemistry we had similar problems and we found them through a similar factor label method. I took 27.5 g of gold I took it through the mole, and divided by the atomic mass (199.9665). Then I multiplied by Avogadro number 6.022 *10^23. I did the same thing for the other then divided, however I got a very different answer. How come this way does not work? Also where did you get the 1.66*10^-27 kg/u?

Thank you!

Greg.

## Giancoli 7th Edition, Chapter 19, Problem 35

Hi jlara4606, thanks for pointing out the textbook mistake, which is indeed a mistake. To check the work in this question, or any "systems of equations" problem, substitute the answers back into the original. Continuing on to find $I_2$ and $I_3$ we can substitute into the original three equations.

Equation 1: $24-23I_1-29I_2 = 0$

Equation 2: $18-23I_1-28I_3 = 0$

Equation 3 (rearranged slightly for convenience): $I_1 - I_2 - I_3 = 0$

Here are the results if $I_1 = 0.5642$:

I1 0.5624

I2 0.38156056

I3 0.18094464

Equation 1: -0.00045624

Equation 2: -0.00164992

Equation 3: -0.0001052

All three equations are expected to be zero, and it's close enough since there was some intermediate rounding in my work.

Compare that if $I_1 = 0.71$, then:

I1 0.71

I2 0.264499

I3 0.059706

Equation 1: -0.000471

Equation 2: -0.001768

Equation 3: 0.385795

And it's the result for Equation 3, $0.385795$ which is the red flag that $0.71$ is not correct, since it's supposed to be zero.

The take-away message is the strategy for checking your work with systems of equations by substituting the answers back into the original equations and check that the equations equal what they are supposed to (zero in each case here).

All the best with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 5, Problem 4

@bbailey1956, I almost forgot to comment on your final answer. I've checked the work, and the answer appears correct, and I can't think of where the mistake would be in your work to have an answer different by a factor of ten, so you'll just have to inspect your work carefully.

All the best,

Mr. Dychko