[Sillybananas]

Hello everyone!!! New to chemistry and I'm having trouble with some conversion help (I know, I'm kinda dumb ): ).
Anyway, I'm not sure what I'm doing wrong. So please, may I get some help? Thanks!

I've made the link public on my google drive with the 2 problems I'm confused with.

https://drive.google.com/a/students.clark.edu/folderview?id=0Bx46YCIAmiYhU0tCUDFLZkF4eGM&usp=sharing

For #12, can you put it with the right amount of sig figs? I'm not really sure how to apply the sig fig rules with conversion so I'll be glad to read any teaching and explanations. Thanks!

[Corribus]

Welcome to the forum, Sillybananas. Not understanding unit conversions does not make you dumb. It means you need help learning the material. Which is not possible for us to do without seeing what exactly you are having trouble with. This is why the forum rules require you to show your work. It's for your benefit as much as for ours.

Also, it's not possible for everyone to access external websites. Therefore, you have a better chance of receiving help if you post the questions in the forum instead of linking to documents elsewhere.

[Sillybananas]

Welcome to the forum, Sillybananas. Not understanding unit conversions does not make you dumb. It means you need help learning the material. Which is not possible for us to do without seeing what exactly you are having trouble with. This is why the forum rules require you to show your work. It's for your benefit as much as for ours.

Also, it's not possible for everyone to access external websites. Therefore, you have a better chance of receiving help if you post the questions in the forum instead of linking to documents elsewhere.

Of course no problem!!!!

Okay so for question #2, the question asks if there is anything wrong with the set-up of the conversion problem of:
16.5*(3.154*10^10 ms/1 year)=5.20*10^11 milleseconds

I answered that there was nothing wrong with the problem set up but there is a formatting error. Having barely any basic knowledge on conversion, I don't see what's the problem. Years is cancelled by the denominator then you multiply the 16.5 with the 3.154*10^10.

The other answers that can be a possibility are shown in the picture I'll be uploading.

Now, with #12, I was very confused. The question asks to find the density in g/mL of a 1.80*10^-2 kg block of metal with the dimensions of 0.4981 in x 0.531 in x 0.5839 in. How I started was by converting 1.80*10^-2 kg to grams which I got 18 g with. Then I multiplied the dimensions to get 2.53075841 (I had a calculation of 2.3 mL in the picture but I redid it and 2.531 idk why). After getting the dimension and volume, I put it in the mass/volume formula to get 7.11 g/mL.

Tin has the closest density to that so would tin be right? And how would I apply the significant rules on here? Since 1.80 from the 1.80x10^-2 has the lowest amount of significant figures (apart from nickel having 8.9 g/mL as a given), will I have an answer with 3 significant figures?

[mjc123]

Like you, I see nothing wrong with the formula in question 2.

For question 12, I agree with your answer of 7.11 g/ml, but you seem to have inserted an answer of 7.71 - just a typo? If it is 7.11, which metal is closest?
The lowest number of significant figures in the problem is 3, for both 1.80 x 10^-2 and 0.531, so you should give your answer to 3 sig figs. But be careful; you say you converted 1.80 x 10^-2 kg to 18 g. The correct conversion is 18.0 g, to 3 sig figs - it is between 17.95 and 18.05, not between 17.5 and 18.5. It makes no difference to the calculation result, but helps you keep track of the significant figures. That final 0 (if used correctly!) is significant.

[Sillybananas]

Like you, I see nothing wrong with the formula in question 2.

For question 12, I agree with your answer of 7.11 g/ml, but you seem to have inserted an answer of 7.71 - just a typo? If it is 7.11, which metal is closest?
The lowest number of significant figures in the problem is 3, for both 1.80 x 10^-2 and 0.531, so you should give your answer to 3 sig figs. But be careful; you say you converted 1.80 x 10^-2 kg to 18 g. The correct conversion is 18.0 g, to 3 sig figs - it is between 17.95 and 18.05, not between 17.5 and 18.5. It makes no difference to the calculation result, but helps you keep track of the significant figures. That final 0 (if used correctly!) is significant.

Thank you for your *delete me*!! Yes, we got the last answer right! (zinc was the nearest to 7.11, obviously). Anyway, for #2, I redid it and it turns out that there was something wrong. It was that the conversions are shoved together, attempting to solve it all in one step. My professor noted how the conversion between years and milliseconds take more than one step. Is this true for all situations? If not, which situations would that answer not apply? Thanks!

[mjc123]

There is only one conversion in that formula, from years to milliseconds. Provided you know the conversion factor (3.154 x 10¹⁰ ms/yr), that's all you need. If you don't know it, and are working it out step by step (seconds per hour, hours per day etc.) it is best to write it out in full to minimise the possibility of mistakes. So 1000 ms/s * 60 s/min * 60 min/hr... (But do you know exactly how many days in a year? It may be better to use a standard tabulated value of s/yr.)

[DrCMS]

3.154 x 10¹⁰ ms/yr - is only true for a year with 365 days.   During the 16.5 years of the question there will be 4 leap years so the average is 365.25 days per year giving 3.156 x 10¹⁰ ms/yr.