[hyeminnie]

can anyone help me with these density type problems?

wire is often sold in pound spools according to the wire gauge number. that number refers to the diameter of the wire, how many meters are in a ten pound spool of 12 gauge aluminum wire? a 12 gauge wire has a diameter of 0.0808 in. aluminum has a density of 2.70 g/cm^3. (V= pi*r^2*l)
went over in class.

and

a metal slug weighing 25.17g is added to a flask with a volume of 59.7 mL. it is found that 43.7 of methanol (d=.791 g/mL) must be added to the metal to fill the flask. what is the density of the metal?

[hyeminnie]

can anyone help me convert this problem, too? it's a lot easier, but i still don't get the coversions.

what would be the volume (in mililiters) of a 3 ounce piece of gold? (16 oz=1 lb) and i think you're supposed to convert the 3 oz into pounds then to grams to get the mass. i just don't know the density to get the volume.
nvm. got the answer.

[Avex]

For

a metal slug weighing 25.17g is added to a flask with a volume of 59.7 mL. it is found that 43.7 of methanol (d=.791 g/mL) must be added to the metal to fill the flask. what is the density of the metal?

I'm only doing this problem because of my time constraints

Now, it is important to think about how to do the problem before you even attempt it. Try to understand logically how to do the problem and what is going on based on the given information. I know it seems like common sense, but sometimes, we don't think in terms of common sense in these situations.

Now, I looked at the given info. It says that 25.17 g of something is added to a volume of 59.7 mL. Also, it says that you need to add some methanol (43.7 g) to fill up the entire flask.

So basically, the volume of the metal slug and the volume of the methanol will equal 59.7 mL.
However, at this point, we can't find the volume of the metal slug because we would need its density in order to do that, based on the information that is given.

But the problem is asking for the density of the metal slug. How should we get to the density of the metal slug?!?!

Well, with the information given, we look around and notice that this problem gives us the amount of methanol and the density of it. The problem says we need 43.7 g of methanol. It also tells us that the density is .791 g/mL.

My idea is, we will attempt to find the volume of the methanol that needs to be filled. Then, we subtract this volume from the total volume of the flask (59.7 mL) to get the volume, which will give us the volume of the metal slug. After getting the number, we use the volume of the metal slug to find its density. Watch:

43.7 g * (1 mL/.791 g) = 55.2 mL

The grams cancel out, giving us the volume of the methanol that needs to fill up the flask.

Volume of entire flask - volume of methanol needs to be filled = volume of metal slug
59.7 mL - 55.2 mL = 4.5 mL

Since density equals mass over volume (logic!!!):
We divide the given mass of the metal slug by the amount of volume the metal slug will be filled up with.

25.17 g / 4.5 mL = 5.6 g/mL <<<< The density of the metal slug

I had to work backwards in this problem. It sounds weird, but think logically about whats up during the problem. Chem is pretty hard to think about, but try to think about it, even if it doesn't make sense.

[hyeminnie]

thanks for helping me with this.
chem would be easier if my teacher would let us solve the problem the way we want instead of putting everything over one or making it equal to something else.