A ring -pure gold has the mass of 50.00 grams & placed in a cylinder. The water in the cyl. will rise 2.9 mL?

Question

This is a true or false question. The graduated cylinder is 100mL and partially filled with water, but doesn't say how much.

Answer 1

They don’t give you all the information. You need to also know the density of Gold (19.3 g/ml) for pure gold.

The math (50.0 g of gold) / (19.3 g/ml) = 2.59 ml of gold.

This is the Archimedes conundrum. How do you tell if a gold artifact (a crown in his case - a ring in yours) is pure gold without destroying it? When Archimedes figured out this answer he jumped out of his bath and ran through the streets yelling “Eureka!”.

In your case the ring is not pure gold, but rather alloyed gold. Now the question is what alloy (or carot), if you assume the alloying metal is copper?

BTW it is OK to rate this as the best answer if the spirit moves you.

Answer 2

Forever, very nearly. You would possibly not be in a position to dissolve gold in natural water. There are just a handful of compounds in the market on the way to dissolve gold. The such a lot identified one is known as "aqua regia" that is a part-and-part blend of hydrochloric acid and nitric acid. And no, Coca-Cola is not going to dissolve gold both.

Answer 3

Hi Cara,

I guess James already got to your question first. Sorry about that. I would've said the same thing: you need to know the density of gold first.

Since he didn't state it in his post, I'm telling you the answer is false, based on his calculations.

Keep up the good work!