how many electrons in the gold
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Answers:http://en.wikipedia.org/wiki/Gold and http://www.webelements.com/gold/
Answers:(1g / cm^3)*(100^3 cm^3 / 1 m3) = 10^6 g of water / m^3 of water, or 10^3 kg water / m^3 of water. Now you can take the mass of ocean water, and see how many cubic meters it is in volume. 41020 kg / (10^3 kg / m^3) = ??? m^3. Now you know how many m^3 of water is in the ocean, so multiply by the amount of gold per 1 m^3. ??? m^3*6x10^-6 = XXX grams of gold, easilly converted to kg of gold. For this part, you need to know 1cm^3 = 1mL in terms of volume, and of course 1000mL = 1L. Then it's just a unit conversion problem: 1 kg = 1000g = (6x10^-6 g / m^3)*??? m^3, where you have to solve for ??? then convert that to L.
Answers:Since you are given the density, you can skip the crystallographic bs. for part a. Find the g-atoms in 19.32 g and multiply that by Avagadro # to get the answer. You should be more specific as to the cell type- is it body centered cubic or face centered cubic? Assuming neither, the number of cells equals the number of atoms. This is so because each atom is common to 8 unit cells, so each is counted as 1/8th in a given cell.
Answers:Conversion : 1 gram = 0.0321507466 troy ounces