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Question:how many electrons , neutron and protons are in a gold atom? where dose gold fit into the periodic chart? and where can i find a picture of golds atomic structure including the shells and valence electron. and one more qustion dose the elemnt gold bond with other elemnts?
Answers:http://en.wikipedia.org/wiki/Gold and http://www.webelements.com/gold/
Answers:http://en.wikipedia.org/wiki/Gold and http://www.webelements.com/gold/
Question:The element gold has properties that have made it much sought after through the ages. A cubic meter of ocean water has 6 106 g gold. If the total mass of the water in Earth's oceans is 4 1020 kg (Assume that the density of seawater is 1 g/cm3.)
and then:
How many liters of seawater would have to be processed to recover 1 kg of gold (which has a value of about $8850 at 2001 prices)?
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:(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.
Question:Many metals pack in cubic unit cells. The density of a metal and length of the unit cell can be used to determine the type for packing. For example, gold has a density of 19.32 g/cm3 and a unit cell side length (a) of 4.08 . (1 = 1 x 108cm).
a. How many gold atoms are in 1 cm3?
b. How many unit cells are in 1 cm3?
c. How many gold atoms are there per unit cell?
Answers:Since you are given the density, you can skip the crystallographic bs. for part a. Find the gatoms 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:Since you are given the density, you can skip the crystallographic bs. for part a. Find the gatoms 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.
Question:1/25 oz means hpw many grams?
Answers:Conversion : 1 gram = 0.0321507466 troy ounces
Answers:Conversion : 1 gram = 0.0321507466 troy ounces