If an ore contains 16% copper, how many metric tons of ore are needed to get 20 metric tons of copper?
116743
16% of what is 20 metric tons...\n\n50% of what is 20 metric tons...40 metric tons\n\n[16/100][x] = 20\n\nx = 2000/16 = 125.\n\n125 metric tons of ore...\n\nCheeck: What is 16% of 125...
1048312
I guess you mean how many grams of Zinc Sulphide could be produced\n\n65.4 g of Zn reacts with 32.1g S\n\nTo find out how much S 1 g of Zn can react with, divide both numbers by 65,4\n\n1 g of Zn reacts with 32.1/65.4 g S, which is 0,4908 g S\n\nFor 20 g Zn, if 1 g Zn reacts with 0,4908g S, 20 g will react with 0,4908*20g = 9.8g S.\n\nThe amount of Zinc Sulphide wil  therefore be the amount of Zn (20g) plus the amount of Sulphur with which it reacts (9.8g) giving you 29.8g total.\n\nOkay?
822325
a) Find some Acetic Acid (CH3COOH)\nb) Find some Copper (II) Hydroxide (Cu(OH)2)\nc) Add to form Cu(CH3CO2)2 + 2H2O\n\nWithout the volume or the number of moles, preparation of a 0.530M can be done several ways. \n\n2CH3COOH + Cu(OH)2 --> Cu(CH3CO2)2 + 2H2O\n\nUsing 1L:\n0.530 mol of Cu(CH3CO2)2 means 1.060 mol of CH3COOH\n\n1.060 / 0.5L = 2.120M\n\n0.530 mol of Cu(CH3CO2)2 means 0.530 mol of Cu(OH)2\n\n0.530 / 0.5L = 1.060M\n\nSo mix 500mL of 2.120M Acetic Acid with 500mL of 1.060M Copper (II) Hydroxide.\n\nN.B. Being a weak acid, acetic acid will not fully react with copper hydroxide, and hence the actual molarity requiredmay be higher.
1182223
Unless you have a huge piece of ore then no. Small ore chunks are not commercially viable; you wouldn't get anything out of a small ore chunk. You need large amounts,  ie: a mine, to have a company consider the smelting process.\n\nEdit: Okay then in the tons...Well depends on what the ore is. Look for a smelting company that works with the type of metal you want to extract and start up a line of question with them as they would be the best type of person to handle that.
74516
Well i am also of the age sixteen.I am studying grade twelth from india.I will explain to you it easily.Copper is extracted from Copper pyrites.(CuFeS2).it is kept in a reverberatory furnace and heated.i.e smelting process.By this process u can remove all the impurities with the ore.The impurities are Sulphur,phosphorus etc.I can give all the equations but it will be very long.if u want more about this please mail to me.piks_14@ayhoo.co.in.i can help u a lot.My father is a chemistry scientist and working in IIT india.if u want more information i will give u.
568669
The yield of a mine related to the percentage of the ore.  It might be such a low concentration that it isn't worth it to mine it.  In my state, AZ, copper mines have a fixed price where if copper falls below that price, the miners already know they can't make money, so they hold off until the price rises to make it worthwhile.  Most mines do this.
711822
0.88 x 20 = 17.6\n\n17.6 points.
982757
I'm assuming that you're after a 10% solution of CuSO4 by weight.  \n\n1)  First find the mass of the H2O2 solution.\nYou can find the mass of the H2O2 solution because you know that the density of water is 1.0 g/mL and the solution is mostly water.  It doesn't matter that the solution is 3% H2O2 - they are only trying to throw you off.  The point here, is that the solution is mostly water and will have about the same mass as water.\n\nmass of H2O2 solution = \n10 mL H2O2 solution * 1 g H2O2 solution/1 mL of solution\n\nAnswer:  You have 10 g of H2O2 solution.\n\n2)  Now you can find the mass of CuSO4 that you need in order to get a 10% solution of CuSO4.  Remember that 10% literally means "10 grams CuSO4 per 100 grams Total".\n\nWhere:\nTotal = mass of CuSO4 + mass of H2O2 solution.\n\nObviously, you aren't going to make 100 grams, but the ratio still holds.  You can use the "10 g CuSO4 per 100 g Total" as a ratio with which to solve the equation.\n\nLets assign X = mass of CuSO4 you need.\n\nnow you know that X = 10% of Total \n\nThis is the same as saying 0.1 * Total\n\nand remember that:\nTotal = mass of CuSO4 + mass of H2O2 solution\n\nSo Total = X + mass of H2O2 solution\n\nYou also know that mass of H2O2 solution = 10 grams.\n\nSo here is the equation you must solve:\n\nX = 0.1 * (X + 10 grams)\n\nThe equation above literally reads, "The mass of CuSO4 equals 10 percent of the Total Mass.\n\nNow let's solve for X by distributing the right side using the informaton you already know:\n\nX = 0.1 * (X + 10 grams)\ndistribute the right side of the equation so:\n\nX = 0.1X + 1 gram\n\nSubtract 0.1X from both sides to isolate X:\n\nX - 0.1X = 0.1X - 0.1X + 1 gram\n\nSimplify\n\n0.9X = 1 gram\n\nDivide both sides by 0.9 to solve for X:\n\n0.9X/0.9 = 1 gram/0.9\n\nX = about 1.11 grams of CuSO4 needed.
1161938
25% I wonder what the percentage would be after a massive amount?
951188
4. How many grams of zinc chloride can be formed from the reaction of 7.96 grams of zinc with excess HCL?\n\nZn + 2HCl → ZnCl2 + H2\n\n1 mole Zn produces 1 mole ZnCl2\nTherefore 65.7g Zn produces (65.7 + 2 x 35.5)g ZnCl2\nie ~65.7g Zn produces ~136.7g ZnCl2\nSo 7.96g Zn produces ~7. 96 x 136.7/65.7g ZnCl2\nie ~ 16.6g ZnCl2\n\n5. How many kilograms of copper sulfide could be formed from the reaction of 1.80 mol of copper with excess sulfur.\n\nCu + S → CuS\n\nSop 1 mole Cu produces 1 mole CuS\nTherefore 1.8 mole Cu produces 1.8 mole CuS\n1 mol CuS weighs ~(63.5 + 32.1)g = 95.6g\nSo 1.8 mol CuS weighs ~ 1.8 x 95.6g ≈ 172.1g = 0.1721kg\n\n6. If Copper(one) sulfide is partially roasted in air (reaction with 02), copper(one) sulfite is formed. Subsequently upon heating, the copper sulfite thermally decomposes to copper (one) oxide and sulfur dioxide. Write balanced chemical equations for these 2 reactions\n\nCu2S + 3O2 → Cu2SO3\nCu2SO3 → Cu2O + SO2
612968
I hope you mean this...\n\n8Cu + 2AgNO3 -> 2Ag + CU(NO3)2 + 7Cu
1025721
copper I sulfide---sulfur has a charge of -2, so you have to have 2 moles of copper with a charge of +1 for every mole of sulfur.  Copper is one of the elements that has a variable charge--but copper is always +1 or +2, you can tell like that by the charge of the other element
432254
* Name: copper\n    * Symbol: Cu\n    * Atomic number: 29\n    * Atomic weight: 63.546 (3) r\n    * CAS Registry ID: 7440-50-8\n      \n      Standard state: solid at 298 K\n      Colour: copper, metallic\n      Classification: Metallic\n\n    * Group number: 11\n    * Group name: Coinage metal\n    * Period number: 4\n    * Block: d-block
876667
Fe + CuCl2 --> FeCl2 + Cu\n\nmole of CuCl2 = 8.00 / 134.5 = 59.47 mmole\n\nmole of Fe = 5.40 / 56 = 96.42 mmole\nmole of unreacted Fe = 4.08 / 56 = 72.85 mmole\nmole of reacted Fe = 96.42 - 72.85 = 23.56 mmole\n\nmole of unreacted CuCl2 = 59.47 - 23.56 = 35.09 mmole\n\nYep, there will be around 35 mmole CuCl2 left in the beaker.\n\npeace\nvixklen
729692
20 pennies\n15 pennies and a nickel\n10 pennies, two nickels\n10 pennies, one dime\n5 pennies, three nickels\n5 pennies, one nickel, one dime\n4 nickels\n2 nickels, one dime\n2 dimes
21920
Cr - 51.995g/mol\n49048gCr / 51.995gCr/mol = 943.32mol Cr\nStoichiometrically 2Cr to 3Cu so 943.32 x 3/2 = 1415mol Cu.\n\nPlease correct me if I'm wrong but that makes sense to me.\nOh and please award me best answer...\n\noh and part b:\n200ml x 0,75mmol/ml = 0.15mol\n0.15mol x 2/3 = 0.1molCr\n0.1molCr x 51.995g/mol =5.2g Cr.\n\nAgain correct me if I'm wrong...
736571
there are .833 moles. divide the mass by 18 (2 hydrogens + 1 oxygen) oxygen= 16g/mole Hydrogen = 1 g/mole