What current, in A, would be needed to produce 10. kg of pure Cu metal[63.546 g/mole], per day (in 24 hrs)?
491209
175.76 Amperes...\nusing m=z*i*t\nz=63.546/96500\nt=24*60*60\nm=10*1000
870147
You really can't unless the metal in question in tested in an actual circuit. \n\nMany metals (nichrome, for instance) increase greatly in resistance as the current heats them up - and, conversely, some decrease in resistance when heated (many semiconductors, for instance).\n\nThe equation for watts is P=IE, where P is the power or watts, I is the current or amps, and E is the applied voltage. \nOhms Law (R=E/I) will determine the actual resistance of the metal once the above quantities are known.
813218
Try the following website if this may be useful :\n\nwww.conexindia.com/copper_casting_alloys_datachart_CDA/copper_casting_alloys
83000
The net ionic equation is:\n\nFe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s)
929154
copper can't produce electricity. it conducts electricity
790092
24g = 4g magnesium + 20g other elements\n1440/24 = 60 so need 60 times the amount\n4g * 60 = 240g magnesium\n20g * 60 = 1200g other elements\n240g + 1200g = 1440g\n\nSo 1200g of other elements
882235
In an electrolytic cell you can plate most any metal that doesn't quickly react with the solvent and its solutes.  The force of the copper ions gaining electrons at a negative charged zinc electrode is certainly possible.\n\nSince the copper would receive two electrons and go from a +2 to a zero oxidation state, you would say that it is being reduced.
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
894420
Here you are:\n\nCu2S + 12HNO3 => Cu(NO3)2 + CuSO4 + 10NO2 + 6H2O
963455
Copper C-145000\n\nCopper  Includes Ag + Te  \nPhosphorus  Other deoxidizers may be used  \nTellurium  0.4 - 0.7
827623
F*m = μ*Joules→ m = .25*40000000/1000 = 10000m = 10 km
52130
E=mc∆T is the equation you want to use for this situation, where E is energy, m is mass, c is the specific heat, and ∆T is the change in temperature.  Now, the heat lost by the alloy is equal to the heat gained by the water, right?  So look what we can do:\nmc∆T=mc∆T\nwhere the left side of the equation is talking about the metal, and the right side of the equation is talking about the water.  Let's start pluggin in values.  We'll leave the left c value as a variable because that's what we're solving for.  \n36.0 g *c*(178 oC -30.6 oC)=185.0 g*(4.184 J/g*K)*(30.6 oC-25.0 oC)\nThe trickiest thing about that step is the temperatures: the metal started at 178 and goes down to 30.6; the water starts at 25 and goes up to 30.6.  \nNow evaluate that equation and solve for c. \nc=.817 J/g*K
328575
reduction:  2e- + Cu2+ --> Cu      Eo = +0.34 V  cathode (+)\n\noxidation:   Zn   --> Zn2+ + 2e-   Eo = +0.76 V    anode (-)\n________________________________________\n\nZn  + Cu2+ --> Cu  +  Zn2+         Eo = +1.10V therefore cell spontaneous as written.
916549
2Cu + O2 ---> 2CuO,          \nThis is copper oxide or cupric oxide or copper( II )oxide\n\n4 Cu + )2---> 2Cu2O\nCopper also combines with oxygen as Cu2O, cuprous oxide , or copper(I) oxide
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...
446103
Part A\n\nReduction: 3e^- + 4H^+ + NO3^- = NO + 2H2O    Eo = \nOxidation:                 Cu  = Cu^2+ + 2e^- Eo =\n\nMultiply the reduction half equation by 2 and the oxidation half equation by 3 (so that you have 6e^- in both) then add together to get:\n\n3Cu + 8H^+ + 2NO3^- = 2NO + 3Cu^2+ + 4H2O\n\nNow Eo cell = Eo cathode - Eo anode = 0.96 - 0.34 = +0.62V\n(Note: you do NOT multiply the Eo values by the number of electrons in the half equations because Eo is a voltage, measure of electrical pressure.\n\n\nPart B\n\nThe Nernst equation can be found at:\n\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/nernsteq.html
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.