Give full description of the thermodynamic $ kinetics of lead reduction $ smelting?
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OK, here goes---no chemical formulas--look that up in your chemistry book.\n\nLead is mined.  The ore comes out of the mine as a Lead Oxide.  Chunks of ore range from about the size of a loaf of bread on down.  Ore goes through jaw crushers to reduce in size.  Golfball size ore goes through a rod mill. (Large revolving drum full of steel rods)  Comes out pea sized.  Ore goes into a ball mill. (Large revolving drum full of steel balls)  Comes out the size and consistency of talcum powder)   The ore/water mixture is called a slurry.  The slurry comes out of the ball mill and goes into flotation chambers where a mixture of water and foaming agents are being agitated.  The lead particles in the slurry attach themselves to the bubbles created by the foaming agent.  They rise to the top of the tank where the foam is scraped off by paddles.  Waste rock called tailings is pumped in a slurry form to a disposal site.  The lead bubble foam is transported to a dryer.  The dryer removes excess water and leaves the final product of the mine and mill--lead concentrate.  (lead oxide)\n\nThe concentrate is hauled to the smelter.  If it was put into the furnace in its current state it would go right up the chimney---the lead is the consistency of talcum powder.  In order to melt the lead, we have to hold it in place long enough to melt so that it doesn't all just blow out the chimney.  This is done by a process called sintering.  Lead concentrate is added to crushed limestone, coke and other chemical reagents.  It is then run through the sintering machine.  The sintering machine roasts the mixture for 20 minutes and when the sinter comes out the end of the machine it resembles the klinkers that you'll see come out of an old coal fired furnace.  The roasting process causes all the ingredients to mold together into a solid chunk---called sinter.  The sinter comes out of the machine in huge chunks---as big as a car---and is run through a crusher to break it up into manageable size--about the size of a basketball on down.  The sinter goes into a storage bin.\n\nNow, its time for the furnace.  The sinter is mixed with more powdered limestone, iron ore, coke, coke breeze, lead scraps from spills.  This mixture is fed into a top feed furnace.  As the mixture falls down into the bottom of the furnace it encounters the fire.  Pure oxygen is pumped into the furnace, feeding the fire.  The coke and coke breeze burn creating enormous amounts of heat.  The sinter holds the lead in place long enough for it to melt and flow into the bottom of the furnace.  The sinter is then consumed by the fire as its remaining ingredients are partially cooked coke.  When the bottom of the furnace fills up with molten lead, it is tapped by the furnaceman.  The furnaceman will drain off a huge kettle full of molten lead.  Once the kettle is full the furnaceman will shut off the flow.  An overhead crane picks up the kettle and takes it over to the refinery area.\n\nA lot of sulfur containing compounds are generated as by-products of the smelting process.  At one plant I worked at they had an acid plant on site.  By scrubbing the furnace's exhaust gasses, they removed most of the SO2 from the exhaust.  The SO2 was then sent to the acid plant where a chemical reaction took place and the SO2 was changed to sulfuric acid---H2SO4.  The acid plant would produce a rail car full of 98% pure sufuric acid every day.\n\nThe scrubbed exhaust gasses went into another scrubbing device called the 'baghouse'.  This was a huge four story building about the size of a city block.  Every floor had dozens of rooms on it.  Each room was filled with bags that stretched from the ceiling of the room to the floor.  This bags were made of cloth---they were filter bags.  The exhaust gasses passed through these bags and the most minute particles were filtered out. The remaining exhaust gasses went up the chimney.\n\nWhoa--that brought back memories!  Believe me, you haven't lived t
1185233
this is so easy... the answer is furnace...\n\n\nFURNACE
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Pb(NO3)2(aq) + H2SO4(aq/l)[depending on concentration] --> PbSO4(s) + 2HNO3(aq)\nPb2+(aq) + SO42-(aq) --> PbSO4(s)\n\nI don't see how an redox reaction could occur.
304984
I would need more information to help you answer that question.\n\nWhat is the mass of the metal being used? What was the original temperature of the metal?\n\nThe formula you need is:\n\nq = mc(Tf - Ti)\n\nq = energy change\nm = mass\nc = specific heat\nTf - Ti = change in temperature
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Pb2+ + 2I- <----> PbI2 (s)? \nMost often this reaction is shown as a double replacement reaction as follows:\n\nPb(NO3)2 + 2 KI ==> PbI2¯ + 2 KNO3   \nSince both solutions have to be heated before mixing, you might think it is an endothermic reaction requiring heat. However, the Lead Iodide precipitate will not form till the mixture is cooled. This would imply the reaction is actually endothermic since cooling or removing heat energy from the solution results in the precipitate.
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This isn't too bad.\n\nFirst, find the enthalpy of formation for each of the reactants and products.\n\nFe2O3(cr) = -272 kJ/mol\nCO(g) = -110.525 kJ/mol\nCO2(g) = -393.509 kJ/mol\nFe = 0 kJ/mol\n\nThen add the enthalpies of formation for the reactants and subtract the enthalpies of formation for the products:\n\nFe2O3 + 3CO - (3CO2 + 2Fe)\n\n-272 kJ/mol + 3(-110.525 kJ/mol) - 3(-393.509 kJ/mol) - 2(0) = -603.575 kJ/mol - -1180.53 = -576.952 kJ/mol \n\nThat should do it.
857176
S(s) + O2(g)------>SO2\n\n2Mg(s) + O2(g) -------> 2MgO(s)\n\nH2(g) + Br2 (g) -------> 2HBr(g)\n\n4Fe(s) + 3O2 -------> 2Fe2O3 (s)\n\nThen there are single replacement \n\nA + BC-------> AC + B\nZn(s) + 2HCl(aq)------->ZnCl2 (aq) + H2 (g)\nwhere one element replaces another\n\nthen there is double replacement \n\nAB+CD -------> AD + BC\n\nIn this one the positive ions in the reacting compounds switch places\n\nBaCl2 (aq) + Na2So4 (aq) --------.BaSo4 (s) + 2NaCl (aq)
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a) 3SiO2 +2 Al --->3 Si + 2Al2O3\n\nb)AgNO3 +HCl---> AgCl +HNO3\n\nc) Mg + CuSO4 ---> MgSO4 + Cu\n\nd) Pb(NO3)4 +4 NaI --->4 NaNO3 + PbI4 (s)
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Here's what I wrote you, hopefully some nice thermodynamics expert out there can comment for you.\nΔG = ΔH - T ΔS is the equation for gibb's free energy, it's a number chemists make up to describe the system as a combination of enthalpy (which is the energy a system has available to heat transfer) and expansion energy, temperature x entropy or ΔS.  The TΔS part indicates how much expansion can take place, because entropy is determined by how many molecules the reaction "sets free" kind of.\nThe pre-equilibrium state can be calculated by finding amounts of gas produced and the temperature at any given moment during the move toward equilibrium.  Then they take a bunch of points and find the slope/rate of change and find the area under the curve.  I suppose that it is possible for that function to be a log of the products/reactants(the usual way to write equilibriums) depending on the order of the reaction and the products.  I think it would have to be gases producing a solid, 1st order.  Then the slope/rate would be a function 1/x and the integral would be logx.  But that wouldn't be true for all reactions.
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44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
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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.
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1. PbO + PbS\nmoles PbO = 0.257 mol\nmass of Pb from PbO: 0.257 mol * 207.2 g/mol = 53.25g\nmoles PbS = 0.141 mol\nmass of Pb from PbS: 0.141 mol * 207.2 g/mol = 29.21g\ntotal mass of Pb: 53.25g + 29.21g = 82.46g\n\n2. BaCl2 + Na2SO4 --->  2NaCl + BaSO4\nmoles Na2SO4: 0.042 L * 0.453 mol/L = 0.019 mol\n- means you also need 0.019 moles BaCl2\nvolume BaCl2 required: \n0.019 mol/ 1.131 mol/L = 0.0168 L or 16.8 mL\n\n3. 4FeCl3 + 3O2 --->  2Fe2O3 + 6Cl2\nbased from the balanced equation above, 4 moles of FeCl3 are needed for every 3 moles of O2 for the oxidation reaction. Hence, the excess reagent is FeCl3. Only 2.66 moles FeCl3 is needed to react 2 moles of O2. There is 0.33 mole of FeCl3 remains unreacted.
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Are you asking about thermal conductivity or specific heat?\nAnd why are you not doing the research you should be doing:\nhttp://www.engineeringtoolbox.com/specific-heat-metals-d_152.html\nhttp://www.engineeringtoolbox.com/thermal-conductivity-metals-d_858.html
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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
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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.
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You need a reference table and some instruction. Go to this site and scroll down to electrochemistry. Good luck!\nhttp://spiderpong.sitesled.com/chemistry_12_handouts.htm
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E = cp*m * ∆T; you have cp (specific heat), m and ∆T.  Find how many moles 1kg of iron is, and subsitute that for the mass in the formula for cp.  You can get that information here http://en.wikipedia.org/wiki/Iron