Which is the specific heat of the cupper sulfate?
95302
Cupric sulfate's heat capacity is described here.
216402
It's:\n\nBaSO4(s)---water <--> Ba2+(aq)  +  SO4 2-(aq)\n\nthe solubility for this compound is very low  per  g/L\n\n\n about 0.0023g/L
374102
176°C to 178°C Atorva Ca\n   it all depends on which polymorph / Salt u r asking for
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
14607
Sulfur - (300 K) (amorphous) 0.205 W/(m·K)\nhttp://en.wikipedia.org/wiki/Sulfur\nGlycerol - (20 C) 0.29 W/(m·K)\nhttp://www.hukseflux.com/thermal%20conductivity/thermal.htm
705505
The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram-°C.  If you want to use 4.18, that is okay.\n\nThe heat added = the specific heat of water (c) times the mass of the water (m) times the change in temperature.\n\nQ = cmT or\n\n= 4.18 joule/g-°C * 125 g * 2.44 °C\n= 4.18 * 125 * 2.44 joules (you can do the math)\n\nThis is for 1.75 g of CaCl2 which has a molecular mass of 111.0 g/mol, so the number of moles is:\n\n1.75 g / 111.0 g/mol = 0.01577 moles of CaCl2.\n\nDivide the heat change (in joules) above by 0.01577 moles to get the heat per mole.
867788
Liquid.  Melting point is 600.65K
200130
definitely sulphate... lol...ne ways, u gotta mix those stuff n leave it over nite to crystalise..then break it to pieces..or, u can jus ask ur chemistry teacher..haha.. all i did in sch was crystalizing copper sulphate usin the solution itself..
723847
I am assuming that the values of H° and G° are the standard enthalpy and Gibbs free energy changes for the formation of phsogene from CO(g) and Cl2(g).\n\nFor part (a), you should already know that:\n\ndelta-G = delta-H - T*delta-S\n\nSo, for the values you are given for 298 K,\n\n-206 kJ/mol = -220 kJ/mol - 298 K * delta-S°\n-46.98 J/(mol*K) = delta-S°\n\nFor part (b), we now have (assuming delta-H and delta-S are independent of temperature) that \n\ndelta-G_601K = -220kJ/mol - 601 K * (-4.698*10^-02 kJ/mol*K)\ndelta-G_601K = -191.8 kJ/mol
975732
Substance: Water\nLatent Heat Melting J/g:  335\nMelting Temp°C: 0\nLatent Heat Boiling J/g:  2272\nBoiling Temp °C: 100 \nHeat of vaporization (kJ/mol): 40.65
505556
you have to look for it at the saturated watter boards or the saturated water graph, then the temperature will tbe the saturation temperature, according to my graph, the temperature would be...\n\n83.79ºC
918146
CuSO4.5H2O => CuSO4.H2O + 4H2O at 98*C (or 110*C)\nCuSO4.H2O => CuSO4 + H2O at 120*C (or 150*C)\nCuSO4 => CuO + SO2 + O2  at 650*C
175159
Solid\nCopper melts at 1083ºC
48785
Boiling Point: 444.6 °C (717.75 K, 832.28 °F)
630844
Barium sulfate is insoluble, so will not form an electrically conducting solution. It will, however, conduct electricity if molten.  It melts at 1350°C (2462°F). \n\nI hope that helps. If not, please re post.
75392
You need to check your units.  The specific heat of water is not 4.184 J/g; it is 4.184 J/g*degrees C (or degrees kelvin).  And the heat of solution should not be in J/degree, but in J/g.\n\nTry this equation instead:\n\n[HoS = J/g] = [(Specific Heat = 4.184 J/g*C)*(mass = 100g)*(Temperature change = Initial C - Final C)] / [Mass of CaCl2 = 3 g]\n\nBy subtracting the final from initial you should get the negative value and not need the -1 at the front.\n\nSometimes enthalpy of solution is in J/mol rather than J/g, in which case you modify the above by calculating the mols of CaCl2 in 3 grams and switching that out with the 3g in the above equation:\n\n(3g) / (110.99 g/mol) = 0.027 mol\n\nTheoretically, you should want to know the heat capacity of the CaCl2 because it is also being heated, but this is assumed to be negligible.  Write that into your error analysis and you teacher will love you.  Also, make sure to check your significant figures; i don't know how many sig figs you used in weighing out your 3 grams, but that will be necessary to know how much to round everything.  Good luck.
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4