write the thermal decomposetion of CuSO4.5H2O on four steps?
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
854997
CuS + H2SO4 ===> H2S + CuSO4
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
564230
This is a selective way to produce cupper of cupper hydroxides by leaching hydroxide and elecrtowinnig as:\nCu(OH)2(s) = Cu+2(aq) + 2 OH–(aq) \nCu+2(aq) + 4 NH3(aq) = Cu(NH3)4+2(aq)\n\nand [Cu(NH3)4]2+ is better agent than [Cu(H2O)6]2+ in (aq) solutions for this.
745889
We can solve this using a system of equations.\n\nLet x be the volume of the 25% solution. Let y be the volume of the 60% solution.\n\nThe sums of the volumes must total 140 ml:\n\nx + y = 140.\n\nThe mass of CuS04 in the desired solution must equal the amounts added through the use of both solutions.\n\n30 percent of 140ml gives 42 ml of "pure" copper sulfate solution.\n\nThe 25% solution would contribute .25x of "pure" copper sulfate solution. The 60% solution would contribute .60y of "pure" copper sulfate solution, so:\n\n0.25x + 0.60y = 42.\n\nHere is the system:\n\nx + y = 140\n0.25x + 0.60y = 42\n\nThere are many different ways to solve the system. I'll use substitution here:\n\nx + y = 140, so y = 140 - x\n\n0.25x + 0.6(140 - x) = 42\n0.25x + 84 - 0.6x = 42\n-0.35x = -42\nx = 120\ny = 20\n\nSince x is the amount of 25% solution and y is the amount of 60% solution, we need 120 ml of 25% solution and 20 ml of 60% solution.\n\nThis appears to match your answer!
951038
Miss Science, you have all the facts but you missed out the most important one, which is that the percentage by mass of water in CuSO4.5H2O is always the same.\n\nthe Mr of CuSO4 is 159.6 and that of 5 water molecules is 90.0. no matter what mass of hydrated copper(ii)sulphate, the masses of CuSO4 and H2O will always be in the ratio of 159.6:90.0\n\nthis means that the percentage mass of water would be 90/(90+159.6)x100% = 36.06%
366358
reaction of conc HNO3 at high temperatures with Cu:\nCu + 4HNO3 = Cu(NO3)2 + 2H2O + 2NO2\n\nreaction of dil HNO3 at ordinary temp with Cu:\n3Cu + 8HNO3 = 3Cu(NO3)2 + 4H2O + 2NO
501191
There is a formula...but this is much easier\n\nhttp://www.onlineconversion.com/temperature.htm
981128
1. synthesis\n2. single replacement\n3. decomposition\n4. single replacement
748872
Cu2+ (aq)  +  Zn (s) -->  Cu (s) + Zn2+ (aq)
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)
48368
2HC2H3O2 + Ba(OH)2 --> Ba(C2H3O2)2 + 2H2O
1123465
Na2S + 2 HCl > H2S + 2 NaCl  \n\nH2SO4 + BaOH > H20 + BaSO4  \n\nNH4Cl + NaOH > NaCl + NH4OH\n\nAgNO3 + NaCl > AgCl + NaNO3\n\nnot sure about complete or reversible..:)
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4
866145
HCl + AgNO3-----> AgCl +HNO3\nPb(OOC-CH3)2  + 2LiCl -------> PbCl2  + 2Li(OOC-CH3)\nNa3PO4 + CrCl3 -------> Cr(PO4) + 3NaCl\nH2SO4 + BaCl2 ----------> BaSO4 + 2HCl\nAl2(SO4)3  + 6NH4OH ------> 2Al(OH)3  + 3(NH4)2SO4\n2AgNO3 + H2S -------> Ag2S  + 2HNO3\nCaCl2   + Pb(NO3)2 -------> Ca(NO3)2  + PbCl2\nCa(NO3)2 + Na2CO3 ------>CaCO3  + 2 NaNO3
424592
H2SO4 + 2NaOH ----> 2H2O + Na2SO4\n\nYOU MUST SEPARATE AND CANCEL THE COMMON IONS (AQUEOUS ONES)
583365
First of all work out the molar mass of CuSO4*5H2O\n Cu + S + (9 * O) * (10 * H)\n\nSorry I don't have a chemical table next to me\n\nCall this A\n\nSecondary work out the mass of 5 moles of water.\nCall this B\n\n(5 * O) + (10 * H)\n\nthe theoretical % is then B/A * 100 = x%