What is the ionic equation and the net equation for CuSO4 +K3PO4?
914017
I will suggest that first you write the equation in a compact mode, balance it and then add all the bells and whistles.\n\nCuSO4 +K3PO4 ---> Cu3(PO4)2 + K2SO4\n\n3CuSO4 +2K3PO4 ---> Cu3(PO4)2 + 3K2SO4\n\n3CuSO4(aq) +2K3PO4(s) ---> Cu3(PO4)2(s) + 3K2SO4(aq)\n\nNow that it is balanced it is easier to break down:\n\n3Cu(+2) + 3SO4(-2) + 2K3PO4(s) --->  Cu3(PO4)2(s) + 6K(+) + 3SO4(-2)
834300
In PbSO4 the Pb is +2 and the SO4 group is -2. \nThe O in SO4 is -2 so the sulphur must be +6 (the four oxygens add up to -8 and for the SO4 group to be -2 the sulphur must be +6)\nThe formula for CaOCl2  IS  correct. The Ca is +2 and the OCl2 group -2 as O is usually -2 it would suggest that each chlorine is 0. This anomaly is because the OCl2 is known as a disproportionate ion (having too much chlorine).\nIn (NH4)2SO4  each NH4 group is +1 and the SO4 is -2.\nSo as each H = +1  the N is -3. The SO4 has the same oxidation numbers as in PbSO4 above.\nIn working out oxidation numbers we assume that certain elements have fixed numbers O is -2,  Cl is -1,  H is +1  and alkali metals like sodium and potassium are +1. The complete compound must add up to zero.
711424
The Equation should be:\nNaCl + H2SO4 = Na2SO4 + HCl\nSo the coefficients are (2,1,1,2)
801118
CuCl2 + 2NH4OH --> Cu(OH)2 + 2NH4Cl\n\nCu(OH)2 (s) + 4NH3 (aq)     ----->     Cu(NH3)4^2+
98695
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)
494638
P3O10 does not exist; it's P4O10.\n\nBalanced equation is:\n\nP4O10 + 6H2O --->  4H3PO4
706351
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
745644
H3PO4 + 3NaOH ----> 3H2O + Na3PO4
975769
Al(s) + MnO4- (aq) -----> MnO2 (s) + Al (OH)4- (aq)\n\nSplit the above into two half equations:\n\n1) Al(s)  --->  Al (OH)4- (aq)  and\n2) MnO4- (aq) -----> MnO2 (s)\n\nBalance 1) as follows:\n\n1) 4OH^-  +  Al(s)  --->  Al (OH)4- (aq) now balance for charge:\n1) 4OH^-  +  Al(s)  --->  Al (OH)4- (aq)  +  3e^-\n\nBalance 2) as follows:\n2) MnO4- (aq) ---> MnO2 (s)  +  2H2O (l) (turn xs O into H2O)\nNow balance with H^+:\n2) 4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)\nnow balance charge:\n2) 3e^-  +  4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)\n\nNow add 1) and 2) (electrons happen to be equal):\n4OH^-  +  Al(s)  +  4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)  Al (OH)4- (aq)\nnow OH^-  +  H^+  ---> H2O:\n\n4H2O  +  Al(s)    +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)  Al (OH)4- (aq)\n\nSubtract 2H2O from both sides:\n2H2O (l)  +  Al(s)    +  MnO4- (aq) ---> MnO2 (s)  +    Al (OH)4- (aq)\n\nI hope that helps. If you need more help, please email me or re post.
424592
H2SO4 + 2NaOH ----> 2H2O + Na2SO4\n\nYOU MUST SEPARATE AND CANCEL THE COMMON IONS (AQUEOUS ONES)
903130
Na2SO4(aq)+ 2 AgNO3(aq)-->Ag2SO4(s)+ 2 NaNO3(aq).
595177
Actually, this is pretty simple.  You have a saturated  solution of lead sulfate (PbSO4) and strontium sulfate (SrSO4).  Probably the biggest hurdle here is to realize that the concentration of sulfate ion will be the same for both salts!  Therefore, the concentration of sulfate ion will be the result of the most soluble of the two salts:\n\nKsp for PbSO4 = [Pb+2]*[SO4-2] = 1.6x10^-8\nKsp for SrSO4 = [Sr+2]*[SO4-2] = 2.8x10^-7\n\nStrontium sulfate is the more soluble of the two salts, so it will control the concentration of sulfate:\n\nSrSO4 ===> Sr+2  + SO4-2, thus [Sr+2] = [SO4-2]\n\nKsp for SrSO4 = 2.8x10^-7 = [SO4-2]^2 , so \n\n[SO4-2] = sqrt(2.8x10^-7) = 5.29x10^-4 = [Sr+2]\n\nNow, to find out the concentration of Pb+2, we use the value we have just calculated for [SO4-2] and the Ksp:\n\nKsp for PbSO4 = [Pb+2] * [SO4-2] = 1.6x10^-8\n\n[Pb+2] = 1.6x10^-8/[SO4-2] = (1.6x10^-8)/(5.29x10^-4)\n[Pb+2] = 3.02x10^-5\n\nand there we have your answers:\n[Sr+2] = [SO4-2] = 5.29x10^-4 moles/liter\n[Pb+2] = 3.02x10^-5 moles/liter
277199
CuSO4 . 5H20 -> CuSO4 + 5H2O\n\nIt is just a hydrate so basically you are just getting water out of the salt. Copper(II) sulfate pentahydrate is blue while CuSO4 is a white salt.
307097
Ksp = [Ag+]^2[CrO4 2-]\nsince [Ag+] = 2*[CrO4 2-]\nKsp = 2*[Ag+]^3\nsolve for [Ag+] then plug it in this equation\n[Ag+] = mass /0.010L H2O
742926
Cu(s) + 2HNO3(aq) = Cu(NO3)2(aq) + H2(g)\n\nCu + 2H+ + 2NO3- = Cu+2 + 2NO3- + H2 (leaving off the s,aq,g)\n\nCu + 2H+ = Cu+2 + H2 (net ionic equation, the NO3-'s just cancel out, they don't change in the reaction).
211591
Ca3PO4 and CuOH\n\n^^those are the answers
283215
[Cu(NH3)4]SO4*H2O (s) + 2HCl(aq) --->\n\nCu(NH3)4 +2 (aq) + H2SO4(aq) + H2O(l) + 2Cl-(aq)