What is the balanced Chemical Equation to the following problem?  Cu+HNO3---->Cu(NO3)2+NO+H2O?
596804
3Cu+8HNO3=3Cu(NO3)2+2NO+4H2O
223710
Start by balancing Al, then Cl. Al is already  balanced, so when you balance Cl everything is easy from their. \nYou final amswer will be:\n3HCl + Al(OH)3 ----> 3H2O + AlCl3
83000
The net ionic equation is:\n\nFe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s)
957775
Ca(OH)2 + CO2 --> C(OH) + CaO\n\nIs unbalanced..
372026
CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g)
901730
Na2S (ag) + 2 AgNO3 (ag) --> 2 NaNO3 (ag) + Ag2S (cr)\n\nB. Double displacement.\n\nA. Ba(NO 3)2 and H2O\n\nB. If there is a double arrow (pointing in both directions), then it is an equilibrium reaction.\n\nC. Ba(OH)2 and HNO3\n\nD. Barium\n\nE. 6 oxygen atoms. \n\nF. Ba(OH)2
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
779077
Please recognize first that this is an oxidation-reduction ("redox") reaction.  Therefore, something gets oxidized, and something gets reduced.  We must identify those first and write their half reactions...\n\nH2O  + 2e- ---->  H2  \n+1                         0 \n\nThe hydrogen in water is reduced to elemental hydrogen gas, resulting in a gain of 2 electrons and a reduction of +2 in oxidation state (one per hydrogen).\n\nAl ---->  AlCl3  \n0           +3\n\nMetallic (ox. st. 0) aluminum gets oxidized to +3 in AlCl3, resulting in a loss of 3 electrons.   \n\nWe must make the electron transfer equal so electrons are conserved (no excess gain or loss).  In order to balance them out, we must multiply the H2's by 3 and the Al's by 2.  Let's see where that leaves the chemical equation at the moment:\n\nCuCl2  + 3H2O and 2Al ---> 3H2 + 2AlCl3 + CuO\n\nThis should result in 6 electrons gained and lost; therefore conserved.  The reaction however is not yet balanced, as the\nCl atoms are not conserved.  Let's multiply the CuCl2 term by 3 (as well as the CuO term to keep the Cu's in balance).\n\n3CuCl2  + 3H2O and 2Al ---> 3H2 + 2AlCl3 + 3CuO\n\nA quick inspection leads us to think that our reaction is now balanced.   However, in order to complete our redox balance we must check TWO  THINGS:\n1.  Mass balance\n2.  Charge balance\n\nMass balance is what we have been doing; making sure there are the same number of atoms of each side of the reaction.\nLet's be sure that is the case:\n3 Cu on both sides\n6 Cl on both sides (3 Cl2 vs. 2Cl3; OK)\n6 H on both sides \n3 O on both sides\n2 Al on both sides\nMass balance is correct.\n\nWe must also see that the net CHARGE on both sides is the same.  (Does not have to be zero!)  We are lucky here that we have no ions, no charged species.  So the charge on both sides is zero (OK).\n\nWe're finished.
750179
NaCl2 + NaCO3 ==> NaCO3 + NaCl2 NOTE: these two reactants will not react because the reactants and the products are identical\n\nCu + O2 ==> CuO2   This is already balenced\nCuSO3 + Fe ==> FeSO3 + Cu   The charge is not given so procuts could be different\n\n2HCl + 2Zn ==> 2ZnCl + H2\nMnO2 + H2O2 ==> No reaction: similar to NaCl2+NaCo3\n\nGood luck
844168
This is very simple\n\nCa+H2O -> Ca(OH)2 + H2\n\nThe reactant has\n1 atom of Ca\n1 atom of H\n2 atoms of O\n\nThe product has\n1 atom of Ca, ok\n2 atoms of H\n2 atoms of O\n\nJust put a coefficient of 2 in the water (H2O)\n\nSo the chem. eqn becomes\n\nCa+2 H2O -> Ca(OH)2 + H2\n\nThat's it.\nThe equation is already balance
749125
The unbalanced equation is:\nCH3-OH + O2 (+ heat) ---> H2O + CO2\n\nThe balanced equation would be:\n2CH3-OH + 3O2 (+ heat)  ---> 4H2O + 2CO2
394775
2Cu(NO3)2 ----> 2CuO + 4NO2 + O2\nOBSERVATION: brown, choking fumes of NO2 evolved; black solid i.e CuO settles at the bottom\n\n2FeSO4 ----> Fe2O3 + SO2 + SO3 \nOBSERVATION: white fumes of SO3 evolved and reddish brown solid i.e. Fe2O3 formed\n\nEDIT: The above equation for iron(II)sulphate may be a supposed characteristic for its thermal decomposition relating to that of copper(II)nitrate above, but Chaos, O2 will not be evolved here, because SO3 will not get reduced to O2, even on strong heating.
934206
2\nthe reaction is\nCa(OH)2+2HNO3=2H2O+Ca(NO3)2
577084
Acid + base = salt + water.\n\nUnbalanced eqn is: Al(OH)3 + H2CO3 --> Al2(CO3)3 + H2O
801118
CuCl2 + 2NH4OH --> Cu(OH)2 + 2NH4Cl\n\nCu(OH)2 (s) + 4NH3 (aq)     ----->     Cu(NH3)4^2+
822606
CuSO4 + NH4S = CuS + NH4SO4\n\n1)Write a rough equation without balancing anything. The 2 after copper, which I presume is to be written as copper (II), indicates that copper has a charge of +2. It is a transition metal and can have a differing charge.\n\n2)Because ammonium has a charge of +1 and the sulfate has a charge of -2, there needs to be 2 ammoniums. SAme thing is for the left side. The ammonium is +1 and the sulfide is -2. There needs to be two ammoniums on the left side. \nCuSO4 + 2NH4S = CuS + 2NH4SO4\n\n3) Because there are two sulfides on the left side, there needs to be two sulfides on the right side. \n\nCuSO4 + 2NH4S = 2CuS + 2NH4SO4\n\n4) Now there are two sulfates on the right side, but not on the left. So we place a two in front of copper (II) sulfate.\n\n2CuSO4 + 2NH4S = 2CuS + 2NH4SO4\n\n5) I believe that we are done with balancing so let's check our work.\n\n6) 2 coppers are on the left and right side.\n2 sulfates are on the left and right side.\n2 ammoniums are on the left and right side. \n2 sulfides are on the left and right side.\n\n7) In most cases, not all compounds will have the same number on each side. It's just a coincidence that there are all 2 of each. If you need anymore help, email me ^_^