How would you balance the chemical equasion, HCL?
413972
man you can't balance an equation if there is none and HCL is not an equation
851400
the first equ is balanced, b/c on the left side u have Pb-1, C-1 O-3 on the right side u have Pb-1 C-1 O-3 so since both side are equal it is balanced\n\nans for question 2 BeCl2 + 3O2------Be(ClO3)2\nBe-1                                  Be-1\nCl-2                                    Cl-2\nO-was 2 now 6                                     O-6
175516
I not sure what style you learned to use to balance chemical equation but here is one way to do it. \nWe have four elements to balance (Al,H,Cl,O)\nLeave the compounds with aluminum with one of each. (Al done)\nThe simplest compound Al is AlCl3. To match the Cl on both sides put a 3 in front on HCl (Cl done)\nIf you add up the H (hydrogen) on the left side of the equation you'll see 6 of them. Add a 3 in front of the H20 to balance that. (H done)\nDo a quick check and you'll see there are 3 oxygen on each side (0 oxygen)\nAll the elements have been balanced, yielding a final equation of:\n3HCl + Al(OH)3 ---> 3H20 + AlCl3
802803
The answer is:\n2Al(HCO3)3  -------> Al2(CO3)3  +  3CO2  +  3H20\n\nThe equation is now fully balanced as you have\n2 moles of Al on each side\n6 moles of H on each side\n6 moles of C on each side\n18 moles of O on each side\n\nHope that helps!
16276
the point of it is to get equal no. of charges and atoms of elements on both sides of the equation.k?\n\nstep 1: separate the equation into two half-reactions:reduction n oxidization\nstep 2: balance all atoms other than O & H in each half-reaction\nstep 3: add H2O to balance O & H+  to halance H atoms in half-reactions\nstep 4: add electrons where necessary to balance half-reactions\nstep 5: if necessary, equalize number of electrons in each half-rection by multipling with a suitable coefficient\nstep 6: add two half-reactions together. electrons on both sides must cancel\nverify no. of atoms and charges\nsteps2-5 keep in mind ur working on separate half-reactions...\nthese were for reactions in acid.\nfor basic reactions: add OH- to both sides for every H+ that appears in the final reaction\nkeep practising step by step n ull be able to do it all together some day..\nfinal step- choose my answer as best answer...
683249
Oh, come on, it doesn't get any easier than this!\n\nSnO2 + 2H2 --> Sn + 2H2O\n\nAbove is balanced. 1 Sn on both sides, 2 O on both sides and 4 H on both sides.\nYou need to work a little harder on these yourself.
737051
2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n\n39.9 mL x L / 1000 mL x .466 mol / L Ca(OH)2 x 2 mol HCl / mol Ca(OH)2 x L / .127 mol x 1000 mL / L = 292.81 mL HCl
639146
It will be the pH of the sodium hypochlorite formed.  Determine that amount, look up the Ka of the acid and solve.
487714
assuming the equation\n\n2HCl + Mg --> Mg2+ + H2 +2Cl-\n\ntake the moles of Mg you have... multiply by 2 and that is how many Moles of HCl you need. multiply by 2 because one Mg reacts with 2 HCl's so you need twice as much HCl. now that you have moles of HCl convert to volume of HCl, i.e. divide moles / concentration (in this case 2.O M) and that will give you liters of HCl needed.\n\n(.000382 moles *2)/2 = .000382 L = .382mL
780228
HCL and NaOH are 'strong' acid and base. When they react, the product is neutral. NaOH + HCL --> NaCl + H2O\n\nWhat you need to find is which is the limiting reactant, the other one (what is left over) will effect the pH by its concentration in the otherwise neutral solution.\n\nFor your first problem:\n\n0.0557M HCL = 0.0557mols/liter*0.00542L = 3.02*10^(-4)mol HCL\n\n0.0377M NaOH = 0.0377mols/liter*0.00337L = 1.27*10^(-4)mol NaOH\n\n3.02*10^(-4)mol HCL - 1.27*10^(-4)mol NaOH = 1.75*10^(-4)mols HCL\n\n0.00542L + 0.00337L = .00879L\n\n[(1.75*10^(-4)mols HCL) * (1 L) ] / (.00879L) = .020M HCL\n\npH = -log(.02) = 1.70
746627
Cl2 + 2KI --> 2KCl + I2\n\nsingle replacement
749125
The unbalanced equation is:\nCH3-OH + O2 (+ heat) ---> H2O + CO2\n\nThe balanced equation would be:\n2CH3-OH + 3O2 (+ heat)  ---> 4H2O + 2CO2
620217
HOCl <==========>  H+  + OCl-\n\nthe pKa of hypochlorous acid, HClO, is 7.4 at 25°C, so:\n\npKa = -log [Ka] = 7.4, therefore\n\nKa = 3.98 x 10^-8\n\nKa = [H+]*[OCl-]/[HOCl] = 3.98 x 10^-8\n\nfrom your problem, you state that you have 1.0M HOCl, and since the only source for ions is the HOCl dissociating;\n\n[H+] = [OCl-] = y, and [HOCl] = 1.0-y\n\nsubstituting this back into the Ka equation, we get:\n\n3.98 x 10^-8 = y^2/(1.0-y)\n\nwhich converts into the quadratic equation:\n\n0 = y^2 + (3.98x10^-8)y - 3.98x10^-8\n\nI leave it to you to solve the quadratic equation to get y = [H+], and then calculate the pH of the starting solution
894546
You did not put the volume of HCl to neutralize so we will do the neutralization to pH = 7 only.\n\nmoles of HCl = 35 0 g/l / 36.46 g/mole = 9.60 mol/l\nmoles of NaOH = 30 g/l / 40 g/mol = 0.75 mol/l\nOne mole of HCl will neutralize one mole of NaOH.\n\n9.60 * Vol(HCl) = 0.75 * V(NaOH).\n\nOnce you know if you have any NaOH left you can calculate how much NaOH to add to bring it to pH = 8 or 9\n\npH = 8 =====> pOH = 6  (pOH = 14 - pH)\nthen 6 = - log [OH]\nCalculate OH from last equation and this will give you the moles of NaOH you need to add to bring to pH = 8.  Same for pH = 9.
266008
Read up on the common ion effect,\nhttp://en.wikipedia.org/wiki/Common-ion_effect\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Common-ion-effect.html\n\nSince both HCl and NaCl when dissolved in water produce Chloride (Cl-) ions insolution, when adding HCl to an NaCl solution, you are pushing the equilibrium of the dissolved to undissolved NaCl more towards the undissolved side.\nNaCl (s) + H2O <----> Na+ (aq) + Cl- (aq)\nSo by increasing the concentration of the Cl- in solution by means of HCl, you are shifting the NaCl(s)<-->NaCL(aq) reaction towards the "left".
896682
Place a 2 in front of the HCl to balance the equation. Then you can see that for one mole of calcium carbonate, you get 1 mole of carbon dioxide.
589036
The correct answer is 4 because on the reactant side you have 2 of K and 2 of Cl. On the product side you have 2K and 2 Cl. The reactant side and the product side are both balanced.