1) In aqueous slution, potassium hydroxide and hydrochloric acid react as follows. KOH(aq) + HCL(aq) + H2O(L).
401299
i already answered parts a and b in the last question like this.\n\nc) enthalpy = mass of water x specific heat of water x change in temp\n\nE=0.1*4.18*3.5\nE=1.463KJ\n\nmultiply by two to get answer per mole = 2.926KJ/mol\n\nd) main source of error is the glass beaker since it is leak out the heat produced by the reaction, the reaction would be better carried out in a calorimeter or in a polystyrene beaker with a lid.\n\ne)  the temperature change would be the same as 1mol of KOH reacts with 1mol of HCl therefore with the extra HCl the KOH will act as a limiting reagent and u will get the same reaction as before.
47023
Sodium Hydroxide (aqueous). It's a solution in water of the compound NaOH.
733447
Br+H20--->HBr+O\n\nNi(II)O+H20--->Ni+H202\n\nKO2+H2O--->KOH + H2\n\nHope that helps.
544064
The two reagents are in aqueous form, which means that they exist in ions. Upon reaction, the H+ & OH- will form water, H2O & the remaining ions will remain aqueous in the solution;\n\nNaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(l)\n\nNo balance required...
215729
K2IC4H6 + HNO3 ---  K2HI + NO3C4H6
664234
NH4ClO4 --> NH4^+ + ClO4^-\n\nNow in a water solution,\nClO4^- + H2O --> HClO4 + OH- --> This doesn't happen because HClO4 is a strong acid.  Therefore the only contribution to the pH is from the NH4^+\n\nNH4^+ <=> NH3 + H^+\nYou start with [NH4^+] = 0.40 M, and [NH3]=[H^+] = 0 M; after undergoing equilibrium [NH4^+] = 0.40 - x, while [NH3]=[H^+] = x; Ka = [NH3][H^+]/[NH4^+] = (x)(x)/(0.40-x) = Kw/Kb = (1 x 10^-14)/(1.8 x 10^5)\nAssume x << 0.40, so x^2/0.40 = 5.56 x 10^-10\nx = 1.49 x 10^-5 M = [H^+]\npH = -log(1.40 x 10^-5) = 4.827
196930
Acid = Hudrochloric acid (HCl), Sulfuric acid (H2SO4)\n\nBase = Sodium Hudroxide (NaOh), Potassium Hydroxide (KOH)
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
804535
Neutralization Reaction:\nHC2H3O2(aq) + KOH(aq) -->H2O(l) + KC2H3O2(aq)\nThe equation is already balanced.\n\na. False KC2H3O2(aq) and H2O are products.\nb. True\nc. False, some ions will cancel out but some will remain.\n\nThe ionic equation is as follows:\n(H^+)(aq) + (C2H3O2^-)(aq) + (K^+)(aq) + (OH^-)(aq) -->H2O(l) + (K^+)(aq) + (C2H3O2^-)(aq)\nRepeated ions are crossed off from both sides and the net ionic equation is:\n(H^+)(aq) + (OH^-)(aq) -->H2O(l)\n\nd. True\ne. False\n\nNote: the + and - signs were written with ^ to indicate that they were raised up (like a power).  Sorry if this notation is confusing.  Good luck and I hope that this helps!
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
412530
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)
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MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
564023
acid + base ---> salt + H2O
400328
Since both HCl and KCl are strong elecrolyes, both should remain totally dissassocitated in solution.  I dont think any HCl will react with KCl in this case.\n\nHowever, have you ever heard the rule "Always Add Acid"?\nWhenever you dilute an acid with water, the resulting solution gives off energy (dissolving HCl in water is an exothermic reaction).  This when you dilsute 1 M HCl to .5 M HCl, energy will be released and will cause a temperature change in the solution.\n\na)  The number of moles of HCl used in this experiement is constant. This value will be equal to the number of moles of HCl present in the initial 100 ml of 1 M solution.\n1 M = x moles / .1 L, thus you have .1 moles of HCL.\n\nb)  The heat of the reaction will be the heat given off when the HCl solution is diluted.\n\nc)  Once you know the heat energy given off by the reaction, you can calculate the change in temperatutre if you know the specific heat of the final solution.  I am not sure if you are suppose to estimate the final solution's specific heat as being equal to that of water (4.186 J/g C), but that would be my guess.\nQ = m*c*(delta T)\n\n\nEDIT:\n"davidchingwong13" is correct in his application of the common ion effect.  By dissolving additional Cl- ions in the solution by adding HCl, the equilibrium between solid and aqeuous KCl is disturbed.  However, KCl is a strong electrolyte, it readily dissolves and breaks in to ions in water.  I doubt the addion of so few moles of Cl- ions to an already unsaturated KCl solution willl effect it that much.\n\nHowever....dissolving KCl in water is a endothermic reaction, so if, indeed, the KCl did precipitate out, it would be exothermic and give off additional energy.
934206
2\nthe reaction is\nCa(OH)2+2HNO3=2H2O+Ca(NO3)2
127210
Hydrochloric acid will react with metals ABOVE Hyrogen in the activity series of metals to produce Hydrogen gas.\nThe metals will displace Hydrogen from the acid and make a metal salt (which is usually in solution).\n\nSo if we say that [A] is a generic metal above Hydrogen in the activity series that has a normal oxidation state of +1, then the reaction would be as follows,\n[A] (s) + HCl (aq) --> H2 (g) + [A]Cl (aq)
1030709
1 H2 +1 Cl2 ------> 2 HCl\n\n     2 - H - 2\n     2 - Cl - 2