water chemistry question?
448790
Alkalinity is usually measured by the presence of bicarbonate (HCO3-) and carbonates (CO3=) and acts to buffer water from acidic inputs, thus is a measure of the buffering capacity of water. Water with high alkalinity have high pH values while water with low alkalinity have low pH values. The hardness of the water is usually determined by the amount of calcium and magnesium salts present in water and is associated with the presence of carbonates. Hard water is generally more productive than soft water waters and can accept more input of salts, nutrients, and acids to their system without change than can soft water waters. So basically, it’s because of the presence of the magnesium. \n\nHope this helps and hope I didn’t confuse you.
386784
Well first of all you have to specify if the H's you want to account for are either comming from HCl alone or both HCl and some H's comming from the dissociation of water.\n\nI suppose that only the number of moles of H's comming from HCl is really worth while working the question for, since the H's of the dissociated water does not change the pH of the 0.10M solution.\n\nTherefore:\n\n0.10moles are in 1000mL\n? moles are in 50mL\n\n50 x 0.1 / 1000 = 0.005moles\n\nTherefore we can say that 0.005moles of HCl in 50mL of water results in a concentration of 0.10M
818083
Rainwater will have a pH determined by its composition.  Its composition will be water (pH=7), O2, CO2, SO2, NOx,... stop me when I've mentioned something that strikes you as a product of human pollution.  There are some ultra left wing radical commie pinko pseudo scientists who LIE about pollution and say it is a GLOBAL problem.  But WE all know that we can fowl our rivers and burn our forests and nothing will be affected more than oh a couple of feet away, right?  RIGHT?!! Tha tha thats all folks.
809082
Water is a combination of 2 hydrogen atoms and 1 oxygen atom: H2O.\nnot H2O2!!!!!
167656
the correct answer is B. A is wrong b/c it isn't an oxidation reduction reaction. Looking at a table of solubility products will confirm the answer is B.
777672
S2O3(2-)-->SO4(2-)          balance S first\nS2O3(2-)-->2SO4(2-)         balance O with water\n5H2O+S2O3(2-)-->2SO4(2-)        balance H with H+\n5H2O+S2O3(2-)-->10H++2SO4(2-)       balance charge with e-\n\n5H2O+S2O3(2-)-->10H++2SO4(2-)\n                                        +8e-\nthe 8e- is with the product side
22647
There's a formula in this article:\nhttp://www.naturodoc.com/library/nutrition/water.htm
424592
H2SO4 + 2NaOH ----> 2H2O + Na2SO4\n\nYOU MUST SEPARATE AND CANCEL THE COMMON IONS (AQUEOUS ONES)
975732
Substance: Water\nLatent Heat Melting J/g:  335\nMelting Temp°C: 0\nLatent Heat Boiling J/g:  2272\nBoiling Temp °C: 100 \nHeat of vaporization (kJ/mol): 40.65
313595
Water... its got a pH=7
291231
H2O...two parts hydrogen and 1 part oxygen in each of the water molecules.
692123
1. pH = -log [H+] which is the hydronium ion concentration.  [H+] = 10^-2.85 = 0.00141 M\n[OH-] = 1 x 10^-14/0.00141 = 7.09 x 10^-12\n\n2.  pH = -log 12 = -1.08\n\n3.  neutral water would be when [H+] = [OH-], so [H+] would be the square root of Kw, so [H+] = 1.58 x 10^-7.  pH = 6.80.
787375
1. AgCl(s) + NaNO3\n\n2. AgOH(s) + NaNO3\n\n3. They would not react as both are nitrates already
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...
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
921558
It will be clearer if you write the reaction as the sum of two reactions.\n\nWhen HN3 is dissolved in water, it reacts with the water to form ammonium ion (NH4+) and hydroxide ion (OH-).  Omitting the reaction NH3(g) = NH3(aq), we have:\n\nNH3 (aq) + H2O (L) = NH4+(aq) + OH-(aq)\n\nThe equilibrium constant for this reaction is:\n\nKb = [NH4+]*[OH-]/[NH3]\n\nThe hydroxide produced in this reaction reacts with the manganese ion to form insoluble manganese hydroxide (Ksp ~ 2*10^-13)\n\nMn++(aq) + 2OH-(aq) = Mn(OH)2(s)\n\nThe equilibrium constant for this reaction is:\n\nK = 1/Ksp = 1/([Mn++]*[OH-]^2) \n\nThe reaction you wrote is the sum of 2 times the first reaction plus the second reacton.  Note that as written, the first reaction produces OH- ions while the second one consumes them.  \n\nIf we add an acid like HNO3 to the solution, we are adding H+ ions.  Adding H+ ions is really like subtracting OH- ions because the concentrations of H+ and OH- are inversely proportional to one another through the water dissociation reaction (H2O = H+ + OH-, [H+]*[OH-] = Kw = 10^-14).\n\nWe can therefore write the two equilbrium constants above in terms of the concentration of H+ instead of OH-:\n\nKb = [NH4+]*[OH-]/[NH3] = Kw*[NH4+]/([NH3]*[H+])\n\nK = 1/([Mn++]*[OH-]^2) = ([H+]^2)/(Kw^2 * [Mn++])\n\nThis means that if we add acid (i.e., HNO3) to the solution, we will drive the first reaction in the forward direction; the concentration NH4+ will increase and the concentration of NH3(aq) will decrease.  Adding H+ will drive the second reaction in the reverse direction; Mn(OH)2(s) will dissolve, and the concentration of Mn++ will increase.
588420
sodium chloride and water