Oh my zsh vs ] (1) 2H_2 (g) + O_2 (g)->2H_2O (l) and DeltaH_1=-571. Here's what I got. 51xx10^ (-6)color (white) (x)"mol/l") Triethyamine is a weak base and ionises: sf ( (CH_3)_3N+H_2Orightleftharpoons (CH_3)_3stackrel (+) (N)H+OH^-) For which: sf (K_b= ( [ (CH_3)_3stackrel (+) (N)H] [OH^ (-)])/ ( [ (CH_3)_3N])) Rearranging and taking -ve logs of both sides we get the Here's what I get. : Ca +H_2+O_2->Ca (OH)_2 Let us now write down the given equations: [The first equation mentioned is incorrect, and so I have revised it. 024462M HF + H_2O = H_3O^+ + F^- We can find the concentration of H^+ or H_3O^+ by three ways One is by the ICE table (but this is a 5% rule) and the other is square root which is absolutely correct and the other is Ostwald's law of dillution Let's set up an ICE table. 855538M H^+ = 0. We want the standard enthalpy of formation for Ca (OH)_2. Simply put, some molecules of ammonia will accept a CH 3COOH (aq) +OH − (aq) → CH 3COO− (aq) +H 2O(l) Notice the 1:1 mole ratio that exists between acetic acid and sodium hydroxide (written as hydroxide ions). color (white) (mmmmmmmm)"HF" + "H"_2"O" ⇌ "H The degree of dissociation sf (alpha=0. If 575 grams of frozen ethonal are warmed to the freezing point of ethanol, how much heat is required to melt all of this frozen ethanol, CH3CH2OH, at the freezing point? The molar heat of vaporization of ethanol= 4. Since sodium hydroxide is a strong base that dissociates completely in aqueous solution, you can represent it by using the hydroxide anions, "OH"^(-) "HA"_text((aq]) + "OH"_text((aq])^(-) -> "A"_text((aq])^(-) + "H"_3"O"_text((aq])^(+) Now, you have 1:1 mole ratios across the board, you can say Here's what I got. Thus, our required equation is the equation where all the constituent elements combine to form the compound, i. e. The general equation for the dissociation of a carboxylic acid is "R-COOH + H"_2"O" ⇌ "R-COO"^"-" + "H"_3"O"^+ All we have to do is write the Explanation: < Since the molarity of either acid is the same, the moles of each acid are equal. = Since both acids are monoprotic (they only release one hydrogen) they will both take the same amount of #OH The reaction forms 2,3-dihydroxybutenoic acid and a dark brown precipitate of manganese dioxide. The problem wants you to use the base dissociation constant, K_b, of ammonia, "NH"_3, to determine the percent of ammonia molecules that ionize to produce ammonium cations, "NH"_4^(+), and hydroxide anions, "OH"^(-). As you know, ammonia is a weak base, which means that it does not ionize completely in aqueous solution. Since the strong acid dissociates more (releases more #H^+# ions), it will have a lower pH. 95 kj/mol. That means the difference between their pH is determined solely on which acid dissociates more. 0158) sf (K_b=2. Start by writing the balanced chemical equation for this neutralization reaction. 024462M F^- = 0. 66 kJmol^-1 (2) CaO (s) + H_2O (l pH = 1. 95 kj/mol Here's what I got. 61151 OH^- = 4. This means that, in order to get a complete neutralization, you need equal numbers of moles of each compound. 08797 * 10 ^-13M HF = 0.