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Question #a721d - Socratic pH = 1 61151 OH^- = 4 08797 * 10 ^-13M HF = 0 855538M H^+ = 0 024462M F^- = 0 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 color (white) (mmmmmmmm)"HF" + "H"_2"O" ⇌ "H
Question #93f99 - Socratic The molarity of the acid is 1 434 Start with the equation: 2HNO_ (3 (aq))+Ca (OH)_ (2 (aq))rarrCa (NO_3)_ (2 (aq))+2H_2O_ ( (l)) This tells us that 2 moles of HNO_3
Question #70577 - Socratic Here's what I got Start by writing the balanced chemical equation for this neutralization reaction 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
If 575 grams of frozen ethonal are warmed to the freezing . . . - Socratic 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 95 kj mol
Question #750c8 - Socratic Here's what I got 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 Simply put, some molecules of ammonia will accept a
Question #0984e - Socratic 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) This means that, in order to get a complete neutralization, you need equal numbers of moles of each compound
Question #d6b18 - Socratic We want the standard enthalpy of formation for Ca (OH)_2 Thus, our required equation is the equation where all the constituent elements combine to form the compound, i e : 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 ] (1) 2H_2 (g) + O_2 (g)->2H_2O (l) and DeltaH_1=-571 66 kJmol^-1 (2) CaO (s) + H_2O (l
Question #18488 - Socratic The degree of dissociation sf (alpha=0 0158) sf (K_b=2 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
Question #97503 - Socratic Here's what I get 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
Question #370a7 - Socratic The sodium ions remain in solution as spectator ions If XS sodium hydroxide is added the precipitate redissolves to give the soluble plumbate (II) ion A simple way of writing this is: (chemguideUK) Ammonia solution can't do this as the concentration of OH^ (-) ions is not high enough