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Question #6f539 + Example - Socratic The effect of strong base on water is to dramatically increase the concentration of OH^- ions and decrease the concentration of H_3O^+ ions Water always contains at least small concentrations of both OH^- (hydroxide) and H_3O^+ (hydronium) ions This is because water can react with itself in a self-ionization reaction: 2 H_2O harr H_3O^+ + OH^- At equilibrium, which is attained quickly for
Calculating the concentration of excess HCL in E - Socratic The acid in excess is then titrated with N aOH (aq) of KNOWN concentration we can thus get back to the concentration or molar quantity of M (OH)2 as it stands the question (and answer) are hypothetical
Question #2e8a6 - Socratic where the combination ratio is 2:1 then 0,12 mol of nitric acid combine with 0,06 mol of #Ca (OH)_2#, but you have not this quantity, then #Ca (OH)_2# is the limitant reactant and 0,056 mol of it combines with 0,112 mol of nitric acid that you have
Question #845d9 + Example - Socratic The inductive effect is the effect on electron density in one portion of a molecule caused by electron-withdrawing or electron-donating groups elsewhere in the molecule In a covalent bond between two atoms of unequal electronegativity, the more electronegative atom draws electron density towards itself This causes the δ⁺ and δ⁻ charges of the bond dipole "H"^(δ+)-"Cl"^(δ-) If the
Question #80fd6 - Socratic 4 27 11 86 Ethanoic acid dissociates: sf (CH_3COOHrightleftharpoonsCH_3COO^ (-)+H^+) sf (K_a= ( [CH_3COO^-] [H^+]) ( [CH_3COOH])=1 8xx10^ (-5)) These are equilibrium concentrations To find the pH we need the sf (H^+) ion concentration: sf ( [H^+]=K_axx ( [CH_3COOH]) ( [CH_3COO^-])) As the base is added we get: sf (CH_3COOH+OH^ (-)rarrCH_3COO^ (-)+H_2O) The initial moles of acid is given by
Question #6b37a + Example - Socratic The longer the alkyl chain attached to the hydroxyl head, usually the more basic the conjugate base is (and the less nucleophilic)
Question #71ce2 - Socratic H^+ + OH^--> H_2O when the acid was added to the resulting solution The H^+ and OH^- react in a 1:1 ratio This tells us that the number of moles of H^+ used will be equal to the number of OH^- moles in solution Likewise, 2 moles of lithium produces 2 moles of OH^- This is also a 1:1 ratio
Question #71b91 - Socratic Since water is in excess, "67 7 g MgO" are needed to produce "98 0 g Mg(OH)"_2 Balanced equation "MgO(s) + H"_2"O(l)"rarr"Mg(OH)"_2("s")" Moles magnesium hydroxide Start with the given mass of "Mg(OH)"_2 and convert it to moles by dividing by its molar mass ("58 319 g mol") Since molar mass is a fraction, "g" "mol", we can divide by multiplying by the reciprocal of the molar mass, "mol" "g
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 #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