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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 #c548d - Socratic Question 1: K_ (sp)= 1 1 xx10^ (-11) Question 2: s= 4 9 xx10^ (-12)M Quest (1) determine the ksp for magnesium hydroxide Mg (OH)_2 where the molar solubility of Mg
Question #9f499 - Socratic Explanation: Your starting point here is the pH of the solution More specifically, you need to use the given pH to determine the concentration of hydroxide anions, #"OH"^ (-)#, present in the saturated solution
6-10. What are the name of the following compounds? a. Ca (OH)2 b. Fe . . . Generally, OH adds "hydroxide" to an inorganic compound's name Moreover, element names aren't capitalized unless at the beginning of a sentence We write iron (II) hydroxide instead of just iron hydroxide as iron takes the form of its +2 oxidation state, out of its 10 oxidation states
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
How many grams of \text {NH}_4\text {OH} do I need to make . . . - Socratic "6 3072 g" >>"Molarity" = "Moles of solute" "Volume of solution (in litres)" "0 45 M" = "n" "0 4 L" "n = 0 45 M × 0 4 L = 0 18 mol" You need "0 18 mol" of "NH"_4"OH" Molar mass of "NH"_4"OH" is "35 04 g mol" Mass of solute = 0 18 cancel"mol" × "35 04 g" cancel"mol" = "6 3072 g"
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 #e7848 - Socratic Similarly, OH^- becomes H_2O, indicating a gain of a H^+ ion So, you can say that NH_4^+ is the acid, and OH^- is the base Conjugates are basically the "other" term For every acid, you have a conjugate base (that no longer has that extra H^+ ion), and for every base, you have a conjugate acid (that has an extra H^+ ion)
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