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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
- 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
- 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 #d9b58 - Socratic
Explanation: #H_3PO_4 (aq)+Ca (OH)_2 (aq) rarr Ca_3 (PO_4)_2+H_2O (l)#
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