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What is the chemical formula for lithium hydroxide? - Socratic
LiOH Lithium is a Group 1 metal and commonly forms a M^+ ion. Hydroxide anion, ""^(-)OH, has a unit negative charge. When they make music together, there is thus 1:1 stoichiometry between ions: Li(s) + H_2O(l) rarr LiOH(aq) + 1/2H_2(g)uarr

What is the formula of #"magnesium hydroxide"#? - Socratic
Mg(OH)_2 Now we know that hydroxides are salts of HO^-, and some metal ion. Now if the parent metal has an electronic configuration of 2:8:2, then there are 12 electrons, and the atomic number of the metal is equal to 12. We look on the Periodic Table, and we find that Z=12, for "magnesium metal". As a Group 2 metal, magnesium forms a Mg^(2+) ion, and hence its hydroxide is Mg(OH)_2.

What is the value of ΔG°cell? - Socratic
They have a cadmium anode, which is oxidized to Cd(OH)2(s) in a basic background electrolyte of concentrated OH–, and a cathode half-reaction in which NiO(OH)(s) is reduced to Ni(OH)2(s). Cd(OH)2 +2e^- -> Cd +2OH- E^o red = -0.403V

What makes for a good leaving group? + Example - Socratic
A good leaving group has to be able to part with its electrons easily enough, so typically, it must be a strong acid or weak base relative to other substituents on the same molecule. It helps to know the pKa of what would be leaving. Let's say you had a mechanism where you are trying to do an E2 reaction to make an -OH (hydroxyl) group leave. Maybe you have this compound on hand, sec-butanol ...

Question #6b9a2 - Socratic
"a." 0.02 "mol" "b." 0.89 "g" In order to answer these questions we first need to know the equation representing this chemical reaction.

If 50.0 milliliters of 3.0 M "H"_3"PO"_4 completely ... - Socratic
Mg(OH)_2(s) +H_3PO_4(aq) rarr Mg^(2+)HPO_4^(-)(aq) + 2H_2O(l) With respect to phosphoric acid, we got a molar quantity of... 3.0*mol*L^-1xx50xx10^-3*L-=0.150*mol And thus we SUPPOSE it to have neutralized 0.0750*mol magnesium hydroxide....which was ostensibly present in a 150*mL volume...

Why does neutralisation of any strong acid in an aqueous ... - Socratic
Because its the same reaction. A monobasic [monoprotic] acid like HCl has one H atom, which fully dissociates to form a hydrogen ion, H^+. The strong alkali fully dissociates to forman hydroxide ion OH^-, which react to from water. This is the ionic equation for neutralisation and occurs for all strong acid/strong alkali reactions.

Molarity question? - Socratic
Step 3: There will be no concentration of #OH^(-)# ions because they are in a solid compound, not dissolved in solution. They will no longer be anions in a solid, neutral compound. Thus, molarity of #OH^(-)# is 0. Part C: Step 1: Calculate #q# for the heat the water absorbed, not the heat released by the reaction. #q=s*m*DeltaT#

Question #bcaa2 + Example - Socratic
Phenol is hydroxybenzene, which has a proton on the hydroxyl ("OH") group. Its "pKa" is about 9.9, which makes it somewhat acidic. When you deprotonate phenol, its anionic form has multiple resonance structures: Because of the electron delocalization, the anion (conjugate base) is stabilized (relative to an aliphatic alcohol), and so the phenol molecule itself is more acidic than an aliphatic ...

Calculating the concentration of excess HCL in E - Socratic
Well, you have got us here mate.... I presume that you performed an acid base reaction with a known amount of HCl(aq)...and this reacts with your base according to the equation.... M(OH)_2(s) + 2HCl(aq) rarr MCl_2(aq) + 2H_2O(l) The acid in excess is then titrated with NaOH(aq) of KNOWN concentration....we can thus get back to the concentration or molar quantity of M(OH)_2...as it stands the ...

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