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- Question #8556c - Socratic
This means that the anilinium ion is a reasonably strong conjugate acid producing higher degree of hydrolytic effect than the Nitrite ion which => # [H^+] > [OH^-]# => pH < 7 (acidic)
- What is the product of the following reaction? 1)CH_3 OH - Socratic
These are ostensibly acid-base reactions For ammonium we could write NH_4^+ +HO^(-) rarr NH_3(aq) + H_2O(l) For methanol, the acid base reaction would proceed
- Question #f2dad - Socratic
1 The yield will decrease; 2 There will be no effect on the yield > 1 Using 80 % propan-1-ol I'm guessing that you are making the propyl ester of a carboxylic acid and then titrating the reaction mixture to determine how much acid remains at the end of the esterification underbrace ("RCOOH")_color (red) ("carboxylic acid") + underbrace ("CH"_3"CH"_2"CH"_2"OH")_color (red) ("propan-1-ol
- Question #9be9c - Socratic
The correct answer is a) hydroxide, carbonate, and hydrogen carbonate Alkalinity is another word for basicity (the concentration of hydroxide ions) a) Hydroxide ions OH⁻ are the strongest base you can have in water Carbonate ions and hydrogen carbonate ions also react with water to form hydroxide ions CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻ HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻ b
- How is the pH of a solution related to the [H_3O^+]? | Socratic
pH=-log_10 [H_3O^+] So given a neutral solution, [H_3O^+] [HO^-]=10^ (-14), and since, by definition, [H_3O^+]= [""^ (-)OH], i,e, [H_3O^+] = 10^-7*mol*L^-1 And thus
- Question #7c802 + Example - Socratic
Oxides that form neither an acid nor a base, in combination with water Acid oxides will react with water to ultimately give off H^+: Sulphur trioxide SO_3 will form sulfuric acid: SO_3 (g) + H_2O (l)->H_2SO_4 which will give off H^+ Basic oxides will accept H^+: Calcium oxide CaO will form a base: CaO (s)+H_2O (l)->Ca (OH)_2 The OH^- part combining with H^+ to form water Ferrous oxide FeO
- Question #cebdb - Socratic
The resultant vector shows the plane moving 528 16 kph bearing 65 9 degrees I always like to sketch out an idea of what we're looking at Let's resolve the vectors and find the x and y components * * * * * * * * * * * * * * * * * * * * * * * Vector A 500 kph 60 degrees We can find the y component using sin (theta)= (opp) (hyp) Lucky for us, we have hyp, 500, and theta, 60 sin (60)= (opp) 500
- Question #22e51 - Socratic
Products formed are; Cu (OH)_2 + HCl -> CuCl_2 + H_2O Since CuCl_2 is soluble in water But; Cu (OH)_2 + HCl -> CuCl_2 + H_2O " [Not Balanced]" Balancing the equation
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