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Question #b8408 - Socratic
The thing to keep in mind here is that nitric acid is a strong acid, which means that it will ionize completely in aqueous solution to produce hydronium cations, H3O+, and nitrate anions, NO− 3. HNO3(aq) +H2O(l) → H3O+ (aq) + NO− 3(aq) Notice that every mole of nitric acid that is dissolved in water produces 1 mole of hydronium cations. This means that the concentration of hydronium ...
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 #5f837 - Socratic
The balanced chemical equation for the partial dissociation of the base looks like this "BOH"_text ( (aq]) rightleftharpoons "B"_text ( (aq])^ (+) + "OH"_text ( (aq])^ (-) By definition, K_b will be equal to K_b = ( ["B"^ (+)] * ["OH"^ (-)])/ ( ["BOH"]) Rearrange this to get K_b/ ( ["OH"^ (-)]) = ( ["B"^ (+)])/ ( ["BOH"]) color (blue) ( ( ["BOH ...
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 ...
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 #a4a33 - Socratic
The added water to reach "100.00 mL" doesn't change the mols of HCl present, but it does decrease the concentration by a factor of 100//40 = 2.5. Regardless, what matters for neutralization is what amount of "NaOH" you add to what number of mols of "HCl". I got "pH"'s of 1.36, 1.51, 1.74, 2.54. You started with "0.1100 M HCl", but it was diluted from "40 mL" to "100 mL". That decreases its ...
Question #50782 - Socratic
sf(pH=7.04) This is a very dilute solution of NaOH so I would expect the pH to be 7 or very slightly above. Because the concentration is so low we must also take into account the ions that are formed from the dissociation of water: sf(H_2OrightleftharpoonsH^++OH^-) sf(K_w=[H^+][OH^-]=10^(-14) at sf(25^@C) If you do a thought experiment you can imagine adding a small amount of NaOH to water ...
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 #70577 - Socratic
Here's what I got. Start by writing the balanced chemical equation for this neutralization reaction. Since sodium hydroxide is a strong base that dissociates completely in aqueous solution, you can represent it by using the hydroxide anions, "OH"^(-) "HA"_text((aq]) + "OH"_text((aq])^(-) -> "A"_text((aq])^(-) + "H"_3"O"_text((aq])^(+) Now, you have 1:1 mole ratios across the board, you can say ...
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