
Question #477c5 - Socratic
On the product side the Carbonic Acid (#H_2CO_3#) is the Conjugate Acid as it is the hydrogen donor to the Conjugate Base (#OH^-#) as it receives the hydrogen ion.
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 …
Question #a52c4 - Socratic
MnO_4^(2-) +4H_2O + 2S^(2-) =2 S+ Mn^(2+) + 8 OH^- Mn reduceds itself from N° of oxidation +6 to +2 buying 4 electrons. To balance the semireaction i write 8 OH^- on the right because the reaction …
6-10. What are the name of the following compounds? a. Ca (OH)2 b.
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 …
Question #019dd - Socratic
Only (C) is fully correct. As you can see, steam typically reacts with alkenes by adding H on one double-bonded atom and OH on the other. But there are two possible choices for which atom gets the single …
Question #74006 - Socratic
(d) is the correct option. CH_3CH(OH)CH_2CH_2CH_3 (pentan-2-ol) being secondary alcohol on oxidation first tansformed into ketone. CH_3CH(OH)CH_2CH_2CH_3stackrel([O])-> …
Question #97503 - Socratic
Here's what I get. The general equation for the dissociation of a carboxylic acid is "R-COOH + H"_2"O" ⇌ "R-COO"^"-" + "H"_3"O"^+ All we have to do is write the ...
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) …
Question #5392e - Socratic
Dec 13, 2017 · Here's what I get. Step 1. Write the molecular equation "Fe" ("NO"_3)_3" (aq)"+ "3NH"_3" (aq)" + "3H"_2"O (l)" → "Fe (OH)"_3" (s)" + "3NH"_4"NO"_3" (aq)" Step 2 ...
Question #0a25a - Socratic
Mar 20, 2017 · Lithium hydroxide has a mass of 24 g per mole, so 2.50 g is 0.104 mole. Dissolve this in 1.50 L, and the concentration is [LiOH] = 0.104 mol -: 1.50 L = 0.0694 "mol"/L Since LiOH is a strong …