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- What are the spectator ions in the reaction between sodium . . . - Socratic
The nitrate and the natrium ions Na_2CO_3(aq) + 2AgNO_3(aq) rarr Ag_2CO_3(s)darr + 2NaNO_3(aq) The net ionic equation is: 2Ag^(+) + CO_3^(2-) rarr Ag_2CO_3(s)darr
- 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
- 0. 10 moles of sodium cyanate dissolved in 250cm^3 distilled . . . - Socratic
"pH" = 9 16 The idea here is that sodium cyanate, "NaOCN", will dissociate completely in aqueous solution to form neutral sodium cations, "Na"^(+), and basic cyanate anions, "NCO"^(-) The cyanate anions will react with water to form isocyanic acid, "HNCO", and hydroxide anions, "OH"^(-) This tells you that you can expect the pH of the resulting solution to be higher than 7 Calculate the
- Could a buffered solution be made by mixing aqueous . . . - Socratic
No, and you've also got two definitions mixed up Two things: The whole point of a buffer is to resist pH change You're not going to do that with two things that dissociate completely "HCl" and "NaOH" are not conjugates of each other What is the definition of a conjugate base or acid? What do they differ by? Any strong acid has a terrible conjugate base that does almost literally nothing in
- 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 #22e51 - Socratic
232 views around the world You can reuse this answer
- Determine the [OH−] in a 0. 22 M solution of HClO4. Answer . . . - Socratic
Well, 14=pH+pOH we get [HO^-]=4 55xx10^-14*mol*L^-1 are we clear on this? And "perchloric acid" is an exceptionally strong acid
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