2017年10月26日 · "aquachlorobis(ethylenediamine)chromium(III) sulfate" What is the electron geometry of this complex? You should have memorized the following ligands: "en" = ethylenediamine, charge = 0 "H"_2"O" = aqua, charge = 0 "Cl"^(-) = chloro, charge = -1 (Remember that "ammine" is for the "NH"_3 ligand, with two "m"'s.) Then you list them alphabetically in front of the metal, along with its oxidation ...

2017年6月28日 · +3 A compound has a zero net charge. so 2 xx Cr + 3 xx O = 0 Oxygen O almost always has a charge of -2 so 2 xx Cr + 3 xx (-2) = 0 2 xx Cr + -6 = 0 add + 6 to both sides 2 xx Cr + -6 +6 = 0 + 6 so 2 xx Cr = +6 divide both side by 2 2/2 xx Cr = + 6/2 equals Cr = + 3

2018年4月20日 · +3 In this compound, CrCl_3, it has a net charge of 0. Chlorine is more electronegative than chromium, and so it will possess its usual state of -1. There are three chlorine atoms in total, and so their total charge will be -1*3=-3. Let x be the oxidation number of chromium. Then we have, x-3=0 x=3 And so, chromium would have an oxidation state of +3.

2017年12月9日 · You need to learn something early on in chemistry, maybe in life. It's not about how little work you can do. If you work hard now, this will be easier in an exam setting. I worked hard at this and this is trivial to me, now—I can do this in my head. Moving on! In reduction-oxidation reactions, one species is reduced and another is oxidized. Usually this is straightforward, having calculated ...

2014年8月2日 · Al: 0 → +3; Change = +3. Step 3. You need 3 atoms of Mn for every 4 atom of Al. This gives us total changes of +12 and -12. Step 4. Insert coefficients to get these numbers. 3 MnO₂ + 4 Al → 3 Mn + 2 Al₂O₃. Every substance now has a coefficient. The equation should be balanced, Step 5. Check that all atoms balance. Left hand side: 3 Mn ...

2017年1月16日 · Yes, they should all work. > The driving force for reactions 1) and 4) is the formation of the weak electrolyte, "H"_2"O". The driving force for reaction 2) is the formation of the gas, "H"_2, which escapes into the atmosphere. The driving force for reaction 5) is the formation of the weak electrolyte, "HNO"_2. I was initially worried about reaction 3) but, if the reaction is carried out in ...

2016年8月21日 · We need a volume of 0.75*L Ni^(2+) + 2OH^(-) rarr Ni(OH)_2(s)darr Alternatively, Ni(NO_3)_2(aq) + 2NaOH(aq) rarr Ni(OH)_2(s) + 2NaNO_3(aq) We need a stoichiometrically balanced equation to inform our calculations. "Moles of nickel ion" = 150.0xx10^-3Lxx0.250*mol*L^-1=0.0375*mol. Because of the stoichiometry, we need 2xx0.0375*mol" hydroxide ion"=0.0750*mol. And thus volume of 0.100*mol*L^-1 ...

2016年3月6日 · "Iron (III) sulfate" or "ferric sulfate" Either name is still commonly used. Of course, it is formulated on the basis of electrical neutrality. 2 equiv Fe^(3+), the ferric ion, and 3 equiv SO_4^(2-) give a neutral charge.

2017年2月26日 · Unknown, you will have to look in a specialist text. This link gives the solubility of "cerium(III) sulfate" as 9.25*g "per 100 mL of water at room temperature".

2016年5月28日 · 3:1 The balanced equation will tell the stoichiometric ratio of reactants and products. Here in the balanced equation the coefficient before NaOH is 6 and that before Al(OH)_3 is 2. So the required mole ratio is 6:2=3:1