2019年4月18日 · $\begingroup$ Well there are basically 2 methods to know, how many apples were taken from the box. 1/ Count apples at the start and at the end, and then calculate the …

2021年5月22日 · $\ce{MnO4^- -> Mn^2+ + 4H2O }$ $\ce{C2O4^2- -> 2CO2}$ We start with O because we are using water to balance it. Each time we add a water, we also add two H, so it …

2018年5月2日 · [tex]sp^2[/tex]single bond to the other carbon, single bond to one of the oxygens and double bond to the other.[tex]sp^2[/tex] hybridization is observed when o…

2018年4月4日 · In table 10, the book specifically lists $\ce{[Co(C2O4)3]}^{3-}$ as low spin and cites to J. Chem. Soc. (A) (1966) 798. Therefore, according to the historical valence bond …

Balance the atoms: $$\ce{C2O4^{2-} -> 2CO2}$$ Now balance the charge by adding electrons: $$\ce{C2O4^{2-} -> CO2 + 2e-}$$ Now for the manganese dioxide reduction: $$\ce{MnO2 -> …

2019年2月15日 · I figured I’d study $\ce{C2O4^2-}$ as an organometallic ligand, and found that it is weak field (pi donor). I hence decided to construct a MO diagram. I hence decided to …

2018年11月7日 · MnO4 - + C2O4 2- + H+ _____> Mn2+ + CO2. Balance the following chemical equation. Get the answers you need, now!

2020年11月21日 · If $\ce{Co}$ is at +II, the charge of the complex is $2$ +, as $\ce{NH3}$ is not charged. If $\ce{Co}$ is at +III, the charge of the complex is $3$ + for the same reason. For …

2017年6月23日 · If so, the compound is a salt and should be chargeless. $$\ce{(NH4)2[Cu(C2O4)]}$$ This is because oxalate is -2 charged and there are two of them …

2017年3月24日 · 2. Separating the reaction into half as it is a combination of reduction and oxidation: Oxidation reaction: Reduction reaction: 3. Balancing atoms in each half of the …