Whats the difference between 2O and O2 [duplicate] $\ce{2O}$ is basically two atoms of oxygen, unbonded and separate On the other hand, $\ce{O2}$ is oxygen molecule , which is more commonly called oxygen gas Share
What is the difference between O O2 - Chemistry Stack Exchange The common confusion here is that two very different things have the same name $\ce O$ is a free oxygen atom and $\ce{O2}$ is two oxygen atoms chemically bound to form an oxygen molecule
diH2O, dH2O, and DI H2O. What do they mean? What do $\ce{diH_2O, dH_2O}$, and $\ce{DI H_2O}$ mean? I'm not asking for a description of deionized water and distilled water I'm asking what the three abbreviations formulas mean
electrochemistry - Why is the water reduction of oxygen favored in . . . $\ce{O2 + 2e^- -> 2O^-}$ Or this? $\ce{O2 + 2e^- -> HOOH}$ 2) From the above we can deduce that water is the product; water is pretty stable in either acidic or basic medium - i e it won't react to a large extent with either hydronium or hydroxide ion Plus the oxygen in water has a -2 oxidation state So if we look at our skeleton half reaction:
How does a cyclization occur with a carboxylic acid and (CF3CO)2O? The product of the reaction below is unsaturated, bicyclic ketone, but I'm confused as to how (CF3CO)2O reacts with it to cause the cyclization There is no base involved so I don't know how the carbon delta to the carboxylic acid attacks it
Why is the iron (III) ion more acidic than the aluminum ion? According to this and other sources, the Ka of $\ce{[Fe(H_2O)_6]^{3+}}$ and $\ce{[Al(H_2O)_6]^{3+}}$ is $\mathrm{6 3×10^{-3}}$ and $\mathrm{1 4×10^{–5}}$ respectively, indicating that the iron (III) ion is more acidic than the aluminum ion
Finding the formula of an alcohol with a given molecular mass $\begingroup$ As I commented 2 years ago, there is no information to show that the alcohol is saturated, to which the formula CnH2n+2O applies $\endgroup$ – user55119 Commented May 9, 2020 at 14:38
inorganic chemistry - Regarding the oxidation of Manganese(II) ion by . . . The manganese ion does get oxidized, but as the oxidation number of sulphur in $\ce{S_2O_8^2-}$ and $\ce{SO_4^2-}$ is the same, i e , equal to +6, $\ce{S_2O_8^2-}$ doesn't seem to undergo reduction For a redox reaction, both oxidation and reduction are necessary to occur simultaneously
Why is 2H2O2 - gt; 2H2O + O2 a first order reaction? My book says that the reaction $\ce{2H_2O_2 -> 2H_2O + O_2}$ is a first order reaction and that it's rate equation is $\ce{k[H_2O_2][I^{-}]}$ However i don't see why that would be first order? Isn't the order of the reaction supposed to be the sum of powers of the concentration of the reactants?