In $NH_4^+$, the $/_H-N-H$ bond angles are $109.5$ $""^@$, as we would expect for a regular tetrahedron. $F_2O$, and $H_3O^+$ share this electronic configuration, but here the central $O$ atom...
......And there are 2 double bonds around the central atom: $O=C=O$. And $OF_2$ is a bent molecule, with 2 lone pairs on the central oxygen atom; $/_F-O-F=103^@$.
Indeed, oxygen can only bond with fluorine to form oxygen difluoride, $"OF"_2$. On the other hand, sulfur can easily form sulfur difluoride, $"SF"_2$, sulfur tetrafluoride, $"SF"_4$, and sulfur hexafluoride,...
Oxidation state is the charge left on the central atom, when all the pairs of electrons are removed from the central atom, with the charge assigned to the most electronegative...
$"Avogadro's number"$ $=$ $6.022xx10^23*mol^-1$. Now, in $44*g$ of gas there are $"Avogadro's number"$ of molecules by definition. And thus $6.022xx10^23*mol^-1$ $xx$ $(20*g)/(44.0*g*mol^-1)$ $~=$ $3.0xx10^23" carbon dioxide molecules"$. You can make...
$2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)$ Molar mass of $NaOH = 40 $ $2xx40 g$ of $NaOH$ produces $6.02 xx 10^23$ molecules of $Na_2CO_3$ $:.$ $10 g$ of $NaOH$...
And the number of nitrogen atoms is $1/14*N_A$. $N_A="Avogadro's number"$, $=6.022xx10^23*mol^-1$. Why use such an absurdly large number? Well, if we got $N_A$ $"hydrogen atoms"$ we have a mass of...
This is a limiting reactant problem. Step 1. Write the balanced equation. $"C"_3"H"_8 + "5O"_2 → "3CO"_2 + "4H"_2"O"$ 2. Identify the limiting reactant One way to identify the...
Oxygen always has an oxidation number of $-2$ unless in peroxides, like $H_2O_2$, where it has an oxidation of $-1$. When $O$ is attached to an even more electronegative element...
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