What is the electron configuration for $Fe_2$?

Here, we use as we would use any other collective number: a dozen eggs; a Bakers' dozen; a Botany Bay dozen. Of course, the mole specifies a much larger quantity,...

Simply cancel out the common items, the ones that do not undergo chemical change in the original equation: $2Fe^(3+) + cancel(6Cl^(-) + 6Na^(+)) + 3CO_3^(2-) rarr Fe_2(CO_3)_3(s)darr +cancel(6Na^(+) + 6Cl^(-))$...

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...

Phosphorus has $15$. As such its electronic configuration in ground state is $1s^2 2s^2 2p^6 3s^2 3p^3$ The following electronic configurations could be excited states $1s^2 2s^2...

Subshell notation of magnesium is 2.8.2. We know that magnesium is in Group 2, Period 3. Hence it has an energy level of 3. By doing this, we can...

If the sodium is ionised, the outer $3s^1$ electron is lost, so the configuration becomes $1s^2 2s^2 2p^6$, which is the same as neon, $Ne$.

Let's eliminate unknowns backwards, starting with $G$. From the equations for $H$, $Mn$, $K$ and $Fe$ we find: $G=C$ $F=B$ $E=1/2 B$ $D=1/2 A$ Substituting these in the...

Water has a $3$ coeeficient. Is the equation stoichiometrically balanced? If not, then it should be. The oxide ion is dibasic, and reacts with 2 equiv $H^+$ to give the...

Start by looking at the most complex compound, and balancing around that. In this case, its $Fe_2(SO_4)_(3(aq))$. We see that we need 2 Fe and 3 $SO_4$ on the...

So we approach the equation by the method of half equations: $Fe(III)$ is reduced to $Fe(II)$: $Fe^(3+) + e^(-)rarrFe^(2+)$ $(i)$ And $"oxalate ion"$ is oxidized to $CO_2$: $C_2O_4^(2-) rarr 2CO_2...