How do i calculate the number of moles of magnesium, chlorine, and oxygen atoms in 7.70 moles of magnesium perchlorate, $"Mg"("ClO"_4)_2$. Express the number of moles of Mg, Cl, and O atoms numerically, separated by commas. moles of Mg, Cl, O =?

The weighted average is the sum of each individual nuclide, multiplied by its isotopic abundance. So, we simply do the arithmetic: $0.7577xx34.969+0.2423xx36.966 = ???$ amu Then look at to...

In a binary compound, $XY$, if $%X$ is given, then, clearly, $%Y=100-X%$ with respect to mass. So we take that figure and multiply it by the mass of the...

Your goal here is to find the that exists between the anhydrous salt lithium perchlorate, $"LiClO"_4$, and the water of crystalization, $"H"_2"O"$. Now, you know that the anhydrous salt...

$"1 mol Mg(ClO"_4)_2"$ contains $"1 mol Mg atoms", "2 mol Cl atoms", and "8 mol O atoms"$. To determine the number of moles of each element in $"7.50 mol"$ of...

The calculation might be easier if we multiply all the atoms inside parentheses by the subscript 2. Then the formula becomes $"MgCl"_2"O"_8$. We see that one formula unit of $"Mg"("ClO"_4)_2$...

The idea here is that you need to use the of the two in the two to find their . Once you have their , you also have the ratio...

The molar mass of $Mg(OH)_2$ is $58.3$ $gmol^-1$. The formula mass contains one mole of $Mg$ $(24.3$ $gmol^-1)$ and two moles each of $O$ $(16$ $gmol^-1)$ and hydrogen $H$ $(1$...

$"Moles of chlorine"=(16.7*g)/(35.45*g*mol^-1)=0.47*mol$ $"Moles of sodium"=(10.8*g)/(22.99*g*mol^-1)=0.47*mol$ $"Moles of oxygen"=(22.5*g)/(16.00*g*mol^-1)=1.41*mol$ We divide thru by the smallest quantity, that of sodium, to give an empirical formula.....of $NaClO_3$, i.e. $"sodium chlorate"$.

We use the Ideal Gas Equation to access the molar quantity of dioxygen gas............ $n=(PV)/(RT)=(0.927*atmxx0.4981*L)/(0.0821*(L*atm)/(K*mol)xx418.4*K)=1.34xx10^-2*mol$. Given the stoichiometric equation.......... $Mg(s) + O_2(g) rarr MgO(s)$ There are $1.34xx10^-2*molxx24.31*g*mol^-1=??*g$ metal required for...

Cl is acting as a catalyst. ClO is an intermediate. A catalyst is a species that is present at the beginning of a reaction and reappears at the end. It...