If you think to magnesium oxide and to concentration of oxygen in MgO the answer is 40,007 33 % oxygen.
The answer is o,5 moles MgO.
MgO
"MgO" is magnesium oxide and "H" is hydrogen, as in "Mg + H(2)O => MgO + H(2)" MgO + H2 ---> H2O + Mg
+2 for Mg and -2 for O
Ionic. because a metal(Mg) + a non-metal (O).
The percent of Mg calculated will be too high. Let's say that you reacted 1.00 g of Mg and made some MgO, but so much MgO escaped as smoke that only 1.00 g of MgO was left. You would then conclude from the numbers that the mass of O in the MgO was zero! This would lead you to conclude that the percent of Mg in the MgO was 100 %, which is silly and clearly in error. Although this is an extreme example, it illustrates that the loss of MgO as smoke from the crucible leads to a percent of Mg (calculated) that is above the expected 60.3 %.
60%
The answer is o,5 moles MgO.
Double bond Mg=O
the Lewis formula for MgO is Mg2+[O]2- Which mean Mg loses 2 electrons and O gains 2 electrons to be stable.
60%
well Mg is an element ( I Magnesium) and O is oxygen hence MgO is definitely a molecule not an atom ...
Pure MgO will absorbe water from the air causing an error in the experiment if that extra mass is not accounted for. By adding water in the very beginning you eliminate this error.
the final formula of magnesium oxide is MgO.
MgO
"MgO" is magnesium oxide and "H" is hydrogen, as in "Mg + H(2)O => MgO + H(2)" MgO + H2 ---> H2O + Mg
Mg + O ----> MgO