By definition, $rho_"metal"="mass of metal"/"volume of metal"$ $=(800*g)/(33.0*cmxx46.0*cmxx48.0*cm)=0.0659*g*cm^-3$ And this is an IMPROBABLY undense metal. The LEAST dense metal, lithium, is approx. $0.5*g*cm^-3$...(at any rate this metal reacts with water)...the...
1 Answers 1 viewsa magnesium atom reacts to form magnesium ion losing 2 electrons and thus has a +2 charge and 10 electrons .
1 Answers 1 views$"Moles of calcium"$ $=$ $(1200xx10^-3*g)/(40.08*g*mol^-1)$ $=$ $0.04008*mol$. And thus we need $0.04008*mol$ $"calcium carbonate"$, i.e. a mass of $0.04008*molxx100.09*g*mol^-1$ $=$ $??g$.
1 Answers 1 viewsThe idea here is that chemical reactions must obey the Law of , which states that when a chemical reaction takes place, the total mass of the reactants must be...
1 Answers 1 viewsOkay, so you've performed the classic magnesium oxide lab. I assume that you have your data, so I'll just walk you through this from a theoretical standpoint. So, you...
1 Answers 1 viewsWe can represent the oxidation so.......... $Mg(s) +1/2O_2(g) stackrel(Delta)rarrMgO(s)$ And thus combustion of one equiv magnesium metal, $24.3*g$, should lead to formation of approx. $40*g$ magnesium oxide. And thus the...
1 Answers 1 viewsSo we have the $"%calcium carbonate, "m/m$. We are not finished yet, because we have to address the percentage by mass of calcium metal in calcium carbonate: $%Ca=(40.08*g*mol^-1)/(100.09*g*mol^-1)=40%$ with respect...
1 Answers 1 views$MgCO_3(s)+DeltararrMgO(s) + CO_2(g)$ And we would have to heat the magnesium salt VERY fiercely to get complete decarboxylation. $"Moles of carbon dioxide LOST"$ $=$ $(4.40*g)/(44.01*g*mol^-1)$ $=0.010*mol$ And therefore in...
1 Answers 1 viewsThe 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 $H$ $(1$...
1 Answers 1 viewsWe 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...
1 Answers 1 views