Mercuric oxide decomposes to the mercury and oxygen. Dividing the weight of the mercury (13.2 g) by the weight of the oxide before heating (14.2 g), and using three significant digits to match the weight measurements, we get 0.930 = 93.0% for the proportion of mercury. The remainder is oxygen.
We use the Ideal Gas equation to access the moles of carbon dioxide evolved: $n=(PV)/(RT)$ $=$ $((750*mm*Hg)/(760*mm*Hg*atm^-1)xx75.0xx10^-3*L)/(0.0821*L*atm*K^-1*mol^-1xx293*K)$ $=3.1xx10^-3mol$. $"Mass of carbon dioxide produced"=3.1xx10^-3molxx44.0*g*mol^-1=??g$ The key to doing this problem was...
1 Answers 1 viewsWe need (i) a stoichiometric equation........ $Pb(NO_3)_2 +Delta rarr PbO(s)+2NO_2(g)+1/2O_2(g)$ This simply must be known; nitrates undergo thermal decomposition to give oxide and nitric oxide and dioxygen gas. And (ii)....
1 Answers 1 viewsTo calculate the mass of aluminum in aluminum oxide, you divide the given mass of aluminum by the given mass of aluminum oxide and multiply times 100. $(4.78color(red)cancel(color(black)("g")))/(6.67color(red)cancel(color(black)("g")))xx100 = 71.7%"$...
1 Answers 1 viewsAs is typical with these questions, we assume $100*g$ of unknown compound, and work out the MOLAR quantities of each element present: $" moles of C":$ $(15.8*g)/(12.011*g*mol^-1)=1.32*mol$ $" moles of...
1 Answers 1 viewsAll we have done is to divide the elemental mass by the compound mass, and made it into a percentage. We could also use these percentages to give an empirical...
1 Answers 1 views$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 of tin in the compound will be equal to the ratio between the mass of tin and the mass of the compound, multiplied by $100$. $color(blue)("% Sn"...
1 Answers 1 viewsIn order to find a compound's you must find the smallest whole number ratio that exists between the that are a part of this compound. You know that your...
1 Answers 1 views1 mole of aluminum oxide has a molecular weight of 102 grams (Al=27 grams and O=16 grams). It means that for 1000 kg of compound you can get 529 kilograms...
1 Answers 1 views$CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)uarr$ The $Delta$ symbol represents heat, and you have to fiercely heat these carbonates to effect decomposition. Note that this reaction is certainly stoichiometrically...
1 Answers 1 views