You know the normal boiling point of water. The normal boiling point of hydrogen sulfide is $-60$ $""^@C$; that of hydrogen selenide is $-41.2$ $""^@C$; that of hydrogen telluride is...
1 Answers 1 viewsThe formula for is $"% by mass" = "mass of a component"/"total mass" × 100 %$ You have $"9.03 g Mg"$ and $"3.48 g N"$, so $"total mass =...
1 Answers 1 viewsIn order to find the of hydrogen in this compound, you must determine how many grams of hydrogen you'd get in $"100 g"$ of compound. In your case,...
1 Answers 1 viewsAnd thus......... $"%C"=(40.8*g)/(95.2*g)xx100%=42.9%$. $"%O"=(54.4*g)/(95.2*g)xx100%=57.1%$. Why should the individual percentage sum to $100%$? Do they...... We can divide the elemental percentages by the atomic masses if we assume $100*g$ of unknown...
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 viewsAnd thus $%C$ $=$ $(5.34*g)/(5.34*g+0.42*g+47.08*g)xx100%$ $=$ $10.1%$ I leave it to you to find $%H$ and $%Cl$. And of course the percentages must sum to $100%$. Why? Given these data,...
1 Answers 1 viewsTrying to present a possible Answer We know that the molar mass of carbon and hydrogen are $C=12"g/mol" and H = 1"g/mol"$ When Carbon Hydrogen mass ratio is 11.89 The...
1 Answers 1 views$%Hg="mass of mercury"/"mass of mercury and bromine"xx100%$ $=(60.2*g)/(60.2*g+24.0*g)xx100%=??%$ In such a binary compound, if you have the percentage by mass of one component, you also have the percentage by mass...
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 viewsWe thus predict that hydrogen sulfide would have a formula of $H_2S$, an analogue of the popular $H_2O$ molecule.
1 Answers 1 views