The chemical formula for ammonia is $2NH_3$ meaning that we have two nitrogen atoms and six hydrogen atoms. To find the ratio of nitrogen atoms to hydrogen atoms...
1 Answers 1 viewsWe always have to read the problems completely to understand the question. In this case even when we have pressure and temperature conditions specified, they don't matter, first because they...
1 Answers 1 viewsRemember to include the coefficients in the change in concentratio, as well as the exponents. I get $K_c = 1.69$. If $K_p = K_c(RT)^(Deltan_"gas")$, what would $K_p$ be if this...
1 Answers 1 viewsAmmonia is a Bronsted base: $NH_3(aq) + H_2O(l) rightleftharpoons NH_4^+ + OH^-$ $pK_b = 4.75$ I am not going to solve this equation using $pK_b$, but most of the ammonia...
1 Answers 1 viewsWe gots $Al(OH)_3$...even tho this is still a basic species. $"Alumina"$, $Al_2O_3$, is an amphoteric oxide...that could act as an acid. The hydroxides of lithium, calcium, or sodium, all give...
1 Answers 1 viewsAmmonia is a weak base and ionises in water: $sf(NH_3+H_2OrightleftharpoonsNH_4^++OH^-)$ For which: $sf(K_b=([NH_4^+][OH^-])/([NH_3])=1.8xx10^(-5)color(white)(x)"mol/l")$ These are equilibrium concentrations. The ICE table in mol/l is: $" "sf(NH_3"...
1 Answers 1 views$NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + HO^-$ And so $K_b=1.8xx10^-5=([NH_4^+][HO^-])/([NH_3(aq)])$ We put in some numbers, and we assume that $x*mol*L^-1$ of ammonia ASSOCIATES, and thus at equilibrium, we know that $x=[NH_4^+]=[HO^-]$,...
1 Answers 1 viewsSo how does this help us? You have an solution that is stoichiometric in ammonia; immediately you add hydrochloric acid, this protonates the ammonia, and now you both $NH_3$ and...
1 Answers 1 viewsNow for $"ammonium ion"$ in water $pK_a=9.24$ And thus, $9.15=9.24+log_10{[[NH_3]]/[[NH_4^+]]}$ Clearly, $log_10{[[NH_3]]/[[NH_4^+]]}=-0.09$ And so, $[[NH_3]]/[[NH_4^+]]=10^(-0.09)=0.813$ And given that $[NH_3]=0.10*mol*L^-1$, $[NH_4^+]=0.123*mol*L^-1$. Because the volume of the solution was specified to be...
1 Answers 1 viewsThere are two heat transfers involved in this problem. $"heat of solution of NH"_4"NO"_3 + "heat lost by water" = 0$ $color(white)(mmmmmmm)q_1color(white)(mmmmml) +color(white)(mmmm) q_2color(white)(mmmm) = 0$ $color(white)(mmmmml)m_1Δ_text(sol)Hcolor(white)(mmmll) + color(white)(mmll)m_2CΔT color(white)(mm)=...
1 Answers 1 views