54+72+192=318 $Al_2=54$ $C_2=24 and then 24*3=72$ $O_4=64 and then 64*3=192$
1 Answers 1 views$"Oxalic acid"$ is the simplest dicarboxylic acid, $HO(O=)C-C(=O)OH$; its salts are called $"oxalates"$, i.e. $C_2O_4^(2-)$, which is a chelating ligand. In the laboratory it is often used as the $"dihydrate"$,...
1 Answers 1 viewsA completely filled outer electron shell is stable because it can not gain any more electrons, it is full. Being completely full it has no desire to lose electrons thereby...
1 Answers 1 viewsSo we approach the equation by the method of half equations: $Fe(III)$ is reduced to $Fe(II)$: $Fe^(3+) + e^(-)rarrFe^(2+)$ $(i)$ And $"oxalate ion"$ is oxidized to $CO_2$: $C_2O_4^(2-) rarr 2CO_2...
1 Answers 1 viewsAssume 100 g of acid. Breaking it down into its components, there are $(26.68*g)/(12.01 *g*mol^-1*C); (2.24*g)/(1.00794 *g *mol^-1*H); (71.08*g)/(16.00*g*mol^(-1)*O)$ $=$ $C:H:O$ $=$ $2.22:2.22:4.44$ $=$ $1:1:2$. Therefore, empirical formula of oxalic...
1 Answers 1 views$sf(H_2C_2O_(4(aq))+2NaOH_((aq))rarrNa_2C_2O_(4(aq))+2H_2O_((l)))$ $sf(n_(H_2C_2O_4)=cxxv=0.1089xx10.00/1000=0.001089)$ $:.$$sf(n_(NaOH)=0.001089xx2=0.002178)$ $sf(c=n/v)$ $:.$$sf([NaOH]-=(0.002178)/(0.01865)=0.1168color(white)(x)"mol/l")$
1 Answers 1 views$"Molarity"$ $=$ $"Moles of solute"/"Volume of solution"$ $=$ $(0.063*g)/(126.07*g*mol^-1)xx1/(0.250*L)$ $=$ $??*mol*L^-1$
1 Answers 1 viewsOxalic acid is a diacid, i.e. $HO(O=)C-C(=O)OH$. Formally each equiv of acid requires 2 equiv of base to achieve neutrality. Of course $pK_(a2)$ will be substantially greater than $pK_(a1)$. You...
1 Answers 1 views$"mass percent"="grams of solute"/"grams of solution"xx100"$ The mass in grams of a mass % solution, is the sum of the mass in grams of the and the mass in grams...
1 Answers 1 viewsThe first thing that you need to do here is to pick a sample of this solution and use its to determine how many moles of it contains. To...
1 Answers 1 views