Questions and answers by

Drbob222

Answers (7)

how to do the question excuse me
Thanks my dude
you fake boi
I have answered three times here and nothing posts. Try again. The balanced equation is Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
..........PbCl2(s) ==> Pb^2+ + 2Cl^- x = solubility..x.......x.......2x Ksp = (Pb^2+)(Cl^-)^2 Ksp = (x)(2x)^2 Solve for x which will give you the moles/L of PbCl2. Mols/L x 0.48 L = moles in 480 mL. Then moles x molar mass PbCl2 = grams PbCl2.
q1 = heat released on cooling liquid water from 25C to zero C. q1 = mass H2O x specfic heat water x (Tfinal-Tinitial)---for example, Tf is zero and Ti is 25) q2 = heat released on freezing liquid water at zero C to ice at zero C. q2 = mass water x heat
I don't know what Ero stands for but these are reduction potentials you have written. Multiply equation 2 by 2, reverse it, add to equn 1, and eliminate ions common to both sides (for example you can eliminate some of the OH and some of the H2O) and that

Ask a New Question