Recent content by ssb

Calculate moles EDTA = 0.040 L * 0.01175 M = 0.0047 mol EDTA Calculate moles of Excess EDTA 0.01007 L * 0.00993 M = .000099995 moles excess EDTA Calculate Moles EDTA reacted with Al^{3+} and Ni^{2+} = 0.0047 - .000099995 = 0.00037 mol total EDTA used Moles Al^{3+} comsumed = 0.02630 L...

Homework Statement 25.00 Ml sample of a solution of Al_2(SO4)3 and NiSO4 was diluted to 500.0 mL and the Al^3+ and Ni^2+ in a 25.00 mL aliquat were complexed by addition of 40.00 mL of 0.01175 M EDTA in a pH 4.8 buffer. The excess EDTA was then back titrated with 10.07 mL of 0.00993 M...

Homework Statement I was given a problem with some calculations that were made by AES. I know that I am suppose to plug in the equations into some of my beers law formulas but the question given says that in one test the AES readings were below the detection limit. Should I put in a 1 or a 0...

It would make it negative wouldn't it? E_{total} = E_{Tl^+/Tl} + E_{ksp} = -0.336 + (\frac {.05916}{-1}) * log(\sqrt {1.2x10^{-22}}) = 0.312 volts with sig figs making the answer 0.31 volts This has to be correct this time. (btw thanks you have been unknowingly teaching me latex as...

omg did i do that honestly E_{total} = E_{Tl^+/Tl} + E_{ksp} = -0.336 + (0.05916) * log(\sqrt {1.2x10^{-22}}) = -0.984 Look good now?

Ok ok ok... E_{total} = E_{Tl^+/Tl} + E_{ksp} = -0.336 + (0.05916/2) * log(\sqrt {1.2x10^{-22}}) ? This would give E_{total} = -0.66 ?

Borek thank you so much for your help on this problem and that other problem I posted. Borek I am going to level with you... I am very confused. I thought I knew what a Kso was (something to do with solubility) but I don't know much more than that. In post # 3 (my first reply) I did something...

Kso formula Tl_2S = 2Tl^+ (aq) + S^- (aq) 1.2x10^{-22} = {[Tl^+]^2[S^-]}/[{Tl_2S}] Tl_2S is a solid so we can remove it from the equation giving me 1.2x10^{-22} = {[Tl^+]^2[S^-]} So this is where I get confused... is the correct way to do it this: 1.2x10^{-22} = {[x]^2[x]}...

moles reacted = (.02425 amps * 287 seconds)/(2 (# electrons) * 9.6482x10^4 (Faraday constant) Moles reacted = 3.607 x 10^{-22} Since its a 1 to 1 ratio of mol As^{3+} to mol I_3 then there are also 3.607 x 10^{-22} mol of I_3 So I will now solve for grams of As_2O_3 Grams...

OK BOREK! You have given me some insight. Thank you! Products over reactants. and because the reactants are solid, they will be in the denominator as the number 1. Ok I was able to come up with the following calculation. I would be much appreciated if someone could confirm for me the accuracy...

Homework Statement A 3.67 gram sample of bug spray was decomposed in acid. Any As^5^+ was reduced to As^3^+ and diluted to 250.0 mL in a volumetric flask. A 5.00 mL sample of this was added to 125.0 mL of 0.0500 M KI buffered to pH 7. A coulmetric titration was carried out with electrically...

Homework Statement From the standard potential Tl^+ + e^- -----> Tl (solid) E^o= -0.336 V Determine the standard potential of Tl_2S (solid) + 2e^- -----> 2Tl (solid) + S^2^- Given that the K_sp for Tl_2S is 1.2x10^-22 ***I cannot get latex to put the -22 in the exponent. The Ksp is...

Homework Statement Determine the response factors for both Na and K with the Li internal standard Concentration Ratio Na : K : Li (in ppm) 5.0 : 5.0: 500 ------------------- Signal Ratio Na : K : Li (relative units) 4.9 : 6.4: 36.2 Homework Equations Ax/x = F(As/s)...

Does A = -log(.12)= .0921 ? and if this is the case E = 0.0921 / [(2.000 cm)(2.482 x 10^-4)] = 185.5359 ? are sig figs correct as well?

Homework Statement Calculate the molar absorptivity of K2Cr2O7 at 455 nm given that 36.5 mg disolved in 500.0 mL exhibits 12.0% transmittance at 455 nm in a 2.000 cm cell. Homework Equations Beers law E = A/bc The Attempt at a Solution I was able to do the following part ( i...