Trufflehog

Wow, that has pretty much cleared away all my confusion about phase diagrams. Once again, thank you.

Thank you so much for your detailed explanation, you have really helped me out allot! Do you also understand why the statements in B and E are false? Is this because both phases are on the boundary so the rate of change from solid to liquid and liquid to solid are equal. If both phases have the same amount of energy than the difference between ΔG for the phases is 0. One other thing, how did you know to draw that type of phase diagram? Off the top of my head I probably would have drawn a water phase diagram which would mean (D) is also correct.

Hi, I am having trouble understanding this question. The answer in the back of the worksheet is (A). I understand that the ΔG = 0 but why is pressure > pt? Consider a pressure – temperature phase diagram with the triple point at a pressure of pt and a temperature of Tt. If a system is on the phase boundary between solid and liquid phases, which of the following statements is always true for the system? (A) ΔG = 0, pressure > pt (B) ΔG ≠ 0, temperature > Tt © ΔG = 0, pressure < pt (D) ΔG = 0, temperature < Tt (E) ΔG ≠ 0, pressure > pt Thank you

Oh, I see what I did there. I should have had Cr2O72- + 14H+ + 6I- → 2Cr3+ + 7H2O + 3I2. Wow that would have saved the strange road where I got the answer from. Thanks for pointing that out and confirming the answer.

Hi I asked the question below on another website and was told that the answer is 66.66666667ml not 600ml. Any chance of a second opinion? The answer in my worksheet states it's 600mL I was just wondering if someone can check over my working for this problem and let me know if I went wrong anywhere or any useful tips that they wish to share? (Everything that I wrote is in the *** triple stars ***) Dichromate ions, Cr2O72- reacts with iodide ions I^‐,in acidic aqueous solution to form Cr3+ ions, iodine molecules I2 and water, according to the following unbalanced equation: Cr2O72- + H+ + I- → Cr3+ + H2O + I2 ***Cr2O72- + 14H+ + 2I- → 2Cr3+ + 7H2O + I2*** A solution of 100 mL of 0.1M K2Cr2O7 in dilute acid is prepared. How many mL of a 0.1M aqueous solution of potassium iodide (KI) would be required for complete reaction? ***I use HCl as the dilute acid*** ***K2Cr2O7 + 14HCl + 2KI → 2CrCl3 + 7H2O + I2 + KCl*** ***Balance this equation*** ***K2Cr2O7 + 14HCl + 6KI → 2CrCl3 + 7H2O + 3I2 + 8KCl*** ***0.1M K2Cr2O7/0.1L = 0.01mol*** ***K2Cr2O7 0.01mol * (KI 6 mol)/(K2Cr2O7 1 mol) = 0.06mol*** ***0.06mol/V = 0.1M → V = 0.06mol/ 0.1M = 0.6L*** ***Therefore 600mL of 0.1M aqueous solution of potassium iodide (KI) is needed for a complete the reaction***