Comments

• Your expression for p(SO3) is not correct; it should be +2X (not 0.5-2X) Kp = p(SO3)^2 / [p(SO2)^2 * p(O2)] p(SO2) = 0.5 - 2X p(O2) = 0.5 - X p(SO3) = 0 + 2X = 2X 0.25 = (2X)^2 / [(0.5-2X)^2 * (0.5-X)] I still end wirh a 3rd degree equa…

  in Algebra for partial pressures....? Comment by skipper June 2021
• Your expression for p(SO3) is not correct; it should be +2X (not 0.5-2X) Kp = p(SO3)^2 / [p(SO2)^2 * p(O2)] p(SO2) = 0.5 - 2X p(O2) = 0.5 - X p(SO3) = 0 + 2X = 2X 0.25 = (2X)^2 / [(0.5-2X)^2 * (0.5-X)] I still end wirh a 3rd degree equa…

  in Algebra for partial pressures....? Comment by skipper June 2021
• Fe is +3 going to 0, a gain of 3 e C is +2 going to +4, a loss of 2 e 2Fe gains 6 e and 3CO loses 6 e Fe2O3 + 3CO --> 2Fe + 3CO2

  in _Fe2O3 + _CO = _Fe + _CO2? Comment by skipper June 2021
• Fe is +3 going to 0, a gain of 3 e C is +2 going to +4, a loss of 2 e 2Fe gains 6 e and 3CO loses 6 e Fe2O3 + 3CO --> 2Fe + 3CO2

  in _Fe2O3 + _CO = _Fe + _CO2? Comment by skipper June 2021
• You can't calculate the molar mass because the acid may contain more than 1 replacable H+. That is why you call it the equivalent mass. 23.90 mL = 0.0239 L 0.0239 L x 0.139 M = 0.003322 moles NaOH 0.003322 moles of NaOH = 0.003322 moles of H+ i…

  in how to do this chemistry problem? Comment by skipper May 2021
• You can't calculate the molar mass because the acid may contain more than 1 replacable H+. That is why you call it the equivalent mass. 23.90 mL = 0.0239 L 0.0239 L x 0.139 M = 0.003322 moles NaOH 0.003322 moles of NaOH = 0.003322 moles of H+ i…

  in how to do this chemistry problem? Comment by skipper May 2021
• P1V1/T1 = P2V2/T2 P1 = 1 V1 = 1.0 L T1 = 253.15 K P1V1/T1 = 0.003950 P2 = ? V2 = 0.50 L T2 = 313.15 K P2V2/T2 = P2 x 0.001597 0.003950 = P2 x 0.001597 P2 = 2.47 P2 = 2.47 x P1

  in Ideal gas problem - Chemistry problem? Comment by skipper May 2021
• P1V1/T1 = P2V2/T2 P1 = 1 V1 = 1.0 L T1 = 253.15 K P1V1/T1 = 0.003950 P2 = ? V2 = 0.50 L T2 = 313.15 K P2V2/T2 = P2 x 0.001597 0.003950 = P2 x 0.001597 P2 = 2.47 P2 = 2.47 x P1

  in Ideal gas problem - Chemistry problem? Comment by skipper May 2021
• ppm = 1 part per 1x10^6 so 1x10^-6 grams per gram 478x10^-6 g / gram 478 micrograms / gram divide 478 by the molar mass of the substance and your answer is micromoles/gram Assume the substance was C6H12O6; molar mass = 180.2 g 478 ppm of C6H1…

  in how do i convert ppm to micromoles/gram? Comment by skipper May 2021
• well i think that u should talk 2 him and his gf about it if she catches u w/ him bhind her back every1 will hate u but if ur upfrount w/ her then she might thank u and after they break up then u can try 2 go out with him trust me i did the same thi…

  in Depressed over a guy!!? Comment by skipper May 2021
• SARà SENZ'ALTRO UN FILO COLLEGATO MALE O LENTO CHE FA CONTATTO A TRATTI DENTRO IL QUADRETTO ELETTRICO DOVE SENTI IL RONZIO. GUARDACI SUBITO PRIMA CHE SI BRUCI IL PUNTO DI CONTATTO o falso contatto che dir si voglia. CIAO SKIPPER

  in problema elettrico : ronzio proveniente dal quadro elettrico? Comment by skipper May 2021
• SARà SENZ'ALTRO UN FILO COLLEGATO MALE O LENTO CHE FA CONTATTO A TRATTI DENTRO IL QUADRETTO ELETTRICO DOVE SENTI IL RONZIO. GUARDACI SUBITO PRIMA CHE SI BRUCI IL PUNTO DI CONTATTO o falso contatto che dir si voglia. CIAO SKIPPER

  in problema elettrico : ronzio proveniente dal quadro elettrico? Comment by skipper May 2021
• well i think that u should talk 2 him and his gf about it if she catches u w/ him bhind her back every1 will hate u but if ur upfrount w/ her then she might thank u and after they break up then u can try 2 go out with him trust me i did the same thi…

  in Depressed over a guy!!? Comment by skipper May 2021
• ppm = 1 part per 1x10^6 so 1x10^-6 grams per gram 478x10^-6 g / gram 478 micrograms / gram divide 478 by the molar mass of the substance and your answer is micromoles/gram Assume the substance was C6H12O6; molar mass = 180.2 g 478 ppm of C6H1…

  in how do i convert ppm to micromoles/gram? Comment by skipper May 2021
• ppm = 1 part per 1x10^6 so 1x10^-6 grams per gram 478x10^-6 g / gram 478 micrograms / gram divide 478 by the molar mass of the substance and your answer is micromoles/gram Assume the substance was C6H12O6; molar mass = 180.2 g 478 ppm of C6H1…

  in how do i convert ppm to micromoles/gram? Comment by skipper May 2021
• P1V1/T1 = P2V2/T2 P1 = 1 V1 = 1.0 L T1 = 253.15 K P1V1/T1 = 0.003950 P2 = ? V2 = 0.50 L T2 = 313.15 K P2V2/T2 = P2 x 0.001597 0.003950 = P2 x 0.001597 P2 = 2.47 P2 = 2.47 x P1

  in Ideal gas problem - Chemistry problem? Comment by skipper May 2021
• You can't calculate the molar mass because the acid may contain more than 1 replacable H+. That is why you call it the equivalent mass. 23.90 mL = 0.0239 L 0.0239 L x 0.139 M = 0.003322 moles NaOH 0.003322 moles of NaOH = 0.003322 moles of H+ i…

  in how to do this chemistry problem? Comment by skipper May 2021
• You can't calculate the molar mass because the acid may contain more than 1 replacable H+. That is why you call it the equivalent mass. 23.90 mL = 0.0239 L 0.0239 L x 0.139 M = 0.003322 moles NaOH 0.003322 moles of NaOH = 0.003322 moles of H+ i…

  in how to do this chemistry problem? Comment by skipper May 2021
• Your expression for p(SO3) is not correct; it should be +2X (not 0.5-2X) Kp = p(SO3)^2 / [p(SO2)^2 * p(O2)] p(SO2) = 0.5 - 2X p(O2) = 0.5 - X p(SO3) = 0 + 2X = 2X 0.25 = (2X)^2 / [(0.5-2X)^2 * (0.5-X)] I still end wirh a 3rd degree equa…

  in Algebra for partial pressures....? Comment by skipper May 2021