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  1. (38.6 + 273.15) K x (0.798 L / 4.64 L) x ((125 kPa x 7.50062 torr/kPa) / 685 torr) = 73.385 K = -200.°C

    2 Answers · Science & Mathematics · 26/07/2019

  2. ... {2+}, Br {-}, Li {+}, NO3 Hg2Br2 will ppt c. Ag {+}, Br {-}, Ca {+2}, SO4 {2-} both AgBr and Ag2SO4 will ppt. d. K{+} NO3 {-}, Na {+}, Br {-} none of the possible procucts will...

    1 Answers · Science & Mathematics · 27/07/2019

  3. ... * (T2 / T1) plug and chug .. P2 = 2.25atm * ( (75+273.15 K) / (30+273.15 K) ) = 2.58 atm

    2 Answers · Science & Mathematics · 24/07/2019

  4. ... = 51,416.8473 ÷ 72.128 The temperature is approximately 713˚K. ˚C = (51,416.8473 ÷ 72.128) – 273 The temperature is approximately 440˚...

    3 Answers · Science & Mathematics · 23/07/2019

  5. P = nRT / V = (95.5 mol) x (0.082057338 L atm/K mol) x (89.00 + 273.15) K / (59.5 L) = 47.7 atm

    1 Answers · Science & Mathematics · 23/07/2019

  6. ...4) plug in the data, chug out the results.. remember T must be absolute.. K or R... never °C nor °F ********** this problem start with .. P1V1 (n1T1) = P2V2...

    2 Answers · Science & Mathematics · 23/07/2019

  7. P = nRT / V = (9.65 g O2 / (31.99886 g O2/mol)) x (62.36367 L torr/K mol) x (29 + 273) K / (1.75 L) = 3246 torr [Adjust the number of significant digits as you see fit.]

    1 Answers · Science & Mathematics · 22/07/2019

  8. T = PV / nR = (740 mmHg) x (6 L) / (0.75 mol x (62.36367 L mmHg/K mol)) = 94.9 K = -178°C

    3 Answers · Science & Mathematics · 22/07/2019

  9. ...x 1480.6mL x 303.65K / 0.491 atm / 250.8 mL T2 = 876 K ............ to three significant digits The second question is...

    2 Answers · Science & Mathematics · 21/07/2019

  10. ... Δ° to J/mol and use R = 8.314 J/mol K. Then, 1.683X10^4 J/mol = - 8.314 J/molK (298...

    1 Answers · Science & Mathematics · 17/07/2019

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