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(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+}, 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

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

**K**) / (30+273.15**K**) ) = 2.58 atm2 Answers · Science & Mathematics · 24/07/2019

... = 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

P = nRT / V = (95.5 mol) x (0.082057338 L atm/

**K**mol) x (89.00 + 273.15)**K**/ (59.5 L) = 47.7 atm1 Answers · Science & Mathematics · 23/07/2019

...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

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

### Calculate the approximate temperature of a 0.75 mol sample of gas at 740 mm Hg and a volume of 6 L.?

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

...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

... 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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