Part A How many moles of Rn are contained in a 4.25 L tank at 155°C and 2.80 atm? O 0.289 moles O 0.339 moles O 0.4…
The following solution is suggested to handle the subject “Part A How many moles of Rn are contained in a 4.25 L tank at 155°C and 2.80 atm? O 0.289 moles O 0.339 moles O 0.4…“. Let’s keep an eye on the content below!
Question “Part A How many moles of Rn are contained in a 4.25 L tank at 155°C and 2.80 atm? O 0.289 moles O 0.339 moles O 0.4…”
Part A How many moles of Rn are contained in a 4.25 L tank at 155°C and 2.80 atm? O 0.289 moles O 0.339 moles O 0.455 moles 1.00 moles 2.51 moles Submit Request Answer Provide Feedback
Answer
From the ideal gas Law PV = 0.25 atm, V= 4.25L, T = 155 C = (237-15+155)K = 428-15+155K = 428-15+155K = 428-15+155K = 428-15+155K = 428-15+155K= 428-15+155K= 428-15+155K= 428-15+155K= 428-15+155K= 429-15+155K= 428-15+155K= 428-15+ volume 02-n. Moles R= Cras constant temperature Ts: Given P = 2.80 atm. V = 4.25L. T = 155 C = (237-15+155).K = 428-15k. – 1 – 1 R = 0.08 L atm Molk. / 428.15 K Putting valves (2.80 atm) (4.25%) = m 10.082 Katyn l mol K .. n = 2.80 x4.25_ moles 0.082 x 428.15 = 10.339 moles| a. option (b) is correct
Conclusion
Above is the solution for “Part A How many moles of Rn are contained in a 4.25 L tank at 155°C and 2.80 atm? O 0.289 moles O 0.339 moles O 0.4…“. We hope that you find a good answer and gain the knowledge about this topic of science.