Problem: To determine the mill kitty of butane. Procedure: First we measurable the mass of the light. Then we fill up a beaker completely with piss and put our hand oer the top. We filled a slideway by half in overflowing with water and turned the beaker round top mastered and placed it in the water. We took our hand turned of the beaker do accredited that no mental strain got into the beaker. Then we placed the lighter downstairs the beaker opening in the water and emptied 400 mL of the butane into the beaker. We whence took the temperature of the water. We dried off the lighter completely and thrifty it afterward the loss of the butane. entropy:                 visual sense of illumination ahead:         17.63 g                         Mass of Lighter after:                 17.00 g                         Mass of Butane released:         0.63 g                         Temperature of water:                 20.00 oC                         flock of Butane                 400 mL                         commit insistency on Nov. 2         101.4 kPa                                         Conclusions:                 Calculations 1) 17.63 ? 17.00 = 0.63 grams of butane used                 2) 101.3 ? 2.33 = 99.
07 kPais the pressure of the butane                 3) P1 x V1 / T1 = P2 x V2 / T2                 99.07 x 36.36 / 293 = 101.3 x V2 / 273         12.29 x 273 / 101.3 = V2         33.12 = Molar pot of Butane at STP 4) n = m / M n = .63 / 58 n = .011 molecules = .011 x (6.02x1023)                 molecules = 6.54                 5) molar al-Quran = .4 / .011                 molar volume = 36.36 L                                                 Questions 1)         M = n / m M = .011 / .63 M = .175                 3) .63 g / .4... If you indirect request to get a full essay, baffle it on our website:
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