21-24 ?! Calculate the density of butane gas, C_4H_8, at 15 degreeC and 0.950 at

Question

21-24 ?! Calculate the density of butane gas, C_4H_8, at 15 degreeC and 0.950 atm. A) 4.85 g/L B) 0.442 g/L C) 1.250 g/L D) 2.26 g/L 18. Calculate the squareroot-mean-square speed of an oxygen molecule at 0 degreeC. A) 2.12 times 10^5m/s B) 1.45 m/s C) 461 m/s D) 265m/s 19. If it takes 30 seconds for 10.00 ml of N_2 to effuse from a porous container, how long will it take for the same amount of H_2 to effuse out of the same center? A) 16 seconds B) 110 seconds C) 8.0 seconds D) 2.1 seconds 20. How much work is done by a system when its volume increases from 4.54 L to 18.30L against a constant external pressure of 1.12 atm? [1 Latm = 101.325 J] A) -1.24 kJ B) -156 kJ C) 15.4 kJ D) -12.3kJ 21. When hydrogen gas reacts with sulfur, H_2S is formed. When 1.20 grams of sulfur are reacted with excess hydrogen, -0.770 kJ of heat are released How much heat is released when 2.50 moles of sulfur are reacted with excess hydrogen gas? A) -103 kJ B) 20.6 kJ C) -22.4 kJ D) -51.5 kJ 22. Part of the process of recovehng zinc metal from zinc sulfide ore is depicted in the following chemical equation. 2 ZnS(s) + 3 O_2(g) rightarrow 2 ZnO(s) + 2 SO_2(g) . What is the work done on this system if 20.0 grams of zinc sulfide are reacted with excess oxygen at 298 K? (R = 8.3145 J mol^-1 K^-1) A) 2.51 J B) 254 J C) 762 J D) 508 J 23. A system at constant pressure produces -5.0 J of heat. What is DeltaH for this system? A) 0.0 J B) it depends on the work performed C) -5.0 J D) 50 J 24. The oxidation of iron metal by oxygen produces iron(III) oxide according to the following equation. 4 Fe(s) + 3 O_2(g) rightarrow 2 Fe_2O_3(s) DeltaH_rxndegree = -1648 kJmiddotmol^-1 Calculate the energy released when this reaction produces 25.0 grams of iron(III) oxide at constant pressure. A) -659 kJ B) -129 kJ C) -258kJ D) -4.12times10^4kJ

Explanation / Answer

17.

P V= nRT

n = m/MW

D = m/V

so

D = P*MW/(RT)

MW butane = 58.12

D = 0.95*58.12 /(0.082*(15+273)) = 2.33799 g/L

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