標題:
:( i need help.. Chem hw.
發問:
1) a sample of oxygen gas is collected over water at 25C ( vp H20 (l) = 23.8 mm Hg). The wet gas occupies a volume of 7.28 L at a total pressure of 1.25 bar. If all the water is removed, what volume will the dry oxygen occupy at pressure of 1.07 atm and a temperature of 37C?2) A balloon filled with nitrogen... 顯示更多 1) a sample of oxygen gas is collected over water at 25C ( vp H20 (l) = 23.8 mm Hg). The wet gas occupies a volume of 7.28 L at a total pressure of 1.25 bar. If all the water is removed, what volume will the dry oxygen occupy at pressure of 1.07 atm and a temperature of 37C? 2) A balloon filled with nitrogen gas has a small leak. Another balloon filled with hydrogen gas has an identical leak. How much faster will the hydrogen balloon deflate? 3) Nitrogen can react with steam to form ammonia and nitrogen oxide gases. A 20.0L sample of nitrogen at 173C and 722mm Hg is made to react with an excess of steam. The products are collected at room temperature 26C into an evacuated flask with a volume of 15.0L a) what is the total pressure of the products in the collecting flask after the reaction is complete? b) what is the partial pressure of each of the products in the flasks? I really need help... :( thanks you guys!
最佳解答:
(1) Converting all pressures into atm: Partial pressure of water = 0.03132 atm Total pressure = 1.23365 atm Therefore, partial pressure of oxygen gas = 1.23365 - 0.03132 = 1.20233 atm which is measured under 298K and volume of 7.28 L. Using the general gas law: P1V1/T1 = P2V2/T2 with P1 = 1.20233 atm, V1 = 7.28 L, T1 = 298 K, P2 = 1.07 atm and T2 = 310 K, we have: V2 = 8.51 L (3a) By ideal gas equation, no. of moles of nitrogen is: n = PV/RT with P = 96258 Pa, V = 0.02 m3, R = 8.314 and T = 446 K, we have: n = 0.5192 moles And after the reaction, V = 0.015 m3, T = 299 K and n = 0.5192 (since all steam has been condensed to water) So with P = nRT/V, P = 86043 Pa which is equivalent to 645.4 mm Hg (b) According to the equation: 7N2 + 2H2O --> 8NH3 + 6NO2 when steam is in excess We can see that the mole ratio (and hence partial pressure ratio) between ammonia and nitrogen dioxide is 4:3. Therefore, Partial pressure of ammonia = 368.8 mm Hg Partial pressure of nitrogen dioxide = 276.6 mm Hg 2009-04-06 15:36:48 補充： For Q2, pls. provide the numerical data, if any.
其他解答:
:( i need help.. Chem hw.
發問:
1) a sample of oxygen gas is collected over water at 25C ( vp H20 (l) = 23.8 mm Hg). The wet gas occupies a volume of 7.28 L at a total pressure of 1.25 bar. If all the water is removed, what volume will the dry oxygen occupy at pressure of 1.07 atm and a temperature of 37C?2) A balloon filled with nitrogen... 顯示更多 1) a sample of oxygen gas is collected over water at 25C ( vp H20 (l) = 23.8 mm Hg). The wet gas occupies a volume of 7.28 L at a total pressure of 1.25 bar. If all the water is removed, what volume will the dry oxygen occupy at pressure of 1.07 atm and a temperature of 37C? 2) A balloon filled with nitrogen gas has a small leak. Another balloon filled with hydrogen gas has an identical leak. How much faster will the hydrogen balloon deflate? 3) Nitrogen can react with steam to form ammonia and nitrogen oxide gases. A 20.0L sample of nitrogen at 173C and 722mm Hg is made to react with an excess of steam. The products are collected at room temperature 26C into an evacuated flask with a volume of 15.0L a) what is the total pressure of the products in the collecting flask after the reaction is complete? b) what is the partial pressure of each of the products in the flasks? I really need help... :( thanks you guys!
最佳解答:
(1) Converting all pressures into atm: Partial pressure of water = 0.03132 atm Total pressure = 1.23365 atm Therefore, partial pressure of oxygen gas = 1.23365 - 0.03132 = 1.20233 atm which is measured under 298K and volume of 7.28 L. Using the general gas law: P1V1/T1 = P2V2/T2 with P1 = 1.20233 atm, V1 = 7.28 L, T1 = 298 K, P2 = 1.07 atm and T2 = 310 K, we have: V2 = 8.51 L (3a) By ideal gas equation, no. of moles of nitrogen is: n = PV/RT with P = 96258 Pa, V = 0.02 m3, R = 8.314 and T = 446 K, we have: n = 0.5192 moles And after the reaction, V = 0.015 m3, T = 299 K and n = 0.5192 (since all steam has been condensed to water) So with P = nRT/V, P = 86043 Pa which is equivalent to 645.4 mm Hg (b) According to the equation: 7N2 + 2H2O --> 8NH3 + 6NO2 when steam is in excess We can see that the mole ratio (and hence partial pressure ratio) between ammonia and nitrogen dioxide is 4:3. Therefore, Partial pressure of ammonia = 368.8 mm Hg Partial pressure of nitrogen dioxide = 276.6 mm Hg 2009-04-06 15:36:48 補充： For Q2, pls. provide the numerical data, if any.
其他解答:
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