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化學問題(英) check answer
It is for making sure...1.A 93-L sample of dry air cools from 145oC to -22 oC while the pressure is maintained at 2.85 atm. What is the final volume?I got 55.86L2.A sample of carbon monoxide occupies 3.65 L at 298 K and 745 torr. Find its volume at -14 oC and 367 torr. I got 2.58L for this 3.The... 顯示更多 It is for making sure... 1.A 93-L sample of dry air cools from 145oC to -22 oC while the pressure is maintained at 2.85 atm. What is the final volume? I got 55.86L 2.A sample of carbon monoxide occupies 3.65 L at 298 K and 745 torr. Find its volume at -14 oC and 367 torr. I got 2.58L for this 3.The density of a noble gas is 2.71 g/L at 3.00 atm and 0 oC. Identify the gas I get M (mol of mass) 20.26 but I don’t know what I need to do next…to Identify the gas.. 4.How many grams of potassium chlorate decompose to potassium chloride and 638 mL of O2 at 128 oC. and 752 torr? I got V=0.638L T= 401.15K P=0.99atm and R is 0.0821, but I don’t know what equation should I use. Do I find the density first? But what we do to find grams? 5.When 35.6 L of ammonia and 40.5L of oxygen gas at STP burn, nitrogen monoxide and water form. After the products return to STP, how many grams of nitrogen monoxide are present? NH3(g)+O2(g)?NO(g)+H2O(l) (unbalanced) Do I do this like Q4? I still don't know when we use mass and whem we use mol of mass
最佳解答:
1. V1 = 93 L, T1 = 273 + 145 = 418 K V2 = ?, T2 = 273 - 22 = 251 K V1/T1 = V2/T2 93/418 = V2/251 Final volume, V2 = 93 x 251/418 = 55.84 L ===== 2. V1 = 3.65 L, T1 = 298 K, P1 = 745 torr V2 = ?, T2 = 273 - 14 = 259 K, P2 = 367 torr P1V1/T1 = P2V2/T2 (745)(3.65)/(298) = (367)V2/(259) Final volume, V2 = 6.44 L ===== 3. ρ = 2.71 g/L P = 3 atm T = 273 K R = 0.082 atm L molˉ1 Kˉ1 PM = ρRT (3)M = (2.71)(0.082)(273) Molar mass = 20.2 g molˉ1 Relative atomic mass = 20.2 Refer to the Periodic Table. In Group 0 (nobles gases), the relative atomic mass of neon (Ne) is 20.2. Hence, the gas is neon. ===== 4. Consider the O2 formed: V = 638 mL = 0.638 L T = 273 + 128 = 401 K P = (752/760) atm R = 0.082 atm L molˉ1 Kˉ1 n = ? mol PV = nRT (752/760)(0.638) = n(0.082)(401) No. of moles of O2 formed 0.0192 mol 2KClO3 → 2KCl + 3O2 Mole ratio KClO3 : O2 = 2 : 3 No. of moles of O2 formed = 0.0192 mol No. of moles of KClO3 used = 0.0192 x (2/3) = 0.0128 mol Molar mass of KClO3 = 39.1 + 35.5 + 16x3 = 122.6 g molˉ1 Mass of KClO3 used = 0.0128 x 122.6 = 1.569 g ===== 5. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) Mole ratio NH3 : O2 = 4 : 5 When 40.5 L of O2 is completely reacted, no. of moles of NH3 needed = 40.5 x (4/5) = 32.4 L Hence, NH3 is in excess, and O2 is the limiting reactant. Consider the O2 used: V = 40.5 L P = 1 atm T = 273 K R = 0.082 atm L molˉ1 Kˉ1 n = ? mol PV = nRT (1)(40.5) = n(0.082)(273) No. of moles of O2 reacted, n = 1.809 mol Mole ratio O2 : NO = 5 : 4 No. of moles of NO formed = 1.809 x (4/5) = 1.4472 mol Molar mass of NO = 14 + 16 = 30 g molˉ1 Mass of NO formed = 1.4472 x 30 = 43.42 g ===== I still don't know when we use mass and when we use molar mass (NOT mol of mass). This is case by case.
其他解答:9B623C59CC17F40C
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發問:It is for making sure...1.A 93-L sample of dry air cools from 145oC to -22 oC while the pressure is maintained at 2.85 atm. What is the final volume?I got 55.86L2.A sample of carbon monoxide occupies 3.65 L at 298 K and 745 torr. Find its volume at -14 oC and 367 torr. I got 2.58L for this 3.The... 顯示更多 It is for making sure... 1.A 93-L sample of dry air cools from 145oC to -22 oC while the pressure is maintained at 2.85 atm. What is the final volume? I got 55.86L 2.A sample of carbon monoxide occupies 3.65 L at 298 K and 745 torr. Find its volume at -14 oC and 367 torr. I got 2.58L for this 3.The density of a noble gas is 2.71 g/L at 3.00 atm and 0 oC. Identify the gas I get M (mol of mass) 20.26 but I don’t know what I need to do next…to Identify the gas.. 4.How many grams of potassium chlorate decompose to potassium chloride and 638 mL of O2 at 128 oC. and 752 torr? I got V=0.638L T= 401.15K P=0.99atm and R is 0.0821, but I don’t know what equation should I use. Do I find the density first? But what we do to find grams? 5.When 35.6 L of ammonia and 40.5L of oxygen gas at STP burn, nitrogen monoxide and water form. After the products return to STP, how many grams of nitrogen monoxide are present? NH3(g)+O2(g)?NO(g)+H2O(l) (unbalanced) Do I do this like Q4? I still don't know when we use mass and whem we use mol of mass
最佳解答:
1. V1 = 93 L, T1 = 273 + 145 = 418 K V2 = ?, T2 = 273 - 22 = 251 K V1/T1 = V2/T2 93/418 = V2/251 Final volume, V2 = 93 x 251/418 = 55.84 L ===== 2. V1 = 3.65 L, T1 = 298 K, P1 = 745 torr V2 = ?, T2 = 273 - 14 = 259 K, P2 = 367 torr P1V1/T1 = P2V2/T2 (745)(3.65)/(298) = (367)V2/(259) Final volume, V2 = 6.44 L ===== 3. ρ = 2.71 g/L P = 3 atm T = 273 K R = 0.082 atm L molˉ1 Kˉ1 PM = ρRT (3)M = (2.71)(0.082)(273) Molar mass = 20.2 g molˉ1 Relative atomic mass = 20.2 Refer to the Periodic Table. In Group 0 (nobles gases), the relative atomic mass of neon (Ne) is 20.2. Hence, the gas is neon. ===== 4. Consider the O2 formed: V = 638 mL = 0.638 L T = 273 + 128 = 401 K P = (752/760) atm R = 0.082 atm L molˉ1 Kˉ1 n = ? mol PV = nRT (752/760)(0.638) = n(0.082)(401) No. of moles of O2 formed 0.0192 mol 2KClO3 → 2KCl + 3O2 Mole ratio KClO3 : O2 = 2 : 3 No. of moles of O2 formed = 0.0192 mol No. of moles of KClO3 used = 0.0192 x (2/3) = 0.0128 mol Molar mass of KClO3 = 39.1 + 35.5 + 16x3 = 122.6 g molˉ1 Mass of KClO3 used = 0.0128 x 122.6 = 1.569 g ===== 5. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) Mole ratio NH3 : O2 = 4 : 5 When 40.5 L of O2 is completely reacted, no. of moles of NH3 needed = 40.5 x (4/5) = 32.4 L Hence, NH3 is in excess, and O2 is the limiting reactant. Consider the O2 used: V = 40.5 L P = 1 atm T = 273 K R = 0.082 atm L molˉ1 Kˉ1 n = ? mol PV = nRT (1)(40.5) = n(0.082)(273) No. of moles of O2 reacted, n = 1.809 mol Mole ratio O2 : NO = 5 : 4 No. of moles of NO formed = 1.809 x (4/5) = 1.4472 mol Molar mass of NO = 14 + 16 = 30 g molˉ1 Mass of NO formed = 1.4472 x 30 = 43.42 g ===== I still don't know when we use mass and when we use molar mass (NOT mol of mass). This is case by case.
其他解答:9B623C59CC17F40C
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