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1031 第二次化學會考考題影音講解  
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 1.20) The law of ________ states that energy that can be neither created or destroyed.A) thermochemistryB) the consecration of energyC) the conservation of energyD) potential energyE) kinetic energyAns:C
演講者: 洪國明
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演講日期: 2014-11-24
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00:15
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 1.19) The titration of 80.0 mL of an unknown concentration H3PO4 solution requires 126 mL of 0.218 M KOH solution. What is the concentration of the H3PO4 solution (in M)?A)0.114 M B) 0.343 M C) 1.03 M D) 0.0461 M E) 0.138 M N1V1 = N2V2 M x 3 x 0.08 = 0.218 x 0.126 M = 0.114 MAns:A
演講者: 洪國明
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演講日期: 2014-11-23
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00:59
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 1.18) How much energy is evolved during the formation of 98.7 g of Fe, according to the reaction below? Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) ΔH°rxn = -852 kJA)241 kJ B) 753 kJ C) 4.20 x 103 kJ D) 482 kJ E) 1.51 x 103 kJ 98.7 ÷ 55.85 = 1.77 mol (1.77 ÷ 2) x 852 = 752.8 Ans:B
演講者: 洪國明
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演講日期: 2014-11-23
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上傳者: energy4104
00:38
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 1.17) For the reaction H2(g) + 1/2 O2(g) → H2O(g) ΔH° = -241.8 kJ/mol, what quantity of heat is liberated by the reaction of 10.0 L of O2 measured at 22.0 °C and 742 mmHg?A)97.5 kJ B) 195 kJ C) 2610 kJ D) 120 kJ E) 1310 kJ PV = nRT (742 ÷760) x 10.0 = n x 0.0826 x (22+273) n = 0.4 mole (0.4 ÷ 0.5) x 241.8 = 195Ans:B
演講者: 洪國明
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演講日期: 2014-11-23
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上傳者: energy4104
00:50
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 1.16) Determine the volume of SO2 (at STP) formed from the reaction of 96.7 g of FeS2 and 55.0 L of O2 (at 398 K and 1.20 atm). The molar mass of FeS2 is 119.99 g/mol. 4 FeS2(s) + 11 O2(g) → 2 Fe2O3(s) + 8 SO2(g)A)36.1 L B) 45.3 L C) 32.9 L D) 18.1 L E) 27.6 L 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 mol 0.80 2.02 -0.73 -2.02 +0.37 +1.46 1.46 x 22.4 = 32.9 Ans:C
演講者: 洪國明
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演講日期: 2014-11-23
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上傳者: energy4104
00:54
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 1.15) You have 10.00 L of a 0.350 M KCl solution, but you need a solution that is 0.450 M. What volume of water, in L, would you evaporate from the solution?A)2.22 L B) 3.50 L C) 7.77 L D) 4.38 L E) 2.85 L Solute: 0.350 x 10.00 = 3.5 mole 3.5 (mol) ÷ X (L) = 0.45 (M) X=7.77 10.00 - 7.77 = 2.23Ans:A
演講者: 洪國明
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演講日期: 2014-11-23
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上傳者: energy4104
00:46
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 1.14) Potassium superoxide (KO2) can simulate a plant-type action by consuming carbon dioxide gas and releasing oxygen gas. The other product is potassium carbonate. When the equation for this process is balanced, it shows that:A) 3 g of oxygen is produced per 2 g CO2 consumedB) moles of reactants equals moles of productC) 3 mol oxygen is produced per mol KO2 consumedD) 2 mol KO2 is consumed per mol carbon dioxideE) moles of products exceed moles of reactants 4KO2 + 2CO2 → 3O2 + 2K2CO3Ans:D
演講者: 洪國明
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演講日期: 2014-11-23
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上傳者: energy4104
00:38
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 1.13) What pressure would a gas mixture in a 10.0 L tank exert if it were composed of 48.5 g He and 94.6 g CO 2 at 398 K?A) 39.6 atm B) 32.6 atm C) 7.02 atm D) 46.6 atm E) 58.7 atm Gases total mole : (48.50 ÷ 4) + (94.60 ÷ 44) = 14.28mole PV=nRT P x 10.00 = 14.28 x 0.0826 x 398 P = 46.9 Ans:D
演講者: 洪國明
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演講日期: 2014-11-23
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上傳者: energy4104
00:30
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 1.12) A 12.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter with a heat capacity of 5.65 kJ/°C. Using the information below, determine the final temperature of the calorimeter if the initial temperature is 25.0°C. The molar mass of ethanol is 46.07 g/mol. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) ΔH°rxn = -1235 kJA)111°C B) 85.7°C C) 28.1°C D) 53.4°C E) 74.2°C 12.80 ÷ 46.07=0.278mole ethanol 0.278 x (-1235) = -343.33 343.33 ÷ 5.65 = 60.77 60.77+25 = 85.7Ans:B
演講者: 洪國明
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演講日期: 2014-11-23
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01:01
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 1.11) This equation is used to calculate the properties of a gas under nonideal conditions.A) van der Waals equationB) Daltonʹs LawC) Charlesʹs LawD) Avogadroʹs LawE) Boyleʹs LawAns:A
演講者: 洪國明
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演講日期: 2014-11-23
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00:22
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