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- ) Listen Given that the reaction of C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H20 (1) ΔΗ - 1411 kJ What would AH be for 2 CO2 (g) + 2 H20 (1) → C,H4 (g) + 3 O2 (g) O -1411 kJ O 1411 kJ O 1411 J O-1411 J 73°F SuPrepare a table identifying several energy transitions that take place during the typical operation of an automobile.Given the following data: C2H2(g) + 5/2 O2(g) ->2CO2(g) + H2O(l) △H=-1300. kJ C(s) + O2(g) -> CO2(g) △H= -394kJ H2(g) + 1/2 O2(g) -> H2O(l) △H= -286 Calculate △H for the reaction 2C(s) + H2(g) -> C2H2(g)
- (-2220.1)=[-1180.5+-1141.2]-[x] (-2220.1)=[-2321.7][2] +23 21.4 +2321.7 -101.6x Днез нд →-101.6 h mol Assuming that all of the heat evolved in burning 30.0 grams of propane is transferred to 8.00 kilograms of water (specific heat = 4.18 J/g-K), calculate the increase in temperature of water. online at www.nms.org. AP is a trademark of the College Entran terial, Used with permission.Calculate AH for the reaction C2H4 (g) + H2 (g) → C2H6 (g), from the following data. C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2H2O (1) AH 1411 kJ / mol C2H6 (g) + O2 (g) → 2 CO2 (g) + 3 H2O (1) ΔΗ - 1560 kJ / mol H2 (g) + O2 (g) → H2O (1) AH = - 285. 8 kJ / molPentaborane BsHg(s) burns vigorously in 02 to give B203(s) and H20(). Calculate AHxn for the combustlon of 5.00 mol of BgHg. AH(B203{s}] = -1,273,5 kJ/mol AH(B5H9(s)) = 73.2 kJ/mol AH(H200] = -285.8 kJ/mol Multiple Choice 9,090 kJ 45,400 kJ - 8,790 kJ - 45,400 kJ - 22.700 kJ
- What is the AH for the reaction below? C2H2(g) +3 H20() → CH4(g) + 02(9) + CH3OH() Use the following information to find AH for the reaction above. CH4(g) +2 02(g) → CO2(g) + 2 H20(I) AH= -890 kJ 2 C2H2(g) + 5 02(g) - 4 CO2(g) + 2 H20(I) AH= -2599 kJ 2 CH3OH() + 3 O2(g) - 2 CO2(g) + 4 H20() AH = -1453 kJ17. Given the following set of reaction data, AH = -285.83 kJ mol- H2 (g)+}02(8) 2B(s) +3H, (8) 2B(s) +O2(8) –→ B,O3(s) → H2O(1) → B,H, (8) AH = 36.4 kJ mol- AH = –1273.5 kJ mol determine the heat of reaction (in kJ mol') for the following reaction. – B2O3 (s) +3H,0(1) w bns maldorg B,H, (8) + 30, (g) - AH = ? kJ mol-1 lo teari done roi noltaupa 189T) of 9je s beau mo A: –2167Consider the following problem. How much energy is evolved during the formation of 197 g of Fe, according to the reaction below? Fe203(s) + 2 Al(s) - Al203(s) + 2 Fe(s) AH°rxn = -852 kJ 21. How do you find the correct answer to the problem question? ODH(rxn of 197 g of Fe) = ratio of (#mol from 197g of Fe in reaction / 2 mol of Fe in equation) x DH(rxn of 2 mol of Fe). #mol from 197 g of is not known and should be determined next. O DH(rxn of 197 g of Fe) = ratio of (2 mol of Fe in equation / #mol from 197g of Fe in reaction) x DH(rxn of 2 mol of Fe). #mol from 197 g of is not known and should be determined next. DH(rxn of 197 g of Fe) = ratio of (#mol from 197g of Fe in reaction / 2 mol of Fe in equation) / DH(rxn of 2 mol of Fe). #mol from 197 g of is not known and should be determined next.
- A bicyclist is stopped at the entrance to a valley, as sketched below: F B Where would the bicyclist have the highest potential energy? |(choose one) Where would the bicyclist have the lowest potenti energy? (choose one) v Where would the bicyclist have the highest kinetic energy? (choose one) ♥ Where would the bicyclist have the highest speed? |(choose one) v Would the bicyclist's kinetic energy be higher at A or B? |(choose one) Would the bicyclist's potential energy be higher at A or B? (choose one) Would the bicyclist's total energy be higher at A or B? (choose one) Suppose the bicyclist lets off the brakes and coasts down into the valley without pedaling. Even if there is no friction or air resistance to slow him down, what is the farthest point the bicyclist could reach without pedaling? (choose one) ▼E) Calculate the AHrxn given the following data CH4 (g) + 202 (g) → cO2 (g) + 2H20 (g) Substance AH°F (kJ/mol) CH-(g) CO (g) CO: (g) CO: (aq) H;O (g) H;O (1) -74.87 -110.5 -393.5 -412.9 -241.8 -285.8Find the deltaH for the reaction below, given the following reactions and subsequent deltaH values: N2H4(l) + H2(g) —> 2NH3 (g) N2H4(l) + CH4O(l) —> CH2O (g) + N2(g) + 3H2(g)—— deltaH= -32.8kj N2(g)+ 3H2(g) —> 2NH3 (g)—deltaH=-57.2 kj CH4O(l) —>CH2O(g) + H2(g) —-deltaH= -65 kj SIG FIGS