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![Consider the following reaction:
C3H13(1) + 25/2 O»(g) → 8 CO2(g) + 9 H20(1)
If d[CO2]/dt = 0.5OOM/s, determine the value of d[O2]/dt,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b16da5a-cc75-4d8b-8a1f-efa2b261e2af%2F9b01529c-74da-4830-87c3-c8df5c0ff6f5%2Fphvpnt5_processed.jpeg&w=3840&q=75)
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- 1:16 AM | 0.2KB/s L A 86 Calculate AGn (KJ) 200 (g) + 2 NO (g) 2C0, (g) + N2 (g); AG"n ? AH (kJ/mol) 5° U/Kmol) CO (g) -110.5 197.7 NO (g) 91.3 210.8 CO2 (g) -393.5 213.8 191.6 N2 (g) O689.6 O+324.5 O689.6 O-324.5 O 1908A propane grill is used to cook ribeye steak and potatoes. About 0.470 L of propane (C3F18) 15 combusted over the course of about 30 min, a typical grilling session time. How much energy (t kJ) was released from the combustion of this amount of propane? (dc.H. = 0.493 g cm ) %3D C3Hs (g) +502 (g8) → 3CO2(8) + 4H;O(1) AH = -2.22 × 10° kJ mol-1 A: 1.17 x 10*Calculate AH° and AE° for the following reactions at 25 °C. Use standard enthalpies of formation from the table below. Round your answers to four significant digits. Substance AH; (kJ mol-1) Substance AH: (kJ mol-1) Co,(g) -393.5 226.75 Ni(g) C,H,(g) (CH3),CO(/) H,O(g) HC(g) NH,(g) -46.19 -248.1 NH.CIG) N,O(g) O(g) -315.4 -241.8 81.57 -92.30 (a) 2C2H2(g) + 502(g)4CO2(g) + 2H20(g) AH° = i kJ AF° = kJ eTextbook and Media (b) C2H2(g) + 5N20(g) → 2CO2(g) + H2O(g) + 5N2(3) AH° = i kJ AE° = i kJ
- 35R) AH for the reaction IF5 (g) → IF3 (g) + F2 (g) is kJ, give the data below. IF (g) + F2 (g) → IF3 (g) AH = -390 kJ IF (g) + 2F2 (g) → IF5 (g) AH =-745 kJ A) +355 B) -1135 C) +1135 D) +35 E) -35. vanA5. Use the Hess' Law to calculate AH° (tot) for the [tot] reaction from the AHo's given: [tot] N2(g) + 3H2(g) → 2NH3(g) AH°(tot) = ? kJ [1] 4NH3(g) + 302(g) → 2N2(g) + 6H;0(g)_ AH°1 = -1268 kJ [2] 2H2(g) + O2(g) → 2H20(g) AH°2 = -484 kJ [tot] = x[1] _x[2] add up equations [1] and [2] scaled by the factors you chose, and cancel the same formulas on both sides: same as [tot] after cancelling? if "Yes", scale the enthalpies AH°1 and AH°2 by the factors you chose for their equations, and combine for the AH°(tot) = _XAH°1 -ΧΔΗ*> k %3Dwhat is delta Hr for the reaction 3NH3 (g) + 3.75 (g) - 3NO(g) + 4.5 H20 (g) delta Hr = -904.7KJ 4NH3 (g) + 5O (g) ----> 4NO(g) + 6 H20 (g) delta Hr= -904.7 JK
- 2) Calculate AH for the reaction below using Hess' law. C2H6(g) → C2H2(g) + H2(g) AH=? C2H2(g) +5/202(g) → 2CO2(g)+H2O(1) AH = -1300 kJ 2H2(g) + O2(g) → 2H2O(1) AH = -572 kJ 2C2H6(g) + 702(g) → 4CO2(g) + 6H2O(1) AH = -3120 kJWhat 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 kJWhich one of the following reactions results in a positive AS sys? A(g) + B(g) → 5C(g) OA.(g) + B.(g)→ 6C(s) O H.(g) + Cl.(0)→ 2HCK(1) O A.(g) + B.(g) → 2AB(1) OH,(g) + Cl.(g) → 2HCK(1)
- The thermagram in Figure below shows the change in mass of a sample of Copper sulfate pentahydrate, CuSO++5H2O (249.5 g/mole) as a function of temperature. The original sample weighing 25.0 mg was heated from room temperature to 1000°C at a rate of 5º C per minute. The following changes in mass and corresponding temperature ranges were observed : Loss of 1.80 mg from 100 – 250° C. Loss of 7.212 mg from 350 – 550° C., Loss of 8.015 mg from 600 – 800° C. Determine the identities of the volatilization products and the solid residue at each step of the thermal decomposition. Given that: H=1 g/mol , S =32 g/mol , 0=16 g/mol. 20.00 - 15.00 10.00 500- 0.00 400 s00 600 700 Temperature C) 100 200 300 00 s00 1000 Thermogram for CUSO4.5H20Hospital patients are often given glucose (blood sugar) through a tube connected to a bottle suspended over their beds. Suppose that this "drip" supplies glucose at the rate of 15 mg per minute, and each minute 20% of the accumulated glucose is consumed by the body. Then the amount y(t) of glucose (in excess of the normal level) in the body after t minutes satisfies the following. y' = 15 − 0.2y (Do you see why?) y(0) = 0 (zero excess glucose at t = 0) Solve this differential equation and initial condition.J 3) If 4.00 g of iron metal are reacted in a calorimeter with a heat capacity of 400. C' what would be the final temperature of the calorimeter if its initial temperature were 23.0°C?