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- Rewrite the Equations and show work C7H9 + HNO3 −→ C7H6(NO2)3 + H2OThe net change in the multistep biochemical process of photosynthesis is that CO₂ and H₂O form glucose (C₆H₁₂O₆)and O₂. Chlorophyll absorbs light in the 600 to 700 nm region.(a) Write a balanced thermochemical equation for formation of1.00 mol of glucose. (b) What is the minimum number of pho-tons with λ=680. nm needed to prepare 1.00 mol of glucose?1.74. What is the ratio of the populations of two energy states whose energies differ by 1000 J at (a) 200 K (b) 500 K (C) 1000 K? What trend do you see in your answers?
- 问题3 The 14 N160 molecule undergoes a transition between its rotational ground state and its rotational first excited state. Approximating the diatomic molecule as a rigid rotor, and given that the bond length of NO is 1.152 Angstroms, calculate the energy of the transition. As your final answer, calculate the temperature T in Kelvin, such that Eshermal = kBT equals the energy of the transition between NO's rotational ground state and first excited state.Identify the systems for which it is essential to include a factor of 1/N! on going from Q to q : (i) a sample of carbon dioxide gas, (ii) a sample of graphite, (iii) a sample of diamond, (iv) ice.i need the answer quickly
- Indicate how to find the energy difference between two levels in cm-1, knowing that its value is 2.5x10-25 joules.Use Hess' law to calculate the AH for the equation below: N2 (e) + 3H, (g) → 2 NH3 (g) →2 NH3 (g) AH= ? + 7 H2 (g) AH = -114 kJ 2 NH; (2) + 4 H,00-2 NO2 () 2 NO2 (2) → N2 (g)+202 (g) 3. H;O → H2 (2) + ½ O2 (g) 1. AH =+66 kJ 2. AH =-35 kJ 13. O-92 kJ O-188 kJ O-83 kJ +92 kJ +83KJ. +188 kJ O-13 kJ +13 kJGiven these two equations, calculate Cv for CO2 and N2 using the velocity of sound. 6. According to statistical mechanics, the contribution to Cy due to the ith vibrational mode of a molecule is c ee;/T Cv,vib,4 = R(7) ³ (e®,/T — 1)² › where Oi = hcv./kB, h is Planck's constant, and c is the speed of light in cm s-¹. v; is the vibrational frequency of the ith vibrational mode in wavenumbers (cm-¹). The quantity O, has dimensions of temperature and is sometimes referred to as the vibrational tem- perature (of the ith vibrational mode). The total vibrational contribution to the molar heat capacity for an N-atom (linear) molecule is then Cy,vib Cv,vib,i Calculate Cy for CO₂ at 300 K, assuming that the translational and rotational con- tributions are (3/2)R and R, respectively, and use the preceding equations to obtain the vibrational contribution. Four vibrational modes for CO₂ are expected (3N- 5). How- ever, three fundamental frequencies are observed at v= 667.3, 1340, and 2349.3 cm-¹.…
- Given the following data PA(s) + 6C12 (9) → 4PCI3 (g) АН — — 1225.6 kJ PA(s) + 502(9) → P,O10 (s) AH = -2967.3 kJ PCI3 (g) + Cl2 (g) → PCI5 (g) АН — — 84.2 kJ PCI, (9) + 02 (9) → CI3PO(9) AĦ = -285.7 kJ Calculate AH for the reaction P,O10 (s) + 6PCI5 (9) → 10C13PO(9) ΔΗ- kJTwo nitro (NO,) groups are chemically bonded to a patch of surface. They can't move to another location on the surface, but they can rotate (see sketch at right). It turns out that the amount of rotational kinetic energy each NO, group can have is required to be a multiple of ɛ, 2 where ɛ = 1.0 × 10 -24 J. In other words, each NO, group could have ɛ of rotational kinetic energy, or 2ɛ, or 3ɛ, and so forth – but it cannot have just any old amount of rotational kinetic energy. Two rotating NO, groups Suppose the total rotational kinetic energy in this system is initially known to be 87ɛ. Then, some heat is removed bonded to a surface. from the system, and the total rotational kinetic energy falls to 66ɛ. Calculate the change in entropy. Round your answer to 3 significant digits, and be sure it has the correct unit symbol. 믐 ?Two nitro (NO,) groups are chemically bonded to a patch of surface. They can't move to another location on the surface, but they can rotate (see sketch at right). It turns out that the amount of rotational kinetic energy each NO, group can have is required to be a multiple of ɛ, where 8= 1.0 x 10 -24 J. In other words, each NO, group could have ɛ of rotational kinetic energy, or 2ɛ, or 3ɛ, and so forth - but it cannot have just any old amount of rotational kinetic energy. Two rotating N0, groups Suppose the total rotational kinetic energy in this system is initially known to be 41ɛ. Then, some heat is removed from the bonded to a surface. system, and the total rotational kinetic energy falls to 30ɛ. Calculate the change in entropy. Round your answer to 3 significant digits, and be sure it has the correct unit symbol. Check 2021 McGraw-Hill Education All Rights Reserved Tems of Uhe Privacy Accesbil Explanation MacBook Air olo
