Two nitro no2 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 no2 group can have is required to be a multiple of ε, where =ε×1.010−24 j. in other words, each no2 group could have ε of rotational kinetic energy, or 2ε, or 3ε, and so forth — but it cannot have just any old amount of rotational kinetic energy. suppose the total rotational kinetic energy in this system is initially known to be 32ε. then, some heat is removed from the system, and the total rotational kinetic energy falls to 18ε. calculate the change in entropy. round your answer to 3 significant digits, and be sure it has the correct unit symbol.

Part b: study of equilibrium on solubility: mg(oh)2(s) ⇌ mg2+(aq) + 2 oh–(aq) cloudy clear (pink) 7. a. b. 8. a. b. 9. 10. 11. 12. when adding concentrated hydrochloric acid, how did the appearance of the equilibrium mixture change? the change in appearance indicated a shift in the point of equilibrium. in which direction did the equilibrium shift? (l) left (r) right explain your answer to question 7a. you should indicate which ion was added to or removed from the equilibrium mixture. when adding edta, how did the appearance of the equilibrium mixture change? the change in appearance indicated a shift in the point of equilibrium. in which direction did the equilibrium shift? (l) left (r) right explain your answer to question 8a. you should indicate which ion was added to or removed from the equilibrium mixture. upon heating in which direction is the equilibrium shifting? upon cooling in which direction is the equilibrium shifting? is the forward reaction a. endothermic explain your answers to questions 9, 10, and 11. (l) left (r) right (l) left (r) right b. exothermic