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Chemistry, 22.06.2019 04:00, mgnbrnne
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.
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Chemistry, 22.06.2019 23:50, josie311251
Be sure to answer all parts. the following equilibrium constants were determined at 1123 k: c(s) + co2(g) ⇌ 2co(g) k'p = 1.30 × 1014 co(g) + cl2(g) ⇌ cocl2(g) k''p = 6.00 × 10−3 calculate the equilibrium constant at 1123 k for the reaction: c(s) + co2(g) + 2cl2(g) ⇌ 2cocl2(g) 4.68 × 10 9 (enter your answer in scientific notation.) write the equilibrium constant expression, kp:
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