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 57ε. Then, some heat is added to the system, and the total rotational kinetic energy rises to 92ε. Calculate the change in entropy.
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How many liters of 3.0 m naoh solution will react with 0.60 liters of 4.0 m h2so4? h2so4 + naoh → na2so4 + h2o 1.2 l 1.6 l 2.4 l 2.8 l
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Consider the reaction: 2al(s) + fe2o3(s) → al2o3(s) + 2fe(s) the δhf for fe2o3(s) = -824.3 kj/mole. the δhf for al2o3(s) = -1675.7 kj/mole. finish the equation. δhrxn = [(1)( kj/mole) + (2)( kj/mole)] - [(1)( kj/mole) + (2) ( kj/mole)]
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Two nitro NO2 groups are chemically bonded to a patch of surface. They can't move to another locatio...
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