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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.

A3.37-mg sample of protein was chemically digested to convert its nitrogen into ammonia and then diluted to 100.0 ml. then 10.0 ml of this solution was placed in a 50-ml volumetric flask and treated with 5 ml of phenol solution plus 2 ml of sodium hypochlorite solution. the sample was diluted to 50.0 ml, and the absorbance at 625 nm was measured in a 1.00-cm cuvette and found to be 0.486. for reference, a standard solution was prepared from 10.0 mg of nh4cl (molar mass = 53.49 grams/mole) dissolved in 1.00 l of water. then 10.0 ml of this standard was placed in a 50-ml volumetric flask, treated in the same manner as the unknown, and the absorbance found to be 0.323. finally, a reagent blank was prepared using distilled water in place of unknown, it was treated in the same manner as the unknown, and the absorbance found to be 0.076. calculate the weight percent of nitrogen in the protein.

How many moles of gas are present if p=11 atm, v=12l, t=185k?