Please answer this question please brianlest

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.

Select the correct value for the indicated bond angle in each of the compounds. o−o−oo−o−o angle of o3 90° 109.5° < 109.5° 120° < 120° 180° f−b−ff−b−f angle of bf3 180° < 109.5° < 120° 120° 109.5° 90° f−o−ff−o−f angle of of2 < 120° 120° 90° 109.5° 180° < 109.5° cl−be−clcl−be−cl angle of becl2 90° 109.5° 180° 120° < 109.5° < 120° f−p−ff−p−f angle of pf3 90° 109.5° < 109.5° 180° 120° < 120° h−c−hh−c−h angle of ch4 90° < 109.5° 180° 120° < 120° 109.5°

Reactions in cells take place at about a. 40°c b. 0° c. 100°c d. 60°c