Please help very easy...

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.

1. if you have a gas at 127 degrees c, what is it's absolute temperature (kelvin)? a. 200kb. 300kc. 400kd. 500k2. if you had a gas whose absolute temperature measured 45 k, what is that temperature in celsius? a. -228 cb. -300 cc. 125 cd. 112 c

Amixture initially contains a, b, and c in the following concentrations: [a] = 0.300 m , [b] = 1.05 m , and [c] = 0.550 m . the following reaction occurs and equilibrium is established: a+2b⇌c at equilibrium, [a] = 0.140 m and [c] = 0.710 m . calculate the value of the equilibrium constant, kc.