What’s the answer please

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.

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

Phosphorous acid, h3po3(aq) , is a diprotic oxyacid that is an important compound in industry and agriculture. p k a1 p k a2 1.30 6.70 calculate the ph for each of the points in the titration of 50.0 ml of 1.5 m h3po3(aq) 1.5 m h 3 po 3 ( aq ) with 1.5 m koh(aq). 1.5 m koh ( aq ) .