Mr. green works as the art teacher at hamburg summer day camp. 600 campers participate in the camp every day. yesterday mr. green 6 campers on an art project. what percent of the campers did mr. green yesterday?

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.

Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. in the first step, nitrogen and hydrogen react to form ammonia: (g) (g) (g) in the second step, ammonia and oxygen react to form nitric acid and water: (g) (g) (g) (g) calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions. round your answer to the nearest .

Consider the following unbalanced equation for the combustion of hexane: αc6h14(g)+βo2(g)→γco2(g)+δh2o(g) part a balance the equation. give your answer as an ordered set of numbers α, β, γ, use the least possible integers for the coefficients. α α , β, γ, δ = nothing request answer part b determine how many moles of o2 are required to react completely with 5.6 moles c6h14. express your answer using two significant figures. n n = nothing mol request answer provide feedback