Ill find photon energy first then solve for the

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I’ll find photon energy first, then solve for the binding energy. E = h ν = (6.626 × 10 34 Js)(1.370 × 10 15 s –1 ) = 9.078 × 10 19 J Then use that and the given KE to get the BE: E photon = BE + KE, so BE = E photon – KE BE = h ν – KE = 9.078 × 10 19 J – 1.488 × 10 19 J = 7.590 × 10 19 J (6 pts) (b). Now consider what would happen if we varied the frequency of the light. On the axes below, sketch the expected form of each graph. (If you understand the process responsible for the photoelectric effect, you should be able to work out what these graphs would look like even if you don’t think you remember them. Try using the equation(s) you used in part (a) as a starting point.) electron kinetic energy light frequency ( ν ) # of electrons emitted light frequency ( ν ) Linear (2 pts) Not through origin – x- intercept at some positive ν (1 pt) Flat line for most ν (2 pts) Step function at threshold ν (1 pt)
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A8 © 2015 L.S. Brown (12 pts) 8. Consider the apparatus shown below, in which two compartments are initially separated by a removable partition (shown by the dashed line). One compartment is filled with nitrogen monoxide and the other with oxygen, with the conditions as shown in the diagram. The partition dividing the container is removed so that the gases mix, the entire container is heated, and the two gases react to produce the maximum possible amount of nitrogen dioxide (NO 2 , which will also be a gas under these conditions) . If the container is at a final temperature of 500. K when the reaction reaches completion, what is the total pressure? (HINT: Be sure you are accounting for the fact that the gases do not just mix, they react with each other!) It’s a good idea to start with an equation for the reaction: 2 NO(g) + O 2 (g) 2 NO 2 (g) Convert initial amounts to moles: n NO = PV RT = (600 torr)(1 L) (62.364 L torr mol K )(400.15 K) = 0.02404 mol NO n O 2 = PV RT = (500 torr)(1 L) (62.364 L torr mol K )(400.15 K) = 0.02004 mol O 2 Compare those and the equation to see that the NO is the limiting reactant. We get 1 mole of NO 2 from each mole of NO, so we will get 0.02404 mol NO 2 formed. We also have some O 2 left over. The amount of O 2 that reacts will be: 0.02404 mol NO × 1 mol O 2 2 mol NO = 0.01202 mol O 2 reacted Subtract that from the original amount to get 0.02004 – 0.01202 = 0.00802 mol O 2 remaining. Add that to the NO 2 and use gas law to get final P. (V is the combined 2 L now!) Total moles = 0.02404 + 0.00802 = 0.03206 mol P = nRT V = (0.03206 mol)(62.364 L torr mol K )(500 K) (2 L) = 500 torr There’s a really elegant way to do this in terms of P × V – ask me if you are curious. NO(g) 1.00 L, 600 torr, 127 ° C O 2 (g) 1.00 L, 500 torr, 127 ° C
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