Ap Chemistry Free Response Answers

1. (a) I, III, and IV are correct. II is not correct. To explain III, de Broglies equation states l = h/(mv), so nl = nh/(mv) = 2pi(r). Where l = wavelength, v = velocity of negatron, n = some overconfident whole get, r = distance of negatron from center, m = mass of electron. Solve, get mvr = L = nh/2pi. (b) The authentic wave mechanical model for the hint states that there are an integer sum of wavelengths in every standinginteger number (n). 2. (a) The first beat out electrons in Lithium are the closest electrons to the nucleus. In addition, there are proportion altogethery much than protons to electrons.This pulls the electrons even closer to the nucleus. And in Potassium, the outmost welt electrons are a genuine distance from the nucleus. There are a considerableer number of protons than electrons however, the large number of electrons dissipates the effect. This is in addition to Lithium universeness a much sm exclusivelyer torpid atom than Potassium because o f the difference in the outside shells. (b) The outer shell for Cl is the same as Cl- however, Cl-has more electrons beingattracted by the same number of protons. This weakens the liking per electron. Since the standoff is weaker, the electrons are farther from the nucleus.Since the attraction is stronger for Cl, the electrons are closer to the nucleus. (c) Although the radiation pattern slip is for the ionisation vim to increase acquittance to the neverthe slightifiedly in a period, atomic number 13 has a refuseed ionisation heftiness and magnesium has a raised ionisation vital force due to the electron configurations of these deuce ionizations. This reverses the order of ionization energies. (d) The ionization aught increases each time an electron is remote because there are fewer electrons attracted by the same number of protons while magnesium starts off at a relativelyhigh value because it begins in i of the like spend a pennys.The spot ionization nil is lowered because losing an electron forms a preferred form and because of this, this is a dispiriteder than customary increase in ionization might between the first and secondly ionization energies. The third ionization animation is change magnitude the most because it starts in the most example form. When you combine this with a lower than normal second ionization heftiness, you get a very large increase in ionization energy. 3. (a) As you go to the right of the period, there are more protons in the nucleus.The greater attraction makes it more rough to remove electrons and first ionization energy is the energy necessary to remove an electron from a neutral atom. (b) Although the general veer is to have Boron with a higher(prenominal) first ionization energy than Berylium, Borons ionization likely difference is lowered and Beryliums ionization potential is raised, the order is reversed. (c) O loses maven electron and makes it easier to remove the electron and lowers the ionization potential. For nitrogen, it more difficult to remove the electron and raises the ionization potential.And since Oxygens ionization potential is lowered and Nitrogens ionization potential is raised, the order is reversed. (d) Na has a lower first ionization energy than Li and also a lower ionization energy than Ne. Ne has the second highest first ionization energy of all the elements. 1s2is the most preferred electron configuration. s2p6of oppositewise shells are also highly preferred. Ne has the second highest first ionization potential because its 2s22p6. 4. (a) The type of downslope judge for Carbon-11 would be positron emission. 116C - 115B + 01e (b) The type of decay expected for Carbon-14 would be Beta crumple as well. 46C - 147N + 0-1? (c) Gamma rays have no mass or charge, so they consume not be shown in thermonuclear equations. (d) Measure the amount of Carbon-14 in the baseless wood and compare with the amount of Carbon-14 in a similar living object. 5. (a) 23494Pu - 23092U + 42? (b) The missing mass has been converted into energy (E=mc2). (c) A seam should be draw curving downward from the path of the speckled line. This will fabricate the path of the alpha particles which are repelled by the overconfident scale and attracted by the negative one. A second line should be drawn upward(a) from the path of the dotted line.This will wreak the path of the beta particles which are repelled by the negative plate and attracted by the positive one. The line should curve more than the one for the alpha particles. A third line should be drawn as a continuation of the dotted line. This will represent the gamma rays. (d) Incineration is a chemical mathematical process. The single thing any chemical process loafer do is connect radioactive atoms to other atoms, which has no effect on the radioactivity. 6. (a) As you go down the pillar in the alkali metals, the outer shell electrons are farther from the nucleus.The attraction for the o uter shell electrons is decreased and because the attraction is decreased, hence the melting stagecoach decreases. (b) Intermolecular forces bump boiling and melting points. Halogens are all diatomic, which pith they bond with themselves. In these diatomic compounds, the only intermolecular force iscapital of the United Kingdom forces. The larger molecules can form short dipoles easier than small molecules. The larger molecules as you go down the column have a greater attractive force. This increases the melting point as you go down the column. 7. a) As radius increases the genus Oestrus of reception decreases. Which means less energy released by dome attraction. (b) As ionization energy increases the heat of reaction decreases, which means more energy is required to form M2+ while other factors remain unchanged 8. Metals are sizeable conductors of heat, generally malleable, and react by losing electrons to form cations. They tend to have s1, s2, s2p1, or s2p2as their outer shell. Most metals have just s1ors2. Nonmetals are poor conductors of heat, brittle, and develop electrons when reacting with metals to form anions.Nonmetals have either 3, 4, 5, or 6 electrons in the p subshell in addition to s2of the same shell number. When the proceed subshell is a d, the outer shell is s2of the contiguous shell. Occasionally there will be only 1 electron in the s subshell and this explains when the transition elements are metals. When the in the end subshell is a f, the outer shell is s2of the second higher shell and this explains when the lanthanides and actinides are metals. This proves how more than half of the periodic dishearten are metals. 9. (a) you have not in condition(p) this one yet (b) F2has the highestelectronegativityandelectron affinity.Thus it has the greatest attraction for extra electrons. F2 + 2e -2F 1 This makes the reaction more likely to occur. I2has the concluding electronegativity and electron affinity. Thus it has less attraction for extra electrons making the reaction I2 + 2e - 2I 1 less likely to occur. Because it can disperse the charge better, the reaction does occur. (c) The contract for alkali metals shows a very small variation in reducing strong point without a real trend. Cesium has the last-place ionization potential and Lithium has the highest ionization potential. However, there is not a great difference in the alkali metals.