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Notice in Table 2-1 that the creation of the reference computer and running sysprep are identical for both WAIK and third-party tools The details of capturing and deploying the image work differently in a third-party tool than they do in the WAIK

When capturing and deploying images, it s important to realize that the 32-bit and 64-bit versions of Vista are significantly different, since they must be deployed to different architectures If you are deploying to both 32-bit and 64-bit architectures, you will need two images Windows Vista images are Hardware Abstraction Layer (HAL) independent In other words, you can deploy a single 32-bit image to all of your 32-bit computers It doesn t matter if some have different graphics cards or different sound cards, or even if some are desktops and some are laptops You can use a single 32-bit image for all of your 32-bit computers Similarly, you can use a single 64-bit image for all of your 64-bit computers EXAM TIP If deploying images to both 32-bit and 64-bit platforms, you ll need at least two images of Windows Vista The other hardware doesn t matter, so you can use a single 32-bit image for all of your 32-bit computers, even if some are desktops and some are laptops, and a second 64-bit image for all of your 64-bit computers Of course, you can have as many custom images of either platform as your environment needs WinPE is a 32-bit application and takes advantage of a 64-bit system to boot into 32-bit mode before switching to 64-bit mode With this in mind, you can use a single WinPE image to boot into any computer 32-bit or 64-bit

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Using Windows Deployment Services (WDS)

You must use the given reactions to add up to this formation reaction Pb is in the product side of the rst reaction, but needs to be on the reactant side in the target reaction The rst reaction must be reversed and the H multiplied by -1 In the second reaction O2 is on the correct side, but must be multiplied by 1 2 to get the correct number of O2 molecules The value of H must also be multiplied by 1 2 Pb(s) + CO(g) PbO(s) + C(graphite) C(graphite) + 1 2O2(g) CO(g) Pb(s) + 1 2O2(g) PbO(s) 13 Answer: C This problem is done in two steps: photon v photon E photon The equation for the rst step is v = c rearranged to v = c/ m 30 108 s = = 033 1015 1 s or 33 1015 1 s 90 10 8 m The next step uses Planck s relationship, E = h E = (66 10 34 J s) (33 10+15) = 22 10 18 J 14 Answer: B The general trend in ionization energy for atoms across a period is that the ionization energy increases from left to right The trend in a column is that the ionization energy increases from the bottom to the top of a column The trend across a period means that O < F The trend in the column means that S < O Putting these two results together gives S < O < F For cations, the ionization energy is always greater than the atom from which the cation is derived, so F < F+ giving a nal order of S < O < F < F+ 15 Answer: A Chlorine is element number 17 so its atom must have 17 electrons This rules out choices C and D, which have 18 electrons Choice B is the ground state electron con guration of Cl Choice A has one of the 3p electrons promoted to the higher 3d level This is the correct answer 16 Answer: C The trend in electronegativity in the periodic table is that electronegativity increases across a period from left to right and up a column from bottom to top The relative order of electronegativity would thus be Si < C < N < O An O-O bond would be nonpolar The greatest difference in electronegativity is between Si and O, and that will be the most polar bond H = 1 (1070 kJ) = 1070 kJ H = 1 2 ( 221 kJ) = 1105 kJ H total = 2175 kJ.

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