9134517409 | Ar: strongest LD IMF, largest # electrons | deviates most from ideal behavior | 0 | |
9134523358 | 36.3 is the average molar mass | 1 | ||
9134548359 | they have the same kinetic energy | If all the balloons are at the same temperature, then | 2 | |
9134559471 | Hydrogen, smallest molar mass | smallest balloon after 12 hours | 3 | |
9134576662 | copper | element | 4 | |
9134585058 | B | similar IMFs to solvent | 5 | |
9134591899 | A and D are similar compounds | 6 | ||
9134596084 | Contains 1.0 M NH₄Cl | 7 | ||
9134830748 | 1s²2s²2p⁶3s²3p⁶4s²3d¹ | 8 | ||
9134871637 | Sc | element | 9 | |
9134877888 | 2 | valence electrons | 10 | |
9134888108 | Galvanic cell | 11 | ||
9134905755 | Anode | Silver | 12 | |
9134912752 | direction of electrons | left | 13 | |
9134922432 | To determine the vapor pressure of the oxygen, you would take the air pressure and subtract the water vapor pressure | 14 | ||
9134974274 | interstitial | B | 15 | |
9135001296 | substitutional | A | 16 | |
9135010835 | 0 | order | 17 | |
9135030609 | Rate = k | rate law | 18 | |
9135033118 | M/time | k units | 19 | |
9135040928 | 1 | order | 20 | |
9135052065 | rate = k[A] | rate law | 21 | |
9135054236 | 1/time | k units | 22 | |
9135057838 | 2 | order | 23 | |
9135063209 | rate = k[A]² | rate law | 24 | |
9135065994 | 1/(M∙time) | k units | 25 | |
9135094848 | q = mcΔT (c of ice) | A | 26 | |
9135109612 | q = mHf | B | 27 | |
9135111051 | q = mcΔT (c of water) | C | 28 | |
9135111052 | q = mHv | D | 29 | |
9135112327 | q = mcΔT (c of steam) | E | 30 | |
9135773104 | A | if the substance is the same, then this shows a lower temperature | 31 | |
9135774059 | B | if temperature is the same, then this substance has the smaller molar mass | 32 | |
9135804048 | 432 | bond energy | 33 | |
9135816183 | 74 | bond length | 34 | |
9135825763 | 2 | number of reaction steps | 35 | |
9135843799 | 1 | faster step | 36 | |
9135855225 | exothermic | 37 | ||
9135859833 | strong acid and weak base | type of titration | 38 | |
9135861051 | 20 mL and pH of 5 | equivalence point | 39 | |
9135876654 | 0.032 | Ka = | 40 | |
9135879330 | methyl red (pKa = 5.0) | indicator | 41 | |
9135899122 | weak acid and strong base | type of titration | 42 | |
9135904726 | 1 x 10⁻⁵ | Ka = | 43 | |
9135906943 | 20 mL and pH of 8.5 | equivalence point | 44 | |
9135909406 | phenopthalein (pKa = 9.3) | indicator | 45 | |
9135928489 | The p orbital is further away from the nucleus so the attraction to the electron in the p orbital is smaller and decreases ionization energy | why B has lower ionization energy than Be | 46 | |
9135929501 | electrons paired up in p orbitals have more repulsion the electrons not paired up. The repulsion makes it easier to remove an electron and lowers the ionization energy | why S has lower ionization energy than P | 47 | |
9135929983 | effective nuclear charge increases | Ionization energy generally increases across the period because | 48 | |
9135930609 | distance between the nucleus and valence electrons increases due to more energy levels | ionizatio energy decreases down a group because | 49 |
AP Review Flashcards
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