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自然科學 大學

大學普化 方智化學第六題

ow Na k 7 CI o se Br Chapter 2 化學鍵結 AH -C=CH 升二技插大 3. Brigma The bond between the carbon atoms in acetylene (HCCH) consists of ) (A)6 pi electrons (B)4 pi and 2 sigma electrons (C)6 sigma electrons (D)2 pi and 4 sigma electrons (E)3 pi and 3 sigma B) B 7.(BCDE) electrons. 【84 成大化學) 14.(B) 20.(C) 4. E ) (A)NO; [83 中興A] Which of the following has a triple bond? (B)O2 (C)C12 (D)CO; (E)CN-. Which one of the following element pairs has the lowest electronegativity (B) 27.(D) 34. (C) C) 41.(CD) 1 G difference and thereby is the least polan? Få 10 323.49 B) (84 5% *ILI) -) 48. (A) 6. NET 2 > 2 D) 58. (D) 62. (B) 69. NO+ D Obind (219) O 发 (A)SeF (B)Seci (QKBr (D/NaCl (E)KAT. 11 The compound Sio does not exist as a discrete molecule while CO2 does. This can be explained because ly 2 ep 3 15 aly zp (A)the Si–O bond is unstable (B)The Lewis structure of SiO has an even number of electrons (C)The SiO2 is a solid while CO, is a gas (D)the 3p orbital of the Si has little overlap with the 2p of the 0 (E)none of the these. V MOJT 515 1 - 4 40 (成大】 Yi (4 0 atom) (86 58*A] 7. In order to explain the valence observed for many of the elements 0 -Si-O in terms of electron configurations, it is often useful to look at electronis staies slightly different from the ground state in comparing 0-41-0 the ground state of boron with the state more useful to describe 0 bonding, the number of unpaired electrons goes from (A)1 to 2 (B)1 to 3 (C)3 to 4 (D)3 to 5 (E)none of the above. (集 (80 清大B) Which of the following compounds should not exist? ( (, 【85 成大A】 alf 566-32 Cle 32-8-24 2-89 cild 14 1 he atoms 83 中興A) mm bove. B 29 p. 1 1 be a ) \CHE > ? (A)Na,P (B)(NH4),PO, CIPO. DPH xH)POCI,, Maspal sput 中興A] Pa zorgt O 15412 = 1) O-P- 5 5+12= 1) 10-4 - 13 13-12=1 -- cl-P-0

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自然科學 大學

請問一下,第45題怎麼算

96d nt” Chemieal EdUUDT LT 人 說 1 ction below ata certam 1 全 同 同 equilibrium ee 說 ii [Ho2] 三 0.0500 TE 、 ipid container are 總 人 2 抽 Ho(g) + 到(tg) 三一 2HF(g) /@巡 > 1 / 1 語全 全胸 : iibri mmIXtUure, 0 2 於 了 人 0.200 mole of Fz is added to this 1 gquilib- thee 2 caJcuJate the'corcentrations 9f a11 gases 1 )。 入 rium is reesfablished- 2 只汪 人 和 沿之 / (9 1100 K, 及p 0.25 for the following reactlon 1 三入 3 、 一一 所 全生 /4 2SQ23(g) 十 Qz(g) 二二 250O3(9 du Ttbe 選娘 an ial pressures of 502, 人2 te 全 了Cajculate the eduilibrium Partial pressure 人 同 葬 也62人 5O3 produced 自om an initfaL_mixrure In 8 oo 三 Po 王 0.50 atm and Pso, 三 0. 2人 三人 1和 宅及林用 生 ” 46. At 22002C 肥 三 0.050 for the reaction 2509 4 2 yy2(9) +O2(g) =二 2NO() 的 What js 和the partial pressure of NO at equilibrtum as 人選 9 suming he N2 and O2 had initial pressures of 0.80 atm 全 // -彈d 0.20 atm, respectively? @ / 32 5 相/ 和pe )f reactiOn_ we wi study is that having a very 汪 補- 六sma上有value (K<< 1 Solving for equilibrium concen- 一條0說人 frations jm an eduilibrium problem usually requires many mathematica| operations to be performed. How- 說 和 even the math involved in solving edquilibrium problems 。。 %6 2 Ina D6909 , for feactions having smal| K values (K << 1) 1s simpli- 527- fied. Whatassumption js made when solving equilibrium 1 Concentrations for reactions having small 太 values? 534 Le 台 戰 Whenever assumptions are made, they must be checked ler ? (吃 hdiry np general the 59 rule> is used to check t cd 旨 Me Vaudlty of assuming thatx (or 2x, 3x, and so on) is ga 只 sma compared to some number Whenx (or 2x, and so 1 less than 和

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