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自然科學 大學

為什麼[A(org)]=n[An(org)]成立 (第三張圖equation 7.9) 感謝

Experiment 7 Experiment 7 The partition of Organic acid between Water and Organic solvent Objectives Understand the partition of a solute between two immiscible solvents. Introduction A chemical analysis that is performed primarily with the aid of volumetric glassware (e.g., pipets, burets, volumetric flasks) is called a volumetric analysis. For a volumetric analysis procedure, a known quantity or a carefully measured amount of one substance reacts with a to-be-determined amount of another substance with the reaction occurring in aqueous solution. The volumes of all solutions are carefully measured with volumetric glassware. The known amount of the substance for an analysis is generally measured and available in two ways: 1. As a primary standard: An accurate mass (and thus, moles) of a solid substance is measured on a balance, dissolved in water, and then reacted with the substance being analyzed. 2. As a standard solution: A measured number of moles of substance is present in a measured volume of solution - a solution of known concentration, generally expressed as the molar concentration (or molarity) of the substance. A measured volume of the standard solution then reacts with the substance being analyzed. The reaction of the known substance with the substance to be analyzed, occurring in aqueous solution, is generally conducted by a titration procedure. The titration procedure required a buret to dispense a liquid, called the titrant, into a flask containing the analyte. A reaction is complete when stoichiometric amounts of the reacting substances are combined. In a titration this is the stoichiometric point. In this experiment the stoichiometric point for the acid-base titration is detected using a phenolphthalein indicator. Phenolphthalein is colorless in an acidic solution but pink in a basic solution. The point in the titration at which the phenolphthalein changes color is called the endpoint of the indicator. Indicators are selected so that the stoichiometric point in the titration coincides (at approximately the same pH) with the endpoint of the indicator.

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自然科學 大學

求解圖裡UICK quiz的第二小題

acit0r 1taneOuls CUrTent Differentiating Equation 27.22 with respectto time gives the inst: as afunction of time: 4 tt號 一Fe ptasafurc 主二 where Q;/有CC 三方 is the initial current. Figure 27.15b 5 in the resistor that we guessedin order to-apply Kirch Equations 27.15 and 27.21. Equation 27.23 shows that the ci ing capacitor is negative, indicating that the current 1s Figure 27.15c. Both the charge on the capacitor and the ct tially at a rate characterized by the time'constant7了三用C. 人 i 一 GOuicK ouiz 27.5 Considerthe circuitin Figure 27.17 and assume the battery has after the switch is closed, what is the currentin the Figure 27.17 no internal resistance. () 」 battery (a) 0 (b) &/2A(G 28/Z04 8/元(e) impossible to determine (ii) After a 有 eyerylong time, what is the Currentin the battery? Choose from the same choices. theswitch is CC (3 前:銜1加信介2放 IntermittentWindshield Wipers nobiles are equipped with windshield wipers that can operate intermittently during alight rainfall. How does ofsuch wipers depend on the charging and discharging ofacapacitor7 民 The wipers are part ofan 及C circuit whose time constant can be varied by selecting different values of及through amultipo- Sition sWitch. As the voltage across the capacitor increases, the capacitor reaches a point at which it discharges and Higgers the wipers. The circuit then begins another charging cycle. The time interval between the individual sweeps of the wiperS 1 determinedjby the value ofthe time constant.

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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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