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.

想知道答案D的更詳細解釋🙏

75. 11 (Nej 25 Na²O² is less stable than [Na₂ ] 0². This lesser stability is due to (A)the very small size of the Na²+ ion (B)the large size of the O²- ion relative to Na²+ (C) the instability of oxide ions in the presence of 2+ ions W (D)more energy is required to ionize sodium than would be realized in the attraction between Na²+ and 0²- (Ethe statement is false, both materials are stable. 【87 成大】

想請問這題🙏，有點看不懂圖三的公式是從何而來，卡住了

EXAMPLE 21.11 FIELD OF A UNIFORMLY CHARGED DISK A nonconducting disk of radius R has a uniform positive surface charge density o. Find the electric field at a point along the axis of the disk a distance x from its center. Assume that x is positive. SOLUTION IDENTIFY and SET UP: Figure 21.25 shows the situation. We rep- resent the charge distribution as a collection of concentric rings of charge dQ. In Example 21.9 we obtained Eq. (21.8) for the field on the axis of a single uniformly charged ring, so all we need do here is integrate the contributions of our rings. EXECUTE: A typical ring has charge dQ, inner radius r, and outer radius r + dr. Its area is approximately equal to its width dr times its circumference 2πr, or dA = 2πr dr. The charge per unit area is σ = dQ/dA, so the charge of the ring is do = o da = 2πor dr. We use dQ in place of Q in Eq. (21.8), the expression for the field due to a ring that we found in Example 21.9, and replace the ring 21.25 Our sketch for this problem. Idr dQ X P dEx dEx radius a with r. Then the field component dEx at point P due to this ring is 1 2πσrx dr 4π€0 (x² + ²)3/2 SOLUTION Continued

請問這題什麼意思

noijasuk 31. Which of the following are exact numbers? a. There are 100 cm in 1 m. exact Wild songnsitib sdT VI sons b. One meter equals 1.094 yards. inexact sowied c. We can use the equation exactam silinside of .05 °F = ²/°C + 32 Insom badtam the to convert from Celsiue to Fahrenheitri edi airlax A alismaiz numbers and 32 exact or inexact?ib sai tedW .SS d. π = 3.1415927. in exact fions

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

一直卡住算不出來 求詳解 謝謝🙏

1. A building locates at x=dſ with a height of h as shown in the right figure. A cannon at origin fires a bullet to bomb the roof of the building, please find out: (a) what is the initial velocity and angle should be to complete this mission? (b) how much time will it take to hit the target? (c) how height the bullet can reach? 2. Similar to the problem #1, now there is a mountain with a height of 2h sitting at d/2. The mission is the same, 2h T firing a cannon to high the roof of the building. Please/10 find out: (a) what is the initial velocity and angle should be to complete this mission by flying the bullet over the mountain? (b) how much time will it take to hit the target? (c) how height the bullet can reach?

求解🙏

The two masses shown in Figure are connected by three massless and frictionless pulleys. What is the magnitude of ai / a2 ? a 411 2 a2

求解答案是否正確

Multiple-choice questions: (70 points) write the letter of statement that is not true: 1. A.B.U The Drude Model: (a) It is a classical approach using concepts from the kinetic theory of gases. (b) A relaxation-time approximation is used where the probability of collision is l/t. (c) No forces act on the electrons in between collisions. (d) It predicts accurately the electronic heat capacity of metals. 2. The Sommerfeld Model: (a) The crystal potential is replaced by an average Coulomb potential. (b) The separation of states is so close that a continuum model can be applied. (c) The electrons are highly localized due to a strong crystal potential. (Q) The energy of the highest occupied state at 0 K is the Fermi energy. 3. Density of states in k-space, N(k): (a) N(k) is determined by the boundary value conditions. (b) For lattices in any dimensions, N(k) is constant in k-space. (C) N(k) depends on the direction and magnitude of k. (d) It is directly proportional to the volume of the crystal. E影響, c 4. The Fermi surface of a metal: (a) is responsible for many of the transport properties of the metal, (b) depends on the nature of the chemical bonds, (c) is always a sphere, (d) can involve mixing of electron orbitals. ✓ 5. The conductivity of a material () increases with the mobility of its carriers, (b) increases with the density of carriers, (C) it is inversely proportional to the energy gap. cogu changes with temperature. 6. BID Plasmons: (a) They are an important characteristic of the Sommerfeld electron gas model. (b) Plasmons occur spontaneously in metals. (c) They are related to charge density oscillations. (d) The Plasmon energy is typically of the order of 10 eV. The Fermi energy Ef in intrinsic semiconductors 8) is usually close to y E, at OK (b) it ACiD. can increase or decrease with temperature depending on the ratio of hole and electron effective masses, (c) it represents the point at which the probability of occupation is 12, even though there may not been any states with that energy, (d) its value relative to the electronic bands must remain unaffected in the presence of defects or interfaces.

謝謝

Equilibrium constant calculations often have as much as 5% error X in them because (A)molar concentration and pressures are often used. (B)there is experimental error involved in measure K. (C)the chemical activity should be used. (D)approximation are often used to solve very difficult calculations. (E)all of the above. 【83 中興B)

想問RC circuit 中 charge 跟 discharge 如附圖內的等式是怎麼得出的呢？ 思考了一段時間也上網查過了

The sum ofthe potential difference across the capacitor and resistoris away Constant, so Vc 十 MRg 三委. As charge flows to the Capacitor, the potential difer: ence across it increases, which means that the Potential difference across及誠 the current through the circuit, must decrease. In this circuit, the current /當 Credyey the charge on the capacitor, and therefore 7 二 十ZO/2/. As the currit decreases, we expect the Tate at which the capacitor charges also to decrease When the potential difference across C Teaches 儿, the potential difference acr0S AR 1s zero: The current Stops flowing, and the charge on the capacitor ha5出 maximum value Oo = Cx, When 7 = +Z0/dr is used in the loop rule, we find