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.

初會ㄉ銀行調節表問題🥲 請問b,c小題要怎麼寫啊？我快居居了

1 Balance b/d " 7 B Man " 16 A Sun " 28 M Bai " 30 K Ng 2,379 The bank account of C Chan's business for June 2009 and the bank statement for the same month are as follows: 2009 Jun Cash at bank $ 2009 Jun 5 D Wong " 158 12 J Yau " 93 16 B Shek " 307 29 Orange Club " 624 30 Balance c/d 3,561 $ 150 433 88 57 2,833 3,561 Bank Statement Date 2009 Jun Details Dr Cr Balance $ $ $ 1 Balance b/d " 8 Cheque " 11 Cheque " 17 Cheque " 17 Cheque " 18 Cheque " 29 Cheque " 29 Standing order - Union Credit " 30 Credit transfer - RS Ltd " 30 Bank charges 2,379 Cr 158 2,537 Cr 150 2,387 Cr 93 2,480 Cr 433 2,047 Cr 88 1,959 Cr 307 2,266 Cr 44 44 2,222 Cr 70 70 90 90 2,312 Cr 2,242 Cr (a) Cross out all the items that appear in both cash book and the bank statement. (b) Prepare a bank reconciliation statement as at 30 June 2009. Start with the balance as per cash book (c) Update the bank account of C Chan's business and prepare a bank reconciliation statement as at 30 June 2009. (-24 (b) Bank Reconciliation Statement as at 30 June 2009 Balance per cash book add credit transfer less: Standing order Bank charges. Balance per bank statement less: check 3447 $ 90 Total 3,5615 40 (44) (70) (114) $3537 $2242 (624) (57)

邏輯設計第6.5題

Es 6.4 找出一個最簡三階NAND 閘電路實現: F(A,B,C,D) = 2m(5,10,11, 12, 13) (四個閘) 利用四個 NOR閘實現Z=A'D+A'C+AB'C'D'。

第二張圖片是b小題的答案 想請問為什麼Retained Earnings 會變₩1994 另外想請問Accounting Titles的順序是固定的嗎 謝謝🙏

E4.7 (LO 2) Je Ju Consultants had the following adjusted trial balance (amounts in thousands). ar boxeqang and oil (EO) Je Ju Consultants Adjusted Trial Balance For the Month Ended June 30, 2025 Prepare closing entries, and prepare a post-closing trial balance. s of boiply anotoupen Instructions Account Titlesi Cash on he Accounts Receivable ady Adjusted Trial Balance Debit of sad ton of Credit t bly 3,712 mos balirahows izud to saylana 3,904 weniged share Supplies Accounts Payable 480 ¥1,382 Unearned Service Revenue Share Capital-Ordinary Retained Earnings Tagbal 160 4,100 Jun1,660 Dividends 550 Service Revenue 4,300 foundent Salaries and Wages Expense bizi 1,260 Miscellaneous Expense 256 Supplies Expense Salaries and Wages Payable P1,900 (0.3) 11. 460 sola ¥12,062 ¥12,062 001-81 a. Prepare closing entries at June 30, 2025. b. Prepare a post-closing trial balance. FEW 00181 008 ona

想請教計算方法，目前求得莫爾比但後續處以0.18（化整數）時感覺很怪，題目要求120克pan的經驗式，但我化完整數後光是C+H前兩項的質量就會超過120克了，想請教各位如何計算

C z Hz N O S o f 3.44 Peroxyacylnitrate (PAN) is one of the components of smog. It is a compound of C, H, N, and O. Deter- mine the percent composition of oxygen and the empirical formula from the following percent composition by mass: 19.8 percent C, 2.50 percent H, 11.6 percent N. What is its molecular formula, given that its molar mass is about 120 g?

想詢問關於系統的觀察性的問題 討論系統的觀察性時 只需討論C矩陣對應喬登方塊的第一行不為0向量即可 但為什麼不會限制特徵值為1時不可觀察 左圖是我的算式，右圖是我對觀察性的理解

X₁₁) = Ax (+) Y(t) = (x(t) [×+ ] [ 6 ] [ 7 ] + BU X₂(t). X₁(t)] 不相干 git) = Cete Xro) At y(t) = [C₁ C₂] X₂(t) t == (C₁ (X₁ (0) + X2 (0)) + (2 X (0) e * = qe² X110), X210)無限多組解

想詢問關於系統的觀察性的問題 討論系統的觀察性時 只需討論C矩陣對應喬登方塊的第一行不為0向量即可 但為什麼不會限制特徵值為1時不可觀察 左圖是我的算式，右圖是我對觀察性的理解

X₁₁) = Ax (+) Y(t) = (x(t) [×+ ] [ 6 ] [ 7 ] + BU X₂(t). X₁(t)] 不相干 git) = Cete Xro) At y(t) = [C₁ C₂] X₂(t) t == (C₁ (X₁ (0) + X2 (0)) + (2 X (0) e * = qe² X110), X210)無限多組解

這是一題單純的R與C並聯，Z是並聯後的阻抗，請問Rx和Cx怎麼推導

(b) Z, remains the same as in Eq. (9.15.4) but R, and X, are in parallel. Assuming an RC parallel combination, Zx = (5.993-j1.356) MQ 1 RK Rx (1-ju = Rx || = jwCx 1 + jwRxCx i+w Rx 5.993 R₁ = By equating the real and imaginary parts, we obtain Real(Z)² + Imag(Zx)² _ 5.993² + 1.356² Real(ZX) Re Zx = 6.3 M2 5.993 Imag(Z) Rx = 5.993² +1.3562 5.993 C₁ = = [Real(Z)² + Imag(Z)²] Cx = - -1356 2Tu₂k[5.993)² + -1.356 2z(2000)(5.9172+1.356²) = = 2.852 μF

想要求解大學初會這一題 練習了很多次 但還是會錯 想請教解題的觀念 因我覺得自己是題目沒有懂

II. The trial balance of Drysdale Company shown below does not balance. DRYSDALE COMPANY Trial Balance June 30, 2025 Errors Correct Debit Credit Debit Credit Cash £2,600 £540 £2,600 Accounts Receivable £7,600 £540 7,600 Supplies 600 Equipment £8,300 + £700 8,300 Accounts Payable £9,766 - £252 -- £225 £9,766 Share Capital-Ordinary 1,952 Dividends £1,500+ £300+ £300 1,500 Service Revenue £15,200+ £279 15,200 Salaries and Wages Expense 3,800 Maintenance and Repairs Expense 1,600 Totals £26,000 £26,918 An examination of the ledger and journal reveals the following errors: 1. Each of the above listed accounts has a normal balance per the general ledger. 2. Cash of £170 received from a customer on account was debited to Cash £710 and credited to Accounts Receivable £710. 3. A dividend of £300 was posted as a credit to Dividends, £300 and credit to Cash £300. 4. A debit of £120 was not posted to Salaries and Wages Expense. 5. The purchase of equipment on account for £700 was recorded as a debit to Maintenance and Repairs Expense and a credit to Accounts Payable for £700. 6. Services were performed on account for a customer, £310, for which Accounts Receivable was debited £310 and Service Revenue was credited £31. 7. A payment on account for £225 was credited to Cash for £225 and credited to Accounts Payable for £252. (@4×10 + 合計1%,41%) 2

為什麼這時積分變數t變換為τ時，積分範圍不會改變呢

例題1-7: Given a function f(t) and its Laplace transform F(s), what is the Laplace transform of the delayed version of f(t) by 2>0 units of time, i.e., f(t−2)? You must give a +b(1) proof of your answer. 解 根據拉氏轉換的定義,F(s)=L[f(t)]=ff(t)e="d,所以 【85 中興電機】 [ƒ (1-2)] = f(t−2) e™ dt = f(t-2)e(-) d(-2) -2s 【大專告 201 0 0 C-TER -λs (a)D -ST =eft-)e-d(1-2) = e | f(t)e dt = eF(s) 0