EDTA滴定，求解🙏

2. Titration of Ca2+ and Mg2+ in a 50.00-ml sample of hard water required 23.65 mL of 0.01205 M EDTA (C10H16N203). Calculate the hardness of the water sample, expressed as ppm CaCO3 Element Ca Mg C H N o Atomic Mass (g/mol) 40.08| 24.31|12.01|1.008| 14.01|16.00

想問5-9 該怎麼算 ？ 第一個選項就可以了 謝謝

the sample size? (b) 250 ing? 211% -6208 *(a) 500 mg. *(c) 150 mg. Xloo to 10.4 ) X100% =0108 / mg 506 016 5-8. A method of analysis yields masses of gold that are low A by 0.4 mg. Calculate the percent relative error caused ylvila by this result if the mass of gold in the sample is 625 mg.6 20 (d) 70 (c) Dit X 106% 8.6 5-9. The method described in Problem 5-8 is to be used for the analysis of ores that assay about 1.2% gold. What minimum sample mass should be taken if the relatived) error resulting from a 0.4-mg loss is not to exceed *(a) -0.1%? (b) -0.4%? 5 * (c) -0.8%? (d) - 1.1%? Slastin 0 (14)。

41.c 答案為4*10＾（-7） 可是此題的有效數字不是應該為兩位嗎？ 求解 感恩🙏

ma史 40.。 Perf 1 | am 記 following mathematical] Operatlons, and expDresSS Tesult to the correct number of s1lgnlficant filgures. @@ 036.089o 2 1.01 DOCX OIO22 2 0 0 XX 50021x 103 X 7.3904X 102 2.00 xX 106 SO ad, 41.。 Perform the following mathematlcal operatlonS, and express the result to the correctnumber of signlflcant filgures. 2.626 祝 0.470 80.705 3隆 0.623 0.4326 b. (6.404 x 2.91)/(18.7 一 17.1) CC eduamaaese2ecero0一9521ye+94 (SO 0記 CC_IO 0 1 fa (Assume that this operation is taking the average of tfour numbers.Thus 4 in the denomlnator 1S eXact.) 人 ao,

一直算都錯😖，求解

Problem: relative to 丸 Is om always smaller than zz, larger than we orthe same Ph as uu, or does it have sorme other shed k 中 Telationship? COTn- TeaS,。 (4 Lisa rushes down onto a subway 7? 1the Platform to find her train already 主 leparting. She stops and watches b the cars go by. Fach car is 8.60 m ( 語 long. The first moves past her in 站 1.50 sand the second in 1.10 5. 實 Find the constant acceleratlion Of 時 則 thetralin. With )eSlt CHALLENGE PROBLEMS np lt hes is 42. Two thin rods are fastened to ~e on the inside of a cCircular TIng as shown in Figure P2.42. One rod of length Dis vertical, and 17th the other of lenoth 了 makes an

求解～答案都是E

例6:絕熱可逆系統 (1) 1.0 mole monoatomic ideal gas at300K and 10.0 atm is expanded adiabatically and reversibly to 1.0 atm. Calculate the final volume (in D. &=0 【107 台綜大(B) (A)$22GB) 1852 。(C) 6 。(DX2芭 (E) 11.2 (2) What's the work (in R) in (1 ? 【107 台綜大(B) (A) -210 (B) -270 (C) -360 (D) -320 (E) -440

第1題和第5 6 7題，有大大能幫解出計算過程嗎？ 謝謝

略語上 點加圖點 Homework 1 1. 如圖,三個電荷分別在直角三角形的三個頂點 ,已知 qs=+1.0 hlC,qi的電量 為6.4 HC ,但不知電性 ,q2的電量和電性都未知 ; 電荷 q3 所受的合力在負x 軸方向。(a)考慮電荷 qm 和 q2 不同的電性 ,畫出四種受力的組合。(b)利用(a) 的結果求出電荷 q 和 q2 的電性? (c)計算出電荷 q2 的電量? (d)求出電荷 qs 所受的合力? 2. 一半徑為 R 的半圓形環上帶有均勻分佈正電荷 Q,求圓環中心的電場。 3. 如圖,兩長度 L 的棍子沿著 x軸方向放置,一根棍子在x=a/2 和x=a/2+L 之間,另一根棍子在x=-a/2-L 和 x=-a/2 的位置 ; 每一根棍子上的電量為Q 均勻分布在棍子上。。(a) 計算第二根棍子在正 x軸上所產生的電場? (b)證 明兩根棍子之間的相互作用力F=kQ7/L7*/g[(a+L) 2/(a*(a+2L))] 。 4. 如圖,一長度為 4cm 距離為 1cm 電場 E=10” N/C 的平行電板 ; 一個電子由 平行電板左邊中點以與水平夾角9=30" 射進平行電板。當電子直接射出平行 電板時,求電子速度範圍為何? 5. 一寬度為 d 的絕緣帶電厚板,它的電荷密度p= Cx ,此處 x為厚板內任一點 至厚板中心的距離,C 為一常數;厚板的y和z 方向延伸至無窮遠。試求空 間的電場(a)厚板外部 (lx| > d/2) (b) 厚板內部 (lx| < d/2)。 6. 半徑為R的無限長實心圓柱體帶有不均勻電荷密度分佈 電荷密度p =po(r/R), 給出的,其中po是常數,r 是距圓柱軸的距離。求出電場的r表達式。 7. 如圖,一個帶有均勻體積電荷密度p且半徑 2a 的非導體球,挖空球內的一個 半徑 a的小球裡的電荷 ; 證明小球內的電場 g =0和蕊=pa/(3&g09)- +lmc. 涼 / 一8一 4 2 K交 / 人 姜- ‧下co一一 六 2.L 二32 92 二32L 題1 題3 4em 生 唔 和 NC em 題4 題5

求解

台 和2人本 5,148。Estimate the distance (in nandQineters) between mol- ecules of water vapor at 100。C and 1.0 atm. Assume ideal behavior. Repeat the calculation for liquid Water at 1005C, given that the density of water 1s 0.96 g/cm at that temperature. Comment on your Tesults. (Assume water molecule to be a sphere With 六 a diameter of 0.3 nm.) (jg/: First calculate the 腑 [numper densityjof water molecules. Next, convert 諷 了he number density to過 je aaa 也at1 18, UU “ofmolecules in one direction. ) 3

想請問為何要減掉兩個電子

13 1 7N->sC+角+v 光INSx GELBSTSTW 之取嗲變過程中 同 產生取粒子之虹 同 (苹核 和全 之最大能量為多少MeY: 玄 原子質量為13.0057388 amu。,子核的原孖質量 13.0033531 amu。) 咕2顆電陸『 以 AnsS: E= mo AM lamu= 931MeY (13.0057388-13.003355 2.22MeV- 2 0.511MeY 了最大能量 三0 時 99年度系輸點陰信全 「輻身坊人師」 得全科 1)amux 931.5MeV/amur 222MeY 三1.198MeY

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

求解55感謝

1 1 WWWVeS livelengths on the 2 GIuerofi>入101m. Galculate he elierpv 1 0X 10 fm radio Waves in units of kilojoules per mole Of 13dio waves. Consider 了出at the bond energy ofacarbone_carbon aingle bond found in 生 說 organic compounds is 347 kJ/mol. Woujld X Tays and/or radio Waves be able to disrupt OTganlc cormpounds by breaking carbon-carbon single bonds? SS。 The Work function of an element is the energy required to 56. ITermove an electron from the surface of the solid element. The Work function for lithium is 279.7 kJ/mol (that 1s, 1 takes 279.7 kKJ of energy to remove one mole of electrons from one mole of Li atoms on the surface ofLi metal). What is the max- lmum wavelength oflight that can remove an electron from an atormn on ffhe surface of lithium metal? It takes 208.4 kJ of energy to remove lmole ofelectrons from an atom on 負e surface of rubidium metal. How much energy does it tfake to remove a single electron from an atom on the surface of solid rubidium? What is the maximum wavelength 說人計生 加計