這五個式子加起來，怎麼等於最後的反應式？圈起來的是不知道怎麼處理的

倒安 CaC₂ Ca(s) + 2C) → Ca+ CaOs 2 CaO(s) + H₂O(e) → -O₂(g) 類題 Ca(OH)2(aq) C₂H2(g) + 2 11 -02(8) ⇒ 升二技插大 " △H°=+1300 2CO2(g) + H₂O(1) X2 2C(石墨) + 2O2(g) 2CO2(g) △H = -393.51×2 最後將這五個式子加起來,恰得CaCzsy+2H2O→Ca(OH)2(aq) + C2H2(g) ...HAH°=AH, °+AH₂° +AH₂° +AH₂°+AH₂°=-713kJ AH°=+ 62.8 △H°=-635.5 △H =-653.1

分析化學 拜託好心人士可以教教我🙏 高職畢業沒有基礎 老師又沒有教怎麼算 自己讀過還是看不懂🥲

22 分析化學 習題五 1.依據下列反應,以酸的式量的分數或倍數表示其當量: UHSOX+KOH-KHSO+HO (b2LOH + H.PO-Li HPO+2HO (c) Ba(OH)+2H1O, Ba(10)+2H.O id Na CO,+2HCICO+ HO+2NaCl 52 計算下列的毫當量數: (fw HSO) (fw H.PO/2) (fw HIO) (fw HC) (a)06498克的HgO,作爲滴定酸的一級標準: HgO+2HO° +4Br=HgBr²+3HO (07644克的AgNO作路硼氫的還原劑: 8Ag¹+BH +80H 8Ag+H.BO, +5H₂O (c)04602克NaH.PO+2HO發生下列反應: H.PO. +OHHPO (295) 53 500ml水溶液含921g K.Fe(CN),用下列方式表示溶液的濃度: (0.06M) (a)莫耳分析濃度 (blkº莫耳濃度 (02M) (c)關於下列反應的當量濃度 MnO+5Fe(CN) +8H" == Mn'" +5Fe(CN)+4HO【006M) 54 將877g KH(10)溶於水配成500ml溶液,用下列方式表示溶液的 (a)莫耳分析濃度 [0045M] 〔0.09M] (BIO 莫耳濃度 (c)用作酸的當量酒度 [0045M] (①用作下列反應之氧化劑的當量濃度 JO"+HNNE+2H' +2C=IC +N+3HO (0361) 5-5配製1000ml的004N KI溶液,需多少克KI? (a)氧化成 (664) (60) 第三篇 第五章 容量分析简介 243 (0.83g) (0.204g) (0.208g) (0.308g) (0.384g) (0.048g) [0153g] (1198) (135g) (1418) (18.8ml) (18.8ml) (22.4ml) (51.69ml) (28.78m1) (bIT 氧化成ION 56在48600m170005N溶液中含有多少克洛買? GalNaHCO (b) Ba(OH) (c)(產物IC) (d) Na.S.O. (S.o/ ) {elNaSO(產物SOP) (NaSO(或物SOF) 57 由下列試劑配製滴定濃度 40mgHNO/ml的溶液25000ml la) KOH UKI(產物NO及L) (KMnO(產物NO"及Mn") 团Ce(SO)(產物NO"及Ce") 58 計算0112N NaOH與下列溶液反應所需之體積 (a)2500ml的 0.084NHCIOL (b) 25.00m10 0.064N H.SO. (c)09783g的KH(IO) (d)03126g的H.COHO(產物CO) e04448g的NaH.POHO (產物HPO)

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

求證明過程~感謝

7. (II) Atwood's machine was an apparatus that allowed a direct verification of Newton's second law. It could also be used to measure g. Two equal blocks of mass M hang on either side of a pulley; see Fig. 5.56. A small square rider of mass m is placed on one block. When the block is released, it accelerates for a distance H till the rider is caught by a a ring that allows the block to pass. From then on the system moves at a constant speed which is measured by timing the fall through a distance D. Show that (2M + m)D 2mH12 where 1 is the time moved at constant speed. m2 M Hotez D M FIGURE 5.56 Problem 7

請問第9題的第四小題，為何B選項1：2：3：4不是正確的答案呢！ 我個人認為B及D選項都是正確的！ 請高手說明，謝謝！

關係圖(a-t圖)應為下列何者? (A) a (m/s) (B) 0 a (m/s) f(s) (C) ( )(1)連續兩影像間的時間間隔為何? (A)0.01秒(B)0.1秒(C) 1 秒(D) 10 秒。 ( )(2)物體落下時的平均加速度為多少 公分/秒?(A)87 (B) 870 (C)175 (D)1750。 ( )(3) x 值為何?(A)32 (B)35.5 (C) 39.3 (D) 42 • 答案:(1)(B);(2)(B);(3)(C) 13.1-4.4 解析:(2) =870(公分/秒) 0.12 [(x-17.5)-13.1] (3) =870,X- t(s) 0 a (m/s) (D) o 0 a(m/s?) t(s) 0 o 13610 0.12 393。 出處:試題 難易度:難 9. 請在閱讀下列敘述後,回答下列問題。 在義大利佛羅倫斯的科學史博物館成立 於西元1927年,經兩年翻修後於西元2010 年6月10日重新開館,更名為伽利略博物 館(Museo Galileo) 館內收藏了一座十八世紀後期的一個演 示儀器,長寬高為544cmx39cmx124cm,作 者不詳,在史上最美的十項科學實驗這本書 裡,作者克里斯(Crease)將此面稱為 Belled Plane。 此演示器材有兩組鈴鐺,第一組鈴鐺是 受懸掛在斜面後方的單擺牽動,每擺到最高 點時會敲響鈴鐺,並發出等時間間隔的鈴鐺 聲。第二組是斜面上5個活動鈴绪組,沿著 斜面設置,當球自高處滾下來經過鈴錯時, 就會發出聲音,演示者可以一邊聆聽單擺敲 出的鈴鐺聲,一邊排列這組活動鈴鐺,使球 滾下來敲響的鈴鐺聲也是等時間間隔的,以 測量各組活動鈴绪的間距比例是否符合「等 加速度」的特性。 ( )(1)當球自斜面高處滾下來的過程屬 於何種運動?(A)等速度運動 (B)等加速度運動 (C)加速度 運動 (D)自由落體運動。 ( )(2)球自斜面高處滾下的加速度-時間 f(s) ( )(3)球自高處滾下來經過鈴鐺時,敲 響的鈴鐺聲也是等時間間隔的意 義為何?(A)經過每鈴鐺的時 間間隔相同(B)鈴鐺間距相同 (C)鈴鐺發出的音高相同(D) 經過每鈴鐺的速度相同。 ( )(4)「等加速度」的特性是每隔相同 時間間格球移動的距離比應為下 列何者?(A)1:1:1:1( B) 1:2:3:4 (C) 1:4:9 : 16 (D)1:3:5:7。 答案:(1)(B);(2)(A);(3)(A);(4 (D) 出處:習作 難易度:難 10. 阿美費盡九牛二虎之力爬上高樓的樓頂, 正在欣賞美景時,不小心口袋中拾圓硬幣 掉出,自樓頂自由落下到地面需時4秒, 若不計空氣阻力(g=9.8 m/s),試回 答下列問題: ( )(1)硬幣著地的速度為多少?(A) 9.8 m/s (B) 19.6 m/s (C) 29.4 m/s (D)39.2 m/s。 ( )(2)阿美爬上多高的樓頂?(A) 78.4 m (B) 44.1 m (C) 29.4 m(D)19.6m。 16 1 4 q 17

求解第五，第六題的明確詳解及意思說明，感謝！

器材:100 公分長布只一捲 方法:請使用步尺連續則量10 次你的步長並記錄在下列表格中 大數 1 2 3 4 5 6 7 8 句 10 平均 26. 32 29. 29. 31 30 29. 28. . 29 30 29。 . 5 5 5 5 5 5 a 69 偏差量 -3.1 。 2.8 -0.1 -0.1 1.8 0.3 -0.1 -0.7 -0.6 0.3 0 9 1 9 日。 1. 1 9 a a 1 (1) 根據上表數據計算對應偏差量:(AV = ="), i = 次數,並填入上表偏差量 對應數據格中。 (2) 計算平均值;_29.69. (3) 計算標準偏差平均值的標準差X): DS Axi N(N − 1)-0.23968(N=TEMBE) (4). $* fto=29.92968 0= (5) 計算變異係數大於5%,若是則將離平均值最遠的數據去除後,再計 算平均值及變異係數直到變異係數小於5%為止 (6) 根據下表不同測量次數之偏差臨界值T = 10 ,將超過“士之数幾除 文數 2 3 4 5 6 7 8 9 10 11 >11 t 1.1 1.3 1.5 1.6 1.7 1.8 1.8 1.9 1.9 2.0 2.0 5 00 N 4 3 o 6 1 6 0 0 偏差臨 界值量

請各位幫我解釋一下第15題🙏🙏

1-5 理想氣體方程式 39 -B612 PV - 4XT - V. I Vi : V2 = 2/3 + 4 三、非選擇題 y14 右圖所示為某 定量的 理想氣體之P-T圖(壓力-溫度圖) 原來 為狀態A,其體積為V,經某過程後成為狀態B,其體積為V2, 见力(P) 0 45° 53 离度(T) 57 cm Hg 注意圖中夾角的度數,則V 與V, 的比值為 中 2 15 圖一所示,一玻璃管長 57 cm,管底封閉,在25°C、latm 下裝滿氮氣後,鉛直倒壓入一水銀槽內,直到管底與水銀槽的 [line 水銀面等高,如圖所示。若外界之壓力及溫度不變,則此時 管內之氮氣壓力應為1.5 atm . 16.51A = 96th) ha h 38 cm 16 試回答下列密閉定容氣體實驗:P-16+ / 11 ord : = 1.5 atm (1)如右圖所示,在20°C、latm下做實驗,未加熱時,p、q兩 水銀面等高,將p高度固定,加熱至50°C時,q比p高出7.6 cm,若繼續加熱,當p、q高度差為19 cm時,燒瓶中氣體的 19cm 溫度為 °C。 (2)此時燒瓶中氣體的壓力為 活塞 17. 如右圖所示,可自由滑動之活塞兩端分別放入4atm之He 及1atm 在等下達平衡時,活塞將右 玻璃管 玻璃管 燒瓶 水 p 水 軟橡皮管 atm. o X He 0

大一普化5.58求解

/r'277e777y 記 的 B 9 5.61 A mixture of gases contains CH4, Co2Heo, and CHS. the total pressure 1s 1.50 atm and the numbers of moles of the gases present are 0.31 mole for CHu, 0.25 mole for CH日c, and 0.29 mole for C3Hs, calcu- ]ate the partial presSures of the gases. A 2.5-L fask at 15”C contains a miXture of three gaseS, N2, He, and Ne, at partial preSSureS of 0.32 atm for N2, 0.15 atm for He, and 0.42 atm for Ne. (a) Cal- culate the total pressure of the miXxture. (b) Calculate 出e volume in liters at STP occupied by He and Ne 議 the N2 1s removed Selectlvely. NN Dry air near sea level has the following composition by volume: N2, 78.08 percent; O,, 20.94 percent; Ar, 0.93 percent; CO,, 0.05 percent. The atmospheric pressure is 1.00 atm. Calculate (a) the partlal pres- Sure of each gas in atm and (b) the COohcentratlion Of each gas in moles per liter at 0。C. (7/7///- Because Volume is proportional to the huber of moljes Present, mole fractions of gases can be eXpresSed as ITatios of volumes at the same temperature and preSSure.) A mixture of ITmHg. If the partial pres- mmHeg, what is the partial pres- QT pressure of water at 285C 三 Sure of helium is 368 erature. (1) Whiceh container has the smallest mole Shi 人

大一普化5.50求解

3.43 AN 失 交 (方 N 全 只 憲 5.49 5,@ Re Problems 5 0t ln the stratosphere itadiation from the sun. and presSsure of ozone in Ths absSQTt Typieal the Stratosi Kandl0人10 am re spectively. How many ozone molecules are present in ]l.OL ofair under these conditions? Assuming that air contalns 78 percent N2, 21 pereemt O>, and 1 percent Ar, all by volume, how many mol- ecules of each type of gas are presentin 1.0Lotair 肚STP? A 2.10-L vessel contains 4.65 g ofa gas at 1.00 atm and 27.0。C. (a) Calculate the density of the gas In grams perliter. (b) What is the molar mass of the gas。 Calculate the density of hydrogen bromide (HBr) gas in grams per liter at 733 mmHg and 467C. A _ certain anesthetic contains 64.9 percent Co, 13.5 percent H, and 21.6 percent QO by masS. At 1205C and 750 mmHg, 1.00 L of the gaseous com- pound weighs 2.30 g. What is the molecular formula 9f 人he compound? Aecompound has the empirical formula SF4. At 20.0 0.100 g of the gaseous compound oceuplesa Volume of22.1 mL and exerts apresSure of 1.02 atm, What 1S its molecular formula? Dissolving 3.00 g of an impure Sample 0f calcium carbonate in hydrochloric acid produced 0.656L 人 carbon dioxide (measured at20.0C and 792 mmHg). Calculate the percent by mass of caleium carbonate in the sample. State any assumptilons, oleulafe the 和ass in orams Ofhydrogen chloride bro-

大一普化 5.36求解

5.38 9389 信4 5,42 sphere (about 30 Where the temperatu lbove the surface of Earth), >and presSure are 一235C : h gd 交 and 3.00 XX 10 atm, respectively. Calculate the final Volume of the balloon. The temperature ot 2.5 L of a gas initially at SUN increased to 2507”C at constant volume. Calculate the final pressure of the gas in atm. The pressure of 6.0 L of an ideal gas in aflexible contalner 1S decreased to one-third of its original presSure, and 1ts absolute temperature 1s decreased by one-half. What 1s the flnal volume of the gas? A gas evolved during the tfermentation of glucose (wine making) has a volume of 0.78 L when mea- sured at 20.1?C and 1.00 atm. What was the volume of this gas at the fermentation temperature of 36.5 C and 1.00 atm presSure? An ideal gas originally at 0.85 atm and 66”C was allowed to expand until its final Volume, preSSure, and temperature were 94 mL, 0.60 atm, and 45 C, respectively. What was its initial volume? The volume of a gas at STP is 488 mL. Calculate 1ts volume at 22.5 atm and 1507C. A gas at 772 mmHg and 35.0”C Qceuples a Volume of 6.85工. Calculate its volume at STEP. Dry ice is solid carbon dioxide. A 0.050-g sample ot dry ice js placed in an evacuated 4.6-L vessel at 30。C. Caleulate the pressure inside the VesSel after all the dry ice has been conVerted to CO gasS, Avolume of 0.280Lofagas at STP weighs 0.400 g. Calculate the molar mass Of the gas. Aquantity of gas Weighing 7.10g at741tor and 445C oeeupies a volume of 5.40L. What is its molar mass? Si人2 5.93 5.54 Dalt 和Ae0U76 5 5.56