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f.4 chemistry
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３４ Ａ正確活潑性高的鎂成功置換銀銀是灰色的 Ｂ不正確活潑性高的鎂雖成功置換鉛，但鉛是灰色的，不是棕色的 Ｃ不正確鎂和溶液沒有反應，而這置換反應不會生成氣體 Ｄ不正確活潑性高的鎂雖成功置換銅，但銅是棕色的，不是灰色的 所以答案是Ａ 35 1正確，容易與熱水反應表示活潑性較高，所以可以置換 ２正確，活潑性越高，代表越難分解 ３正確，可以與熱水反應，證明可以與反應性更高的水蒸氣反應 答案是Ｄ１，２，３都正確 ３６ 活潑性高的金屬才可以與水反應，所以Ｙ活潑性最高 活潑性越高提取方法越難，熱分解為較不活潑金屬氧化物的提取方法 Ｘ氧化物不能用熱分解提取，而Ｚ可以 證明Ｘ的活潑性高於Ｚ 所以Ｙ＞Ｘ＞Ｚ 答案是Ｃ ３７ １摩爾的Ｏ３有Ｙ個原子設Ｌ為亞佛加得羅常數 原子粒子數目＝摩爾數ｘ粒子數ｘＬ １ｘ３ｘＬ＝Ｙ－－－１ 設２摩爾的Ｏ２有Ｘ個原子 ２ｘ２ｘＬ＝Ｘ－－－２ 把２式除以１式 得出Ｘ＝Ｙ（４）／３ 所以答案是Ｃ ３８ Ｃａ的摩爾質量＝４０．１Ｓ的摩爾質量＝３２．１Ｏ的摩爾質量＝１６ Ｈ的摩爾質量＝１ ＣａＳＯ４．１／２Ｈ２Ｏ＋３／２Ｈ２Ｏ生成ＣａＳＯ４．２Ｈ２Ｏ 從方程ＣａＳＯ４．１／２Ｈ２Ｏ和３／２Ｈ２Ｏ１：３／２ ０．５００ｋｇｘ１０００＝５００ｇ ＣａＳＯ４．１／２Ｈ２Ｏ摩爾質量＝（４０．１＋３２．１＋１６Ｘ４＋０．５Ｘ１８）＝１４５．２ Ｈ２Ｏ的摩爾質量＝１８ ＣａＳＯ４．１／２Ｈ２Ｏ摩爾數＝５００／１４５．２＝３．４４ｍｏｌ 水的質量＝３．４４ｘ３／２ｘ１８＝９２．９ｇ 所以答案Ｂ應該是９２．９ｇ吧 所以答案是Ｂ ３９ 反應的方程式 Ａｇ＋２ＨＮＯ３生成ＡｇＮＯ３＋ＮＯ２＋Ｈ２Ｏ ＫＣｌ(aq)+ＡｇＮＯ３(aq)生成ＡｇＣｌ(s)+ＫＮＯ３(aq)把這方程寫成離子方程 Ｃｌ－＋Ａｇ＋生成ＡｇＣｌ 銀摩爾質量：１０７．９氯的摩爾質量＝３５．５ ＡｇＣｌ：ＡｇＮＯ３摩爾比１：１ ＡｇＮＯ３：Ａｇ摩爾比１：１ 所以ＡｇＣｌ摩爾數＝Ａｇ摩爾數 ＡｇＣｌ摩爾數＝１４．４／１４３．４＝０．１００ｍｏｌ 銀的質量＝０．１ｘ１０７．９＝１０．８ｇ 銀質量百分比＝１０．８／１２ｘ１００％＝９０．３％ 所以答案是Ｄ ４０ 反應的方程式 Ｎａ２ＣＯ３．ＸＨ２Ｏ生成Ｎａ２ＣＯ３＋ＸＨ２Ｏ Ｎａ摩爾質量＝２３Ｃ的摩爾質量＝１２Ｏ的摩爾質量＝１６ Ｈ的摩爾質量＝２ 方程中Ｎａ２ＣＯ３．ＸＨ２Ｏ和Ｎａ２ＣＯ３的摩爾比１：１ Ｎａ２ＣＯ３．ＸＨ２Ｏ的摩爾數＝Ｎａ２ＣＯ３的摩爾數 66.9/23x2+12+16x3+18x=24.8/23x2+12+16x3 66.9/106+18x=24.8/106 66.9=0.234(106+18x) 66.9=24.8+4.21x 42.1=4.21x x=10 答案是Ｄ 2010-02-08 18:13:35 補充： 多謝幫手＝．．＝ 本人是讀中文的 還以為哪是解氫氣 2010-02-08 18:18:03 補充： 修正 ３５的答案是Ａ只有１，２正確，雖然它們和水蒸氣有反應 可是只會形成氧化物，而不會形成氫氧化物
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35 . 1 X displaces Y from the solution of nitrate of Y 2 It is more difficult to reduce the oxide of X when compared to the oxide of Y 3 Both X and Y react with steam to form hydroxides 第3個 Metal react with steam 係form Metal Oxide 而唔係 Hydroxides 第3個錯 A 先真呀
f.4 chemistry
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34 In each of the four solution shown below, a strip of Mg is added.Tube I: silver nitrate solution + MgTube II: lead(ll) nitrate solution + MgTube III: magnesium sulphate solution + MgTube IV: copper(II) sulphate solution + MgWhich of the following combinations is correct? Tube ... 顯示更多 34 In each of the four solution shown below, a strip of Mg is added. Tube I: silver nitrate solution + Mg Tube II: lead(ll) nitrate solution + Mg Tube III: magnesium sulphate solution + Mg Tube IV: copper(II) sulphate solution + Mg Which of the following combinations is correct? Tube Observation A I Grey coating on Mg B II Brown coating on Mg C III Bubbles are give off D IV Grey coating on Mg 35 Metal X reacts with hot water more readily than metal Y does. Which of the following statements are probably correct? 1 X displaces Y from the solution of nitrate of Y 2 It is more difficult to reduce the oxide of X when compared to the oxide of Y 3 Both X and Y react with steam to form hydroxides A 1 2 B 1 3 C 2 3 D 1 2 3 Question 36 refer to the results of experiments for three metals X, Y and Z, and their compounds. Metal Reaction with cold water Action of heat on metal oxide X No reaction No reaction Y Hydrogen liberated No reaction Z No reaction Metal and oxygen formed 36 The descending order of reactivity of the three metals is A X,Y,Z B X,Z,Y C Y,X,Z D Y,Z,X 37 The formula of ozone is O3. If 1 mole of ozone contains y atoms, how many atoms will 2 moles of oxygen gas contain? A 2y/3 B y C 4y/3 D 4y 38 Plaster of Paris is used for setting broken limbs. Its formula is CaSO4。1/2H2O. When water is added, it sets to give gypsum, CaSO4。2H2O. What is the minimum mass of water needed to set 0.500kg of plaster of Paris? A 62.0kg B 93.0kg C 124g D 290g 39 A silver coin, with a mass of 12.0g, was dissolved completely in concentrated nitric acid to give silver nitrate solution. When excess potassium chloride solution was added to the resulting solution, 14.4g of a white ppt were obtained. What is the percentage by mass of silver in the coins? A 45.5% B 60.0% C 75.0% D 90.0% 更新: 40 The formula of hydrated sodium carbonate crystals is Na2CO3。XH2O. When 66.9g of the hydrated crystals are heated, 24.8g of anhydrous sodium catbonate are produced. What is the value ofx? A 3 B 5 C 7 D 10最佳解答:
３４ Ａ正確活潑性高的鎂成功置換銀銀是灰色的 Ｂ不正確活潑性高的鎂雖成功置換鉛，但鉛是灰色的，不是棕色的 Ｃ不正確鎂和溶液沒有反應，而這置換反應不會生成氣體 Ｄ不正確活潑性高的鎂雖成功置換銅，但銅是棕色的，不是灰色的 所以答案是Ａ 35 1正確，容易與熱水反應表示活潑性較高，所以可以置換 ２正確，活潑性越高，代表越難分解 ３正確，可以與熱水反應，證明可以與反應性更高的水蒸氣反應 答案是Ｄ１，２，３都正確 ３６ 活潑性高的金屬才可以與水反應，所以Ｙ活潑性最高 活潑性越高提取方法越難，熱分解為較不活潑金屬氧化物的提取方法 Ｘ氧化物不能用熱分解提取，而Ｚ可以 證明Ｘ的活潑性高於Ｚ 所以Ｙ＞Ｘ＞Ｚ 答案是Ｃ ３７ １摩爾的Ｏ３有Ｙ個原子設Ｌ為亞佛加得羅常數 原子粒子數目＝摩爾數ｘ粒子數ｘＬ １ｘ３ｘＬ＝Ｙ－－－１ 設２摩爾的Ｏ２有Ｘ個原子 ２ｘ２ｘＬ＝Ｘ－－－２ 把２式除以１式 得出Ｘ＝Ｙ（４）／３ 所以答案是Ｃ ３８ Ｃａ的摩爾質量＝４０．１Ｓ的摩爾質量＝３２．１Ｏ的摩爾質量＝１６ Ｈ的摩爾質量＝１ ＣａＳＯ４．１／２Ｈ２Ｏ＋３／２Ｈ２Ｏ生成ＣａＳＯ４．２Ｈ２Ｏ 從方程ＣａＳＯ４．１／２Ｈ２Ｏ和３／２Ｈ２Ｏ１：３／２ ０．５００ｋｇｘ１０００＝５００ｇ ＣａＳＯ４．１／２Ｈ２Ｏ摩爾質量＝（４０．１＋３２．１＋１６Ｘ４＋０．５Ｘ１８）＝１４５．２ Ｈ２Ｏ的摩爾質量＝１８ ＣａＳＯ４．１／２Ｈ２Ｏ摩爾數＝５００／１４５．２＝３．４４ｍｏｌ 水的質量＝３．４４ｘ３／２ｘ１８＝９２．９ｇ 所以答案Ｂ應該是９２．９ｇ吧 所以答案是Ｂ ３９ 反應的方程式 Ａｇ＋２ＨＮＯ３生成ＡｇＮＯ３＋ＮＯ２＋Ｈ２Ｏ ＫＣｌ(aq)+ＡｇＮＯ３(aq)生成ＡｇＣｌ(s)+ＫＮＯ３(aq)把這方程寫成離子方程 Ｃｌ－＋Ａｇ＋生成ＡｇＣｌ 銀摩爾質量：１０７．９氯的摩爾質量＝３５．５ ＡｇＣｌ：ＡｇＮＯ３摩爾比１：１ ＡｇＮＯ３：Ａｇ摩爾比１：１ 所以ＡｇＣｌ摩爾數＝Ａｇ摩爾數 ＡｇＣｌ摩爾數＝１４．４／１４３．４＝０．１００ｍｏｌ 銀的質量＝０．１ｘ１０７．９＝１０．８ｇ 銀質量百分比＝１０．８／１２ｘ１００％＝９０．３％ 所以答案是Ｄ ４０ 反應的方程式 Ｎａ２ＣＯ３．ＸＨ２Ｏ生成Ｎａ２ＣＯ３＋ＸＨ２Ｏ Ｎａ摩爾質量＝２３Ｃ的摩爾質量＝１２Ｏ的摩爾質量＝１６ Ｈ的摩爾質量＝２ 方程中Ｎａ２ＣＯ３．ＸＨ２Ｏ和Ｎａ２ＣＯ３的摩爾比１：１ Ｎａ２ＣＯ３．ＸＨ２Ｏ的摩爾數＝Ｎａ２ＣＯ３的摩爾數 66.9/23x2+12+16x3+18x=24.8/23x2+12+16x3 66.9/106+18x=24.8/106 66.9=0.234(106+18x) 66.9=24.8+4.21x 42.1=4.21x x=10 答案是Ｄ 2010-02-08 18:13:35 補充： 多謝幫手＝．．＝ 本人是讀中文的 還以為哪是解氫氣 2010-02-08 18:18:03 補充： 修正 ３５的答案是Ａ只有１，２正確，雖然它們和水蒸氣有反應 可是只會形成氧化物，而不會形成氫氧化物
其他解答:
35 . 1 X displaces Y from the solution of nitrate of Y 2 It is more difficult to reduce the oxide of X when compared to the oxide of Y 3 Both X and Y react with steam to form hydroxides 第3個 Metal react with steam 係form Metal Oxide 而唔係 Hydroxides 第3個錯 A 先真呀
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