(1)若A為Na2CO3.C為溫室氣體.E為一種常有調(diào)味品.A固體與少量B溶液反應(yīng)的化學(xué)方程式為 .(2)若A為MnO2.C為黃綠色有毒氣體.則C與澄清石灰水反應(yīng)的離子方程式為 : 【查看更多】

 

題目列表(包括答案和解析)

（Xinhu）The Expo 2010 Shanghai China formally opened its door to the highly avid public this morning.

Top Chinese political advisor Jia Qinglin and International Exhibitions Bureau President Jean-Pierre Lafon activated（有活性的） the opening device together at a commencement（開始；開端） ceremony.

The Expo, carrying a theme of "Better City, Better Life", reflects the crystallization （結(jié)晶化）of wisdom about urban（城市的）construction and vision of a better future life, Jia, chairman of the National Committee of the Chinese People's Political Consultative Conference, said while addressing the ceremony.

BIE Secretary Teneral Vicente Gonzalea Loscertales said the Shanghai World Expo would be "the most splendid and unforgettable" and would help the world better understand the present, have a glimpse（一瞥；一看）into the future and strengthen cooperation between countries and organizations.

Visitors, from home and abroad, are thronging（成群；擠滿） to the gates of the Expo site, waiting for security checks in long queues. Around 300,000 tickets have been sold or distributed（分發(fā)） for the opening day, organizers said.

The Expo is expected to attract 70 million visitors from China and abroad. The Expo site covers an area of 5.28 square kilometers along both sides of the Huangpu River, a tributary（支流）of the Yangtze River.  

51. What is the theme of the EXPO 2010 Shanghai China?

A. Better City, Better Future.

B. Better City, Better World.

C. Better City, Better Life.

D. Better City, Better Future.

52. What does the underlined word “avid”in paragraph 1 probably mean?

A. active.           B. brave.         C. energetic.       D. eager.

53. How many visitors is the Expo expected o attract from home and abroad?

A. 7000000.          B. 70000000.      C. 70000.           D. 700000.    

54. What is the best title for this passage?

A. Expo site opens to public.

B. Expo site covers an area of 5.28 square kilometers.

C. The Expo is wonderful.

D. The Expo is unforgettable.

55. According to the passage which of the following statements is NOT true?

A. The Expo reflects the crystallization of wisdom about urban construction

and vision of a better future life.

B. The Shanghai World Expo would be "the least splendid and unforgettable".

C. Expo site covers an area of 5.28 square kilometers.

D. Visitors, from home and abroad, are thronging to the gates.

 

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At present, in many American cities especially, many teachers in the public schools say they are underpaid. They point to jobs such as secretary or truck driver, which often pay more to start than that of a teacher. In many other fields, such as law, medicine, computer science, a beginning worker may make more than a teacher who has taught for several years.

Teaching has never been a profession that attracted people interested in high salaries. It is by history a profession that has provided rewards in addition to money—the satisfaction of sharing knowledge, of influencing others, of guiding young people. But in the past several years, there are more difficulties in teaching, for many, than there are rewards.

Unruly students, especially in big cities, large classes and a lack of support from the public in terms of money and understanding have led many public school teachers to leave the profession.

As a result, many of the best students, who would have chosen teaching as their life career in the past, are going into other fields.

Another reason for this change in teacher candidates is the changing status of women in the United States. Until the late 1960s and 1970s, one of the most popular choices for women was teaching. But as other professions, such as law and medicine opened up to women, women stopped pouring into teacher training programs. Thus, a major pool of excellent candidates for the teaching profession dwindled.

Bit by bit government officials and others realized that the status of the teacher had suffered. They talked about change. But the change in a vast society like the United States is not easy. People’s attitudes have formed over many years, and sometimes change takes many years.

The underlined word “that” in Paragraph 1 refers to “      ”.

A. money       B. job     C. secretary    D. truck driver

What is the present situation of the teaching?

A. Teachers work harder and get underpaid.

B. Teachers have no opportunities to work in other fields.

C. Teaching can attract best students to work as a teacher.

D. Teaching can provide rewards as well as high salaries.

Many public school teachers turn to other professions because        .

A. the unruly students and large classes

B. the difficulties of teaching as well as a lack of money, support and understanding

C. the attracting power of other jobs

D. a lack of satisfaction of sharing knowledge and influencing others

The author believes that change in teachers’ status in the United States       .

A. is not great        B. impossible

C. influences people’s attitude       D. needs time

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從A, B, C, D四個(gè)選項(xiàng)中，找出其劃線部分與所給單詞的劃線部分讀音相的選項(xiàng)，

例：have       A. gave   B. save     C. hat   D. made       答案是C。

1． wrist       A. sword       B. onwards      C. otherwise      D. runaway

2.  sympathy   A. worthy       B. thus       C. youths          D. breakthrough

3.  chorus     A. achieve       B. march       C. butcher       D character         

4.  gentle      A. gallon      B. genius         C. guest        D. gramme

5.  widespread  A. steady       B. leap         C. feast         D. league

 

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My favorite English teacher could draw humor out of the driest material. It wasn’t forced on us either. He took Samuel Johnson’s dictionary, Addison’s essays, and many other literary wonders from the eighteenth century and made them hilarious, even at eight o’clock in the morning. The thing that amazed me most was that the first time I read these works on my own, some of them seemed dead, but the second time, after his explanation, I couldn’t believe that I hadn’t seen the humor. The stories and poems and plays were suddenly filled with allusions(典故) and irony and hilarious moments. I learned more from him than from any other teacher.

My least favorite English teacher also made people laugh. Some students found him to be wonderfully funny. Many others did not. He assigned journals over a six-week period, to be written every day. At the end of the six weeks I had a notebook full of bits and pieces about my ideas, short stories, reactions to what we had read, and so on. Our teacher announced that we would be grading each other’s journals. Mine was passed to Joe, that class clown, who always behaved in a funny or silly way. He saw it fit to make a joke of and said, “This writing isn’t fit to line the bottom of a birdcage.” Our teacher laughed at that funny remark. It hurt me so much that the anger from it has driven my writing and teaching ever since.

So what makes the difference? Humor is one of the most powerful tools teachers or writers have. It can build up students and classes and make them excited about literature and writing, or it can tear them apart. It is true that humor is either productive or counter-productive and self-defeating.

1.The passage mainly discusses ________.

A. teaching                     B. literature                    C. humor                D. knowledge

2.The underlined word “hilarious” in Paragraph 1 probably means ________.

A. funny                B. tiring                          C. inspiring                       D. brilliant

3.The English teacher the writer disliked most ________.

A. was not able to make students laugh                   B. hurt his student’s feelings

C. didn’t let his students do the grading                  D. had no sense of humor

 

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閱讀理解（共20題； 每小題2分， 滿分40分）

閱讀下列短文，從每題所給的A、B、C 和D項(xiàng)中，選出最佳選項(xiàng)，并在答題卡上將該項(xiàng)涂黑。

They like using the Internet．They have lots of pocket money to spend．And they spend a higher proportion of it online than the rest of us．Teenagers are just the sort of people an online seller is interested in, and the things they want to buy－games, CDs and clothing－are easily sold on the Web.

But paying online is a tricky business for consumers who are too young to own credit cards．Most have to use a parent’s card．They want a facility that allows them to spend money.

That may come sooner than they think: new ways to take pocket money into cyber (網(wǎng)絡(luò)的) space are coming out rapidly on both sides of the Atlantic．If successful, these products can stimulate online sales.

In general, teenagers spend huge amounts: $153bn (billion) in the US last year and ￡20bn annually in the UK．Most teenagers have access to the Internet at home or at school－88 percent in the US, 69 percent in the UK．According to the Jupiter Research, one in eight of those with Internet access has bought something online－mainly CDs and books.

In most cases, parents pay for these purchases with credit cards, an arrangement that is often unsatisfactory for them and their children．Pressing parents to spend online is less productive than pressing on the high street．They are more likely to ask “Why?” if you ask to spend some money online.

One way to help teenagers change notes and coins into cybercash is through prepaid cards such as InternetCash in the US and Smart cards in the UK．Similar to those for pay-as-you-go mobile telephones, they are sold in amounts such as ￡20 or $50 with a concealed 14-digit number that can be used to load the cash into an online account.

1.What does the word “They” in paragraph 1 refer to?

    A. Sellers.     B. Buyers.      C. Teenagers.      D. Parents.

2. According to the passage, which of the following statements is TRUE?

    A. More than half of the teenagers in the US and the UK have Internet access.

    B. Teenagers pay for goods online with their own credit cards.

    C. Most teenagers in the US and the UK have bought something online.

    D. Teenagers found it easier to persuade parents to buy online than in a shop.

3. A new way to help teenagers shop online is to use _________.

    A. a new machine                  B. special coins and notes

    C. prepaid cards                      D. pay-as-you-go mobile phones

4.What is the passage mainly about?

    A. Online shopping traps.             B. Internet users in the US and the UK.

    C. New credit cards for parents.       D. The arrival of cyber pocket money.

5. Which of the following words can best describe the writer’s attitude towards the phenomenon?

    A. Proud and satisfied.            B. Worried and anxious.

    C. Objective and informative.          D. Concerned and sad.

 

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1．D【解析】β－珠蛋白DNA探針、RNA聚合酶結(jié)合位點(diǎn)、大腸桿菌質(zhì)粒的化學(xué)本質(zhì)都是DNA；胰島素是蛋白質(zhì)；HIV的遺傳物質(zhì)是RNA；生長素是吲哚乙酸；2，4－D是一種生長素類似物。

2．D【解析】作物“燒心”是由于缺乏如鐵、鈣等在細(xì)胞中以穩(wěn)定化合物形式存在的礦質(zhì)元素，有別于因缺水而造成的“燒苗”。

3．B【解析】疫苗的作用是在機(jī)體不患病的情況下發(fā)生免疫反應(yīng)，產(chǎn)生的抗體與抗原結(jié)合，發(fā)揮免疫效應(yīng)；流感病毒的遺傳物質(zhì)是單鏈的RNA，其結(jié)構(gòu)穩(wěn)定性不如天花病毒的雙鏈DNA，容易發(fā)生變異，所以流感疫苗的研究的難度較天花疫苗研究的難度大很多；正是由于流感病毒極易發(fā)生變異，所以某種特定的單克隆抗體不一定對其它的抗原起作用。

4．C【解析】生態(tài)系統(tǒng)的成分除生產(chǎn)者、消費(fèi)者和分解者外，還包括了非生物的物質(zhì)和能量；生物圈的自給自足表現(xiàn)在物質(zhì)上，能量來源于太陽能；草原上的牛和羊同屬于第二營養(yǎng)級，共獲得生產(chǎn)者固定太陽能的10％～20％。

5．B【解析】考查分泌蛋白的形成和分泌過程這一知識點(diǎn)和圖形分析能力。蛋白質(zhì)分泌以細(xì)胞膜的外排方式實(shí)現(xiàn)，所經(jīng)過的膜結(jié)構(gòu)順序?yàn)椋簝?nèi)質(zhì)網(wǎng)→高爾基體→細(xì)胞膜，所以結(jié)果是：內(nèi)質(zhì)網(wǎng)面積減小，高爾基體膜面積不變，細(xì)胞膜面積增加。

6．B　【解析】有機(jī)分子的特點(diǎn)一般滿足C四鍵、H一鍵、N三鍵原則，三聚氰胺分子中含有3個(gè)－NH2，則其余3個(gè)C、3個(gè)N形成一個(gè)六元環(huán)，故三聚氰胺的結(jié)構(gòu)簡式為。。根據(jù)三聚氰胺的結(jié)構(gòu)簡式，該物質(zhì)不是高聚物，也不是氨基酸，但分子中含有不飽和鍵，在一定條件下能發(fā)生加成反應(yīng)。六元環(huán)不變且除自身外的三聚氰胺的異構(gòu)體有3種。

7．B　  【解析】選項(xiàng)A，Na2O2中含有的陽離子為Na＋，陰離子為O22?,0.1mol  Na218O2中含有的陰陽離子總數(shù)為0.3NA。選項(xiàng)B，C2H4、C3H6的化學(xué)式均為CH2，1.4g二者含有的C原子數(shù)的物質(zhì)的量＝1.4g/14g?mol－1＝0.1mol，即0.1NA。選項(xiàng)C，0.1mol Cl2參與反應(yīng)電子轉(zhuǎn)移0.1mol。選項(xiàng)D，D2O為10電子微粒，18g D2O物質(zhì)的量小于1mol，含有的電子數(shù)也小于NA。

8．A　【解析】選項(xiàng)A，ClO－具有氧化性，SO2具有還原性，SO2被氧化為SO42?，ClO－被還原為Cl?，正確。選項(xiàng)B，加入少許H＋時(shí)，CO32?優(yōu)先與H＋結(jié)合生成HCO3?。選項(xiàng)C，離子方程式兩邊電荷不守恒。選項(xiàng)D，Mg2＋能與電解產(chǎn)生的OH?結(jié)合生成難溶性Mg(OH)2沉淀：Mg2＋ + 2Cl? + 2H2O＝Mg(OH)2↓+ H2↑+ Cl2↑。

9. B 【解析】加入稀H2SO4出現(xiàn)白色混濁說明一定存在Ba2＋，因SO32?、CO32?能與Ba2＋結(jié)合生成難溶物BaSO3、BaCO3，所以原溶液中不存在SO32?、CO32?。由于溶液呈電中性，剩下的一種HCO3?，一定存在。無法判斷是否含有K＋，所以原溶液中一定含有Ba2＋、HCO3?，可能含有K＋。溶質(zhì)可能是Ba(HCO3)2或Ba(HCO3)2和KHCO3。

10．D 【解析】根據(jù)題設(shè)條件可推知A為NH4＋，B為OH?，C為NH3，D為H2O。NH3能與H2O反應(yīng)生成NH3?H2O。NH4＋為離子，不是分子，所以NH4＋不是非極性分子。選項(xiàng)D，固態(tài)H2O分子間存在氫鍵，其熔沸點(diǎn)高于固態(tài)H2S，與H－O和H－S鍵強(qiáng)弱無關(guān)。NH4Cl ，NH4＋水解溶液的pH<7。

11．C【解析】分析反應(yīng)①②中各元素的價(jià)態(tài)變化可知，反應(yīng)①中，SO2為還原劑，F(xiàn)e3＋為氧化劑，且還原性：SO2>Fe2＋，氧化性：Fe3＋>SO42?。反應(yīng)②中，F(xiàn)e2＋為還原劑，Cr2O72?為氧化劑，且還原性：Fe2＋>Cr3+ ，氧化性：Cr2O72? > Fe3＋。由此可見選項(xiàng)A、B錯(cuò)誤。選項(xiàng)C，由于Cr2O72?具有氧化性，Na2SO3具有還原性，故Cr2O72? 能將Na2SO3氧化成Na2SO4。選項(xiàng)D，反應(yīng)①中Fe2(SO4)3為氧化劑，反應(yīng)②中Fe2(SO4)3為氧化產(chǎn)物。

12．C【解析】選項(xiàng)A，NaHS、Na2S溶液又因?yàn)橛蠬S?、S2?的水解，使其溶液呈堿性，但S2?的水解能力大于HS?，故溶液的pH值：③>②。H2S溶液呈酸性，H2S和NaHS混合液中，由于HS?抑制的H2S的電離，故溶液的pH：④>①，4種溶液pH大小順序?yàn)椋孩?gt;②>④>①。選項(xiàng)B，由于HS?抑制H2S的電離，所以H2S溶液中的c(H2S)小于H2S和Na2S混合液中的c(H2S)。選項(xiàng)C，c(Na＋)＝0.1mol?L-1，根據(jù)物料守恒有：c(H2S) + c(HS?) + c(S2?)＝0.2mol?L-1，故2c(Na＋)＝c(H2S) + c(HS?) + c(S2?)。選項(xiàng)D，根據(jù)電荷守恒有：c(H＋) + c(Na＋)＝c(OH?) + c(HS?) +2 c(S2?)。

13．B　【解析】設(shè)達(dá)平衡時(shí)生成SO3(g)物質(zhì)的量為2x ，則剩余SO2(g)的物質(zhì)的量為(3－2x)，O2(g)(2－x)，混合氣體總的物質(zhì)的量為（5－x），根據(jù)阿伏伽德羅定律有5/（5－x）＝1/0.9，解得x＝0.5mol，再結(jié)合熱化學(xué)方程式可知，放出的熱量為196.6kJ/2＝98.3kJ。選項(xiàng)B，起始物質(zhì)的量改為 4mol SO2 、 3 mol O2 、2SO3 (g) 相當(dāng)于加入6mol SO2、4mol O2，n(SO2)/n(O2)＝3/2，故與第一次平衡是等效平衡，兩次平衡中SO2的轉(zhuǎn)化率、SO3的體積分?jǐn)?shù)相等，故選項(xiàng)B正確，C錯(cuò)誤。選項(xiàng)D，題目沒有告訴達(dá)平衡時(shí)的時(shí)間，無法計(jì)算反應(yīng)速率。

14．AD【解析】由狀態(tài)方程知溫度升高而壓強(qiáng)增大體積必增大，故狀態(tài)I時(shí)氣體的密度比狀態(tài)II時(shí)氣體的密度大，A正確，平衡態(tài)II的溫度比狀態(tài)I高，故狀態(tài)I時(shí)分子的平均動(dòng)能比狀態(tài)II時(shí)分子的平均動(dòng)能小，B錯(cuò)誤，由熱力學(xué)第一定律知從狀態(tài)I到狀態(tài)II過程中溫度升高內(nèi)能變大，體積增大對外界做功，氣體要從外界吸熱，故C錯(cuò)，D正確，故選AD。

15.答案：BD【解析】：在同一介質(zhì)中紅光傳播速度最大，從AB面射入到BC面射出，紅光用的時(shí)間最短，故選項(xiàng)A錯(cuò)．由于玻璃對紅光折射率最小，對紫光的折射率最大，即紫光的偏折本領(lǐng)最大，所以彩色光帶右邊緣的色光為紅光，左邊緣的色光為紫光，且紫光的頻率比紅光的要高，當(dāng)紅光能讓某金屬板發(fā)生光電效應(yīng)，紫光也一定能夠，故選項(xiàng)B正確．在同樣條件下做雙縫干涉實(shí)驗(yàn)，波長越長，相鄰干涉條紋間距越大，而彩色光帶左邊緣的色光是紫光，其波長最短，故選項(xiàng)C錯(cuò)．對玻璃而言，在七色光中，紅光的臨界角最大，當(dāng)∠MNB逐漸變小時(shí)，射到AC面上的光的入射角變小，且紅光入射角小得更多，故紅光最先從從AC面透出，所以選項(xiàng)D正確．

16、答案 D 【解析】 燒斷細(xì)線后，無論是彈簧將A+B和C彈開過程，還是A和B分離后，系統(tǒng)始終動(dòng)量守恒、機(jī)械能守恒；彈簧將A+B和C彈開過程，A+B和C動(dòng)量大小相等，動(dòng)能跟質(zhì)量成反比，因此A+B的總動(dòng)能是E/3，其中A的動(dòng)能是E/6；當(dāng)時(shí)C的動(dòng)能是2E/3；前3個(gè)選項(xiàng)都錯(cuò)，可判定D正確。證明：A和B分離時(shí)，B+C的總動(dòng)能是5E/6，B、C共速時(shí)彈性勢能最大，當(dāng)時(shí)A和B+C動(dòng)量大小相等，動(dòng)能跟質(zhì)量成反比，因此B+C的動(dòng)能是E/12，該過程B+C的動(dòng)能損失就是此時(shí)的彈性勢能，因此Ep=5E/6- E/12=3E/4。

17.答案：ACD【解析】：根據(jù)波的傳播方向，可以判斷b質(zhì)點(diǎn)此時(shí)刻振動(dòng)方向沿y軸負(fù)方向，離開平衡位置，速度正在變小，A對；由圖象可知該波的波長是4m，根據(jù)發(fā)生明顯衍射現(xiàn)象的條件判斷D正確；根據(jù)得，由波干涉條件知C正確；經(jīng)過0.01s即半個(gè)周期，任何質(zhì)點(diǎn)通過的路程都是2個(gè)振幅，即0.4m，B錯(cuò)誤.

18．C【解析】從圖(甲)到圖(乙)的過程中，根據(jù)動(dòng)能定理有：，所以；從拋出后到落地，根據(jù)動(dòng)能定理得：，代入上式可得：。

19.答案： BC　【解析】此模型為類雙星模型，兩電荷做圓周運(yùn)動(dòng)的角速度相等；兩個(gè)電荷之間的庫侖力充當(dāng)各自做圓周運(yùn)動(dòng)的向心力，所以向心力大小相等，A錯(cuò)，B對．由知，線速度大小與質(zhì)量成反比，運(yùn)動(dòng)半徑與質(zhì)量成反比，C對，D錯(cuò)．

20.答案：ABD 【解析】天然放射性元素的半衰期與溫度改變無關(guān)。根據(jù)質(zhì)能方程計(jì)算可知D答案正確。

21.答案：.D  【解析】地球同步衛(wèi)星是指與地球自轉(zhuǎn)同步的人造衛(wèi)星，它的周期是24小時(shí)，它的軌道平面只能在赤道，得軌道也是固定的，但并不是說同一赤道平面內(nèi)的、或是周期與地球自轉(zhuǎn)周期相等的就都是同步衛(wèi)星，故A、C是錯(cuò)的、D是正確的；同步衛(wèi)星做圓周運(yùn)動(dòng)時(shí)，內(nèi)部的儀器是處于失重狀態(tài)而不是超重狀態(tài)，B錯(cuò)

22.答案．（1）(g＋a)× (OM－ON)＝(g－a)× OP    （4分）

【解析】利用紙帶分析得m1帶動(dòng)m2的加速度為a ,又由牛頓第二定律得a＝解得

⑴3000；（2分）

⑵丙(1分)；乙圖中電流表的示數(shù)太小，誤差太大。丙圖中R的阻值與電壓表阻值接近，誤差小。（3分）。

⑶實(shí)物圖連接如右圖所示：(4分)

⑷實(shí)驗(yàn)步驟：

①閉合K1．再閉合K２，讀得電壓表示數(shù)Ｕ１；再斷開K２，讀得電壓表示數(shù)Ｕ２.（2分）�、冢�V＝。（2分）

23.【解析】：（1）負(fù)電……(2分)∵mg =E×……(5分)

∴E=４(r+R)dmg/Rq…………(2分)

（2）mg+q v0B=……………(5分)    ∴v0=mg/qB…………(2分)

24.【解析】：（1）ab通過最大電流時(shí)，受力分析如圖甲，此時(shí)靜摩擦力最大，，方向沿斜面向下，由平衡條件得：

水平：

（3分）

豎直：（3分）

以上兩式聯(lián)立得出

（3分）

（2）通以最小電流時(shí)，ab受力分析如圖乙，此時(shí)ab受靜摩擦力，方向沿斜面向上，與（1）類似，由平衡條件得：（3分）

（3）當(dāng)ab中電流最小時(shí)，變阻器阻值為：（3分）

當(dāng)ab中電流最強(qiáng)時(shí)，變阻器阻值為：，（2分 ）

為保持ab靜止，R的調(diào)節(jié)范圍為0.91~10．（1分）

25.【解析】：（1）設(shè)A物塊碰撞B物塊前后的速度分別為v1和v2，碰撞過程中動(dòng)量守恒，

  代入數(shù)據(jù)得：        （4分）

（2）設(shè)A、B兩物塊碰撞前后兩物塊組成的系統(tǒng)的機(jī)械能分別為E1和E2，機(jī)械能的損失為，根據(jù)能的轉(zhuǎn)化和守恒定律：   

     %     （4分）

（3）設(shè)物塊A的初速度為v0，輪胎與冰面的動(dòng)摩擦因數(shù)為µ，A物塊與B物塊碰撞前，根據(jù)動(dòng)能定理：                 （3分）

碰后兩物塊共同滑動(dòng)過程中根據(jù)動(dòng)能定理：

         （3分）

由、  及（1）、（2）得：    （2分）

設(shè)在冰面上A物塊距離B物塊為L′時(shí)，A物塊與B物塊不相撞，

則：                             （4分）

26．（15分）（1）（1）KNO3 （2分）CuSO4（2分）

(2) Na2CO3  （2分）

（3）Al3＋ + 3OH?＝Al(OH)3↓（3分） Al(OH)3 + OH?＝AlO2? + 2H2O （3分）

（4） Al3＋ + 3H2O Al(OH)3（膠體）+ 3H＋ （3分）

【解析】根據(jù)實(shí)驗(yàn)①可知，D中含有Cu2＋；根據(jù)實(shí)驗(yàn)②可知C中含有Al3＋，E可能是KOH或NaOH，再根據(jù)③，只有B、C中含有K＋，故E為NaOH。根據(jù)實(shí)驗(yàn)③A中含有HCO3?，故A為NaHCO3，C、D中含有SO42?，故D為CuSO4，C為KAl(SO4)2。

最后可判定B為KNO3。等物質(zhì)的量的NaHCO3與NaOH反應(yīng)生成Na2CO3和H2O。 NaOH溶液加入到KAl(SO4)2溶液中，首先是Al3＋與OH?反應(yīng)生成Al(OH)3，Al(OH)3沉淀又溶解在過量的NaOH溶液中：Al3＋ + 3OH?＝Al(OH)3↓，Al(OH)3 + OH?＝AlO2? + 2H2O。KAl(SO4)2中的Al3＋水解生成具有吸附作用的Al(OH)3膠體而凈水。

27．（14分）（1）Na2CO3 ＋HCl＝NaHCO3 ＋ NaCl    （3分）

（2）Cl2 + 2OH?＝Cl? + ClO? + H2O （3分）

（3）①Na2O或Na2O2  （4分，每空各2分）②Na或NaOH （4分，每空各2分）

【解析】（1）根據(jù)題設(shè)條件可知，B為鹽酸，C為CO2，D為H2O，E為NaCl。

（2）根據(jù)題設(shè)條件可知B為濃鹽酸，C為Cl2。

（3）若C為O2，D、E的焰色反應(yīng)均為黃色，D、E中含有Na＋，含有Na＋的能產(chǎn)生O2的固體為Na2O2，E能與鹽酸反應(yīng)生成的氣體能使澄清石灰水變混濁，該氣體為CO2，B、E可相互轉(zhuǎn)化，故可推知B溶液為NaHCO3溶液，E為Na2CO3溶液，D為NaOH溶液。Na2O2與NaHCO3溶液反應(yīng)可分解為：2Na2O2 + 2H2O＝4NaOH + O2↑，NaHCO3 + NaOH＝Na2CO3 + H2O。amol NaHCO3→a mol Na2CO3，只要增加a mol Na＋，同時(shí)用OH? 將HCO3?轉(zhuǎn)化為CO32? ，所加物質(zhì)所產(chǎn)生的Na＋和OH?的物質(zhì)的量相等，才能不產(chǎn)生雜質(zhì)，故X為Na2O或Na2O2，Y為Na或NaOH。

28．（16分）（1）B（2分）  銅與HNO3反應(yīng)前，應(yīng)排凈裝置內(nèi)的空氣，防止NO與O2發(fā)生反應(yīng) （2分）

（2）將C中的溶液加適量水稀釋（2分）

（3）③④⑤⑨ （3分）

（4）第二，打開a，通足量N2，排凈裝置中的空氣（2分）

（5）向下移動(dòng)乙管，使甲、乙兩管液面在同一水平面上（2分）
（6）(V－11.2n)/33.6n （3分）

【解析】根據(jù)實(shí)驗(yàn)?zāi)康�，要求得m值，需測定出Cu與HNO3反應(yīng)生成的NO2和NO的物質(zhì)的量，為此需將產(chǎn)生的氣體首先通入裝置C中，H2O吸收NO2生成NO和HNO3，用裝置E測定出NO的體積，如果裝置中有空氣，空氣中的O2會將NO氧化成NO2導(dǎo)致實(shí)驗(yàn)誤差，裝置B通入N2能將裝置中空氣趕走，防止NO被氧化，為此需要的裝置為B、C、E，裝置接口連接順序?yàn)棰邰堍茛帷Ｓ捎跐釮NO3具有強(qiáng)氧化性，能將指示劑氧化而影響實(shí)驗(yàn)，可加水稀釋降低其氧化性，而溶質(zhì)HNO3的量不變，便于觀察指示劑顏色變化。用裝置E測定NO的體積時(shí)，如甲的液面高于乙的液面，測出的NO體積偏小，如甲的液面低于乙的液面，測出的NO體積偏大，故應(yīng)向下移動(dòng)乙管，使甲、乙兩管液面在同一水平面上，從而減少誤差。根據(jù)反應(yīng)：3NO2 + 2H2O＝2HNO3 + NO，混合氣體中含有NO21.5nmol，NO總的物質(zhì)的量為V/22.4mol，其中屬于Cu與HNO3反應(yīng)生成的NO為（V/22.4－n/2）mol，故M＝（V/22.4－n/2）：1.5n＝(V－11.2n)/33.6n。

29．（15分）（1）取代（或水解）、中和反應(yīng) （2分） （2）HCOOCH3  （2分） HOCH2CHO （2分）

（3）（2分）

  （4）+ 3NaOH+ CH3COONa + 2H2O （3分）

（5） （2分） （2分）

 

【解析】A的分子式為C9H8O4，A能與醇發(fā)生酯化反應(yīng)，說明A中含有－COOH，且A在NaOH溶液中發(fā)生水解生成CH3COONa，說明A中酯的官能團(tuán)，該官能團(tuán)與－COOH處于苯環(huán)上的鄰位，再結(jié)合A的分子式推知A的結(jié)構(gòu)簡式為，B為，由于H2CO3的酸性大于酚而小于羧酸，故在溶液中通入CO2時(shí)，只有酚的鈉鹽反應(yīng)生成D（）和NaHCO3。CH3COONa與H＋反應(yīng)生成E（CH3COOH）。CH3COOH的同分異構(gòu)體中R中含有－CHO和－OH：HOCH2CHO，Q中含有－CHO而沒有－OH：HCOOCH3。對照和結(jié)構(gòu)可知，首先用與酚羥基反應(yīng)，然后再用酸性KMnO4氧化－CH3為－COOH即可得到A。

30.【解析】（1）新陳代謝是生物最本質(zhì)的特征。（2）人體內(nèi)水的來源包括：飲水、食物中的水、代謝產(chǎn)生的水，人體代謝產(chǎn)生水的途徑有：核糖體上的氨基酸脫水縮合、線粒體中的有氧呼吸等。（3）異化作用類型包括需氧型、厭氧型和兼性厭氧型，根據(jù)材料提供信息，氣性壞疽的異化作用類型為厭氧型。（4）本題考查的是細(xì)胞的選擇透過性，細(xì)胞的功能特性決定于細(xì)胞膜上的載體的種類和數(shù)量

【答案】（12分，每空各2分）（1）新陳代謝現(xiàn)象  （2） 代謝產(chǎn)生水   

氨基酸脫水縮合（或有氧呼吸）      （3）  厭氧型

（4）選擇透過性    載體蛋白 

【解析】由反應(yīng)式：CO2+C5→C3和 2C3學(xué)科網(wǎng)(Zxxk.Com)C5+CH2O可知：一定范圍內(nèi)的CO2增加可以使細(xì)胞中C3增加；一定范圍內(nèi)的光照增強(qiáng)會導(dǎo)致細(xì)胞內(nèi)C5增加。甲圖信息顯示：只考慮光照的影響，光照強(qiáng)度為由n增加到k時(shí)，光合速率幾乎不變，而乙圖的變化應(yīng)是光照之外的其他因素引起，比如溫度等

【答案】（10分，每空各2分）（1）低、高、約等于   （2）溫度   

（3）光照強(qiáng)度、CO2濃度、溫度

31．（20分）【解析】：（1）該6個(gè)品系玉米的基因型分別為：①：AABBCCDDEE  ②：aaBBCCDDEE  ③：AAbbccDDEE  ④：AABBCCddEE  ⑤：AABBCCDDee  ⑥：aabbccddee

基因分離定律適用于一對等位基因控制的相對性狀的遺傳，基因自由組合定律適用于2對（及以上）的同源染色體上的2對（及以上）等位基因控制的性狀遺傳。具有兩對相對性狀的純合親本雜交，F(xiàn)1自交。若F2中出現(xiàn)性狀分離比為：雙顯∶單顯1∶單顯2∶雙隱＝9∶3∶3∶1，則控制這兩對相對性狀的基因位于2對同源染色體上，反之則位于同一染色體上

讓F1側(cè)交，若F2中出現(xiàn)性狀分離比為：雙顯∶單顯1∶單顯2∶雙隱＝1∶1∶1∶1，則控制這兩對相對性狀的基因位于2對同源染色體上，反之則位于同一染色體上

【答案】（1）②與①（③或、④、⑤） （1分）  不行  （1分）  品系①和⑤只有一對相對性狀   （2分）不行 （1分）  控制花色和種皮顏色的基因位于同一對同源染色體（Ⅰ）上，而控制子葉味道的基因位置未知（2分）

（2）D（1分）

①若綠色非甜子葉∶綠色甜子葉∶黃色非甜子葉∶黃色甜子葉＝9∶3∶3∶1，則控制子葉顏色和味道的基因不是位于同一染色體上。（3分）

②若綠色非甜子葉∶綠色甜子葉∶黃色非甜子葉∶黃色甜子葉≠9∶3∶3∶1（答綠色甜子葉：綠色非甜子葉：黃色非甜子葉＝1：2：1也可），則控制子葉顏色和味道的基因位于同一染色體上。（3分）

（3）

①若綠色非甜子葉∶綠色甜子葉∶黃色非甜子葉∶黃色甜子葉＝1∶1∶1∶1，則控制子葉顏色和味道的基因不是位于同一染色體上。（3分）

②若綠色非甜子葉∶綠色甜子葉∶黃色非甜子葉∶黃色甜子葉≠1∶1∶1∶1（答綠色甜子葉：黃色非甜子葉＝1：1也可），則控制子葉顏色和味道的基因位于同一染色體上。（3分）

 

 

 


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