(3)若C為O2.D.E的焰色反應(yīng)均顯黃色.B.E可相互轉(zhuǎn)化.E與鹽酸反應(yīng)生成的氣體能使澄清石灰水變渾濁.現(xiàn)有只含B的溶液.欲使其變成只含E的溶液.可采取下列方法.試寫出有關(guān)物質(zhì)的化學(xué)式. 【查看更多】

 

題目列表(包括答案和解析)

One morning Mrs Smith was driving home after she had done shopping. When she drove near a rubbish dump, she noticed a microwave oven(微波爐) not far from the side of the road.“John is a good electrician!”she said to herself.“Perhaps he can repair this. I'll take it home and let him try.”She picked up the oven and put it in the boot of her car. Then she drove on happily. A few kilometers later, she heard the siren (警報(bào)器)of a police car behind her. She looked in the driving mirror and saw a policeman waving to her to tell her to pull over and stop.

　Mrs Smith was very puzzled. She slowed down at the side of the road. A traffic policeman got out of the police car and walked up to her.　

“Can I see your driving license and insurance certificate(保險(xiǎn)證)，please? ”he asked her. He copied down details of her name, address and the number of the car.“What's wrong, officer?” Mrs Smith asked. The policeman did not reply. He looked in the car and then at the back.      “Open the boot, please.”he said to Mrs Smith.

　Mrs Smith was still puzzled. She opened the boot and pointed to the microwave oven. "I found this old microwave oven a few minutes ago," she said. "I'm just taking it home to see if my husband can repair it." The policeman stared at her for a moment to see if she was telling the truth. "That's not a microwave oven." he said at last. "That's our radar set(雷達(dá)裝置). It was the start of a speed trap. Do you mind if we have it back?"   Mrs Smith's face turned red. "Oh", she said," I'm very sorry. I wouldn't have touched it if I'd known what it was."

1.Why did Mrs Smith pick up the police's radar set and want to take it home?

A．She had no microwave oven and wanted one.

B．She took it for a waste microwave oven.

C．She saw nobody was looking.

D．She just wanted to steal it.

2.The underlined word "boot" in the third paragraph means_____ .

A．the outer covering for the foot

B．the outer covering for the car

C．the place for luggage at the back of a car

D．the place for metal equipment for protection

3.Choose the right order of the events(事件) given in the passage.

a. The policeman wrote down Mrs Smith's name, address and the car number.

b. Mrs Smith picked up a radar set and put it in the boot of her car.

c. The policeman took back the radar set.

d. Mrs Smith went shopping.

e. A policeman signed Mrs Smith to stop her car.

f. The policeman found the radar set in the boot of Mrs Smith's car.

A．b,d,e,f,c,a         B．d,b,e,f,c,a          C．b,d,e,a,f,c          D．d,b,e,a,f,c

 

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The following table shows some results of a survey （調(diào)查）in which 800 Japanese school pupils were asked to give their impressions（印象）of their classroom teachers. The pupils’ impressions were found to differ depending on whether the teacher was new (with less than three years’ experience), middle-standing(ten to twenty years), or veteran（有經(jīng)驗(yàn)的）（twenty to thirty years）. The numbers in the table show the percentage of the pupils who answered “very satisfied” or “extremely satisfied” for each question item （項(xiàng)目）

Question Items
New
Middle-standing
Veteran
1.Shows sense of humor in class
2.Explains clearly
3.Teaches in a relaxed（放松的）manner.
4. Writes neatly on the blackboard
5. Lets pupils ask questions in class
6. Makes checks in notebooks
7. Speaks loudly and clearly
8. Treats pupils equally
9. Cares about pupils opinions
10. Spends time with pupils between classes
42
33
30
9
18
22
45
43
47
25
56
58
46
43
30
30
85
58
43
10[來(lái)源:學(xué),科,網(wǎng)]
70
68
65
56
47
43
54
42
17
6
 
1. The pupils’ evaluation（評(píng)價(jià)）can be said to rise at a steady rate（比率）as their teachers’ experience increases in Question Items______.
A. I and 4    B. 3 and 5     C. 6 and 8    D. 9 and 10
2. In contrast（對(duì)比）to the new teachers, the middle-standing and veteran teachers seem to have made a remarkable（顯著的）improvement in their ability to _______.
A. be fair to any pupil         B. evaluate pupils’ progress
C. present materials clearly     D. understand and play with pupils
3. Pupils seem to regard the new and the middle-standing teachers as being more_____.
A.relaxed in class than the veterans
B.interested in pupils’ ideas than the veterans
C.neat in appearance than the veterans
D.skilful at explaining than the veterans
4. Which of the following statements is true according to the table?
A.In the new teachers’ classes, pupils seem to ask questions more freely.
B.Pupils seem to be quite satisfied with the amount（量）of time their teachers spend with
them between classes.
C.The evaluation of the middle-standing teachers is lower than that of the veterans in seven
items.
D.Though veterans do not play games with pupils during breaks, their teaching is rated
highly.

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1.In which bookstore can you buy used books?

    A. The Bargain Books Store              B. The 4-for-3 Books Store.

C. The Significant Seven.               D. The Textbook Store

2.If you order two 9-dollar books, one 5-dollar book and one 3-dollar book in the 4-for-3 Books Store, then you can get       free.

A. none                         B. the 3-dollar book

C. the 5-dollar book                D. the two 9-dollar books

3.In the Significant Seven, you can find         .

    A. your favorite books 40% off the list price

B. the reader’s favorite books 30% off the list price

C. the new titles chosen by the editors as the must-read books of the season everyday

D. the best-selling books chosen by editors on show every Monday

4. The underlined word “categories” means “       ’.

    A. stores           B. shelves      C. types            D. sports

 

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List Price: $22.99

Our price: $13.57

You save: $9.42(41%)

Bargain Books Store

Big savings in Books

   Find great titles at fantastic prices in our Bargain Books Store.

1. Art and culture  ◆ Art and Photography    ◆ Entertainment & Pop Culture

2. Cooking and Home  ◆ Cooking, Food, and Wine  ◆ Home and Garden

3. Health, Mind, & Body  ◆ Health, Mind and Body   ◆ Parenting

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On the first Monday of every month we show The Significant Seven: the new titles chosen by our editors as the must-read books of the season.

The Textbook Store

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Save up to 30% on over 100000 new textbooks, and up to 90%off the list price of millions of used listings(列表), in our Textbook Store.

New Textbooks: Take advantage of great savings on hundreds of thousands of new textbooks, And, shop for the latest reference books(參考書), and the most popular study guides.

In which bookstore can you buy used books?

    A. The Bargain Books Store              B. The 4-for-3 Books Store.

C. The Significant Seven.               D. The Textbook Store

If you order two 9-dollar books, one 5-dollar book and one 3-dollar book in the 4-for-3 Books Store, then you can get       free.

A. none                         B. the 3-dollar book

C. the 5-dollar book                D. the two 9-dollar books

In the Significant Seven, you can find         .

    A. your favorite books 40% off the list price

B. the reader’s favorite books 30% off the list price

C. the new titles chosen by the editors as the must-read books of the season everyday

D. the best-selling books chosen by editors on show every Monday

The underlined word “categories” means “       ’.

    A. stores           B. shelves      C. types            D. sports

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     Yinxu (Ruins of Yin) is the ruins of the last capital of China’s Shang Dynasty (1600 BC - 1046 BC). The capital served 255 years for 12 kings. It shows the golden age of early Chinese culture, crafts (工藝品) and science, a time of great prosperity (繁榮) during the Chinese Bronze Age (青銅時(shí)代).
Discovered in 1899, Yinxu is one of the oldest and largest archeological sites (考古遺址) in China and is one of the historical capitals of China and is listed as a UNESCO World Heritage Site. It lies in central Henan Province, near the modern city of Anyang, and is open to the public as the Garden Museum of Yinxu. It is famous as the source of oracle bone script (甲骨文), the earliest recorded form of Chinese writing. The oracle bone script has recorded almost everything from dream-explaining to events such as harvests, birth of a child, the weather and the success of military campaigns.
Over 3,000 tombs, 2,200 pits, and 200 houses have been dug out at Yinxu. The large number of burial accessories found there shows the high level of the Shang crafts industry.The site includes a main palace and an ancient tomb. Besides, there are a number of large buildings, at least 53 of which have been dug out.
Yinxu has seen many years of research, first researched by the Academia Sinica in the late 1920s to the early 1930s and most recently by the Chinese Academy of Social Sciences.
41. During which period may Yin have been the capital of Shang Dynasty?
A. 1600 BC - 1500 BC
B. 1500 BC - 1200 BC
C. 1200 BC - 1050 BC
D. 1050 BC - 850 BC
42. From the passage, we can know that Yinxu ______.
A. was the last capital of China’s Shang Dynasty
B. proved the failure of the Shang Dynasty
C. was discovered in the 1920s
D. can be visited by the public now
43. Which of the following is NOT true about the oracle bone script?
A. It was discovered in Yinxu.
B. It is the earliest form of Chinese writing.
C. It has recorded many things.
D. It is important for studying the Shang Dynasty.
According to the passage, which of the following belongs to all the things   
that are dug out from Yinxu?
a. crafts  b. gold  c. oracle bone script  d. tombs 
e. sites  f. houses  g. accessories
A. afg  B. bcd    C. bdg         D. aec
45. We can infer from the passage that ______.
A. Anyang is a city of the Shang Dynasty
B. there are 200 houses in Yinxu
C. many things are still to be dug out
D. no research has been carried out on Yinxu

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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(zhǎng)素是吲哚乙酸；2，4－D是一種生長(zhǎng)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(duì)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(lái)源于太陽(yáng)能；草原上的牛和羊同屬于第二營(yíng)養(yǎng)級(jí)，共獲得生產(chǎn)者固定太陽(yáng)能的10％～20％。

5．B【解析】考查分泌蛋白的形成和分泌過(guò)程這一知識(shí)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(guò)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(jiǎn)式為。。根據(jù)三聚氰胺的結(jié)構(gòu)簡(jiǎn)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(yáng)離子為Na＋，陰離子為O22?,0.1mol  Na218O2中含有的陰陽(yáng)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(shuō)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(wú)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(wú)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(jiàn)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(méi)有告訴達(dá)平衡時(shí)的時(shí)間，無(wú)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(guò)程中溫度升高內(nèi)能變大，體積增大對(duì)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(duì)紅光折射率最小，對(duì)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(zhǎng)越長(zhǎng)，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(zhǎng)最短，故選項(xiàng)C錯(cuò)．對(duì)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(wú)論是彈簧將A+B和C彈開(kāi)過(guò)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(kāi)過(guò)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(shì)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(guò)程B+C的動(dòng)能損失就是此時(shí)的彈性勢(shì)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(kāi)平衡位置，速度正在變小，A對(duì)；由圖象可知該波的波長(zhǎng)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(guò)0.01s即半個(gè)周期，任何質(zhì)點(diǎn)通過(guò)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(guò)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(kù)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(duì)．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(duì)，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(wú)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(shuō)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(kāi)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(guò)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(guò)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(guò)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(zhǎng)′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(guò)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(kāi)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(cè)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(cè)定出NO的體積，如果裝置中有空氣，空氣中的O2會(huì)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(cè)定NO的體積時(shí)，如甲的液面高于乙的液面，測(cè)出的NO體積偏小，如甲的液面低于乙的液面，測(cè)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(shuō)明A中含有－COOH，且A在NaOH溶液中發(fā)生水解生成CH3COONa，說(shuō)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(jiǎn)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(méi)有－OH：HCOOCH3。對(duì)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(lái)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(guò)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(guò)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(huì)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(duì)等位基因控制的相對(duì)性狀的遺傳，基因自由組合定律適用于2對(duì)（及以上）的同源染色體上的2對(duì)（及以上）等位基因控制的性狀遺傳。具有兩對(duì)相對(duì)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(duì)相對(duì)性狀的基因位于2對(duì)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(duì)相對(duì)性狀的基因位于2對(duì)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(duì)相對(duì)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(duì)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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