95%的医学参考值范围的计算公式为（）

A: B: C: D: E:

某工程，业主在招标文件中规定：工期T(周)不得超过80周，也不应短于60周。 某施工单位决定参与该工程的投标。在基本确定技术方案后，为提高竞争能力，对其中某技术措施拟定了三个方案进行比选。方案一的费用为C1=100+4T；方案二的费用为C2=150+3T；方案三的费用为C3=250+2T。 这种技术措施的三个比选方案对施工网络计划的关键线路均没有影响。各关键工作可压缩的时间及相应增加的费用见表4-16。 表4-16 各关键工作可压缩的时间及相应增加的费用 关键工作 A C E H M 可压缩时间/周 1 2 1 3 2 压缩单位时间增加的费用/(万元/周) 3.5 2.5 4.5 6.0 2.0 假定所有关键工作压缩后不改变关键线路。 问题 如果该工程的施工网络计划如图4-11所示，则压缩哪些关键工作可能改变关键线路压缩哪些关键工作不会改变关键线路

以下不发生碘仿反应的是（）。

A:CH3CHO B:CH3CH2OH C: D:

如图所示的几何体的俯视图是（　　）

A: B: C: D:

实验室有一瓶标签残缺并且瓶盖没有完全密封的无色溶液（如图），为确认其中的溶质，同学们进行了如下的探究活动．请你参与到其中去，回答有关问题．

【进行猜想】猜想Ⅰ：该溶液的溶质是NaCl；

猜想Ⅱ：该溶液的溶质是NaOH；

猜想Ⅲ：该溶液的溶质是Na2CO3；

猜想 IV：该溶液的溶质是NaHCO3．

【查阅资料】常温下物质的相关信息如表：

从物质的相关信息可知，该溶液的溶质不是NaHCO3，因为常温下NaHCO3溶液溶质的质量分数最大是　 　（精确到0.1%）．

【进行实验】（1）测定溶液的pH大于7，该溶液的溶质不是　 　（填化学式）．

（2）同学们又进行了如下实验，实验过程如图：

①产生白色沉淀的化学方程式是　 　．

②实验可以判断原无色溶液中含有的溶质有　 　．

【获得结论】

该瓶原来装的无色溶液的溶质是NaOH，并且　 　（填“没有”、“部分”或“全部”）变质．

【拓展】

提纯该无色溶液的步骤是：先向无色溶液中加入适量的　 　，然后过滤．反应化学方程式是　 　．

解：

【查阅资料】根据计算可以知道在20℃时饱和碳酸氢钠的质量分数为：[0ecc01f6e57c7135.png]×100%＝8.8%＜10%，所以能够排除碳酸氢钠；

【进行实验】（

（1）测定溶液的pH大于7，该溶液的溶质不是，而氯化钠溶液呈中性，pH等于7所以不可能是氯化钠，故本题答案为：NaCl；

（2）①过量的氯化钙可以检验并除尽碳酸钠，碳酸钠可以和氯化钙反应生成碳酸钙沉淀和氯化钠，产生白色沉淀的化学方程式是Na2CO3+CaCl2＝CaCO3↓+2NaCl；

②滴加无色酚酞，滤液变红，说明溶液呈碱性，还有氢氧化钠，实验可以判断原无色溶液中含有的溶质有NaOH、Na2CO3；

【获得结论】该瓶原来装的无色溶液的溶质是NaOH，并且部分变质；

【拓展】提纯该无色溶液的步骤是：先向无色溶液中加入适量的Ca（OH）2，然后过滤．反应化学方程式是Ca（OH）2+Na2CO3═CaCO3↓+2NaOH．

答案：

【查阅资料】8.8%；

【进行实验】（1）NaCl；

（2）①Na2CO3+CaCl2＝CaCO3↓+2NaCl； ②NaOH、Na2CO3；

【获得结论】部分；Ca（OH）2Ca（OH）2+Na2CO3═CaCO3↓+2NaOH．

碱式碳酸钴[COx（OH）y（CO3）2]常用作电子材料，磁性材料的添加剂，受热时可分解生成三种氧化物．为了确定其组成，某化学兴趣小组同学设计了如图所示进行实验．

（1）请完成下列实验步骤：

①称取3.65g样品置于硬质玻璃管内，称量乙、丙装置的质量；

②按如图所示装置组装好仪器，并检验装置气密性；

③加热甲中玻璃管，当乙装置中　 　（填实验现象），停止加热；

④打开活塞a，缓缓通入空气数分钟后，称量乙、丙装置的质量；

⑤计算．

（2）步骤④中缓缓通入空气数分钟的目的是　 　

（3）某同学认为上述实验装置中存在一个明显缺陷，为解决这一问题，可选用下列装置中的　 　（填字母）连接在　 　（填装置连接位置）．

（4）COCl2•6H2O常用作多彩水泥的添加剂，以含钴废料（含少量Fe、Al等杂质）制取COCl2•6H2O的一种工艺如下：

已知：

①净除杂质时，加入H2O2发生反应的离子方程式为　 　．

②加入COCO2调PH为5.2～7.6，则操作Ⅰ获得的滤渣成分为　 　．

③加盐酸调整PH为2～3的目的为　 　．

④操作Ⅱ过程为蒸发浓缩、　 　（填操作名称）、过滤．

解：（1）加热甲中玻璃管，当乙装置中不再有气泡产生，即碱式碳酸钴分解完毕，

故答案为：不再有气泡产生；

（2）步骤④中缓缓通入空气数分钟，将装置中生成的CO2和H2O全部排入乙、丙装置中，以免影响测量结果．

故答案为：将装置中生成的CO2和H2O全部排入乙、丙装置中；

（3）在活塞a前，加装装置D，装置中盛放的碱石灰容易吸收空气中的水蒸气和二氧化碳．

故答案为：D；活塞a前；

（4）向含钴废料中加入过量稀盐酸，Fe、Al和稀盐酸反应生成FeCl2、AlCl3、CoCl2，向溶液中加入双氧水和CoCO3，双氧水具有强氧化性，能将亚铁离子氧化为铁离子，离子反应方程式为2Fe2++H2O2+2H+＝2Fe3++2H2O，加入CoCO3，调节溶液的pH至7.6，使Fe（OH）3、Al（OH）3生成沉淀，然后过滤，滤渣为Fe（OH）3、Al（OH）3，滤液中含有CoCl2，然后向滤液中加入稀盐酸，抑制CoCl2水解，然后采用蒸发浓缩、冷却结晶和过滤方法得到CoCl2•6H2O；

①双氧水具有强氧化性，能将亚铁离子氧化为铁离子，离子反应方程式为2Fe2++H2O2+2H+＝2Fe3++2H2O，

故答案为：2Fe2++H2O2+2H+＝2Fe3++2H2O；

②加入CoCO3调pH为5.2～7.6，则操作I获得的滤渣成分为Fe（OH）3、Al（OH）3，

故答案为：Fe（OH）3、Al（OH）3；

③CoCl2为强酸弱碱盐，阳离子水解导致溶液呈酸性，加入稀盐酸能抑制水解，所以加入稀盐酸的目的是抑制CoCl2水解，

故答案为：抑制CoCl2水解；

④操作Ⅱ过程为蒸发浓缩、冷却结晶、过滤，

故答案为：冷却结晶．

Almost everybody in America will spend a part of his or her life behind a shopping cart. They will, in a lifetime, push the chrome-plated contraptions many miles. But few will know—or even think to ask—who it was that invented them.
Sylvan N. Goldman invented the shopping cart in 1937. At that time he was in the supermarket business.Every day he would see shoppers lugging groceries around in baskets they had to carry.
One day Goldman suddenly had the idea of putting baskets on wheels. The wheeled baskets would make shopping much easier for his customers, and would help to attract more business.
On June 4, 1937, Goldman’s first carts were ready for use in his market. He was terribly excited on the morning of that day as customers began arriving. He couldn’t wait to see them using his invention.
But Goldman was disappointeD. Most shoppers gave the carts a long look, but hardly anybody would give them a try.
After a while, Goldman decided to ask customers why they weren’t using his carts. “Don’t you think this arm is strong enough to carry a shopping basket?” one shopper replieD.
But Goldman wasn’t beaten yet. He knew his carts would be a great success if only he could persuade people to give them a try. To this end, Goldman hired a group of people to push carts around his market and pretend they were shopping! Seeing this, the real customers gradually began copying the phony customers.
As Goldman had hoped, the carts were soon attracting larger and larger numbers of customers to his market. But not only did more people come—those who came bought morE. With larger, easier-to-handle baskets, customers unconsciously bought a greater number of items than beforE.
Today’s shopping carts are five times larger than Goldman’s original model. Perhaps that’s one reason Americans today spend more than five times as much money on food each year as they did before 1937—before the coming of the shopping cart.

4.What do the underlined words “ chrome-plated contraptions” in the first paragraph refer to?
A.Baskets. B.Private cars.
C.Suitcases. D.Shopping carts.
5.What was the purpose of Goldman’s invention?
A.It was to prove him to be a good inventor.
B.It was to reduce the burden of his employees' work.
C.It was to make shopping easier and attract more business.
D.It was to help the disabled make shopping easily in his market.
6.Goldman hired a group of people to push carts around his market in order to
A.attract people to buy things in his market
B.encourage people to use his shopping carts
C.make his market different from the others
D.keep the groceries from being stolen
7.What can we infer from the last paragraph?
A.Goldman will become very famous because of his invention.
B.Goldman's invention will be regarded as the greatest one in the worlD.
C.Supermarket business has benefited a lot from Goldman's invention.
D.There will be nothing that can replace Goldman's invention.

特技演员从高80m的大楼楼顶自由下落，在演员开始下落的同时汽车从60m远处由静止向楼底先匀加速运动t₀ =3s，再匀速行驶到楼底，演员刚好能落到汽车上（不计空气阻力，人和汽车看作质点，g取10m/s²），求:

（1）汽车开到楼底的时间t；

（2）汽车匀加速运动时的加速度a和汽车匀速行驶的速度v

  解：⑴由 得＝4s

    ⑵对车：   

联立解得8 m/s²      v=24m/s