2020~2022英语周报七年级下册。答案

26.答案(1)2Al+2Na0H+2H,0=2NaA102+3H2↑(2分)(2)NaCo02中钴为+3价，硫酸和硝酸无法将其还原为Co2+(合理即可，2分)温度升高，消耗的能量增多，钴的浸出率提高不明显，且盐酸挥发损失严重（合理即可，2分）2NaCo0,+8H++2Cl-一2Na++2Co2++Cl2↑+4H20(2分)(3)4.6≤pH<6.6(2分)(4)NH4Cl、NaCl(2分)在空气中煅烧（合理即可，2分）命题透析本题以用钴锂膜废料制取C0,0,为素材，考查元素化合物及物质基本转化规律等知识，意在考查考生的信息获取加工与逻辑推理等能力，证据推理与模型认知的核心素养。思路点拨(1)由题干知，铝以单质形式存在，Al能溶于强碱溶液生成NaAl02和H2。(2)硫酸、硝酸及盐酸均是强酸，但盐酸中C1~具有较强还原性，由流程知其可将+3价的C0还原为C02+,但硫酸和硝酸无法将+3价的C0还原为C。2+;由图知，升高温度有利于钴的浸出，温度大于80℃时，钴的浸出率没有明显的提高，但温度升高也有利于HCl气体逸出；由流程图知，NaCo0,应是难溶物，NaCoO,与盐酸反应生成NaCl、CoCL2、CL2,反应的离子方程式为2NaCo02+8H++2Cl—=2Na++2Co2++Cl2↑+4H20。(3)要确保A13+和Fe3+沉淀完全，而Co2+不沉淀，根据表中数据可控制溶液pH范围为4.6≤pH<6.6。(4)由HCl和NaCo02的反应及反应CoCl2+(NH4)2C204一CoC204↓+2NH4Cl可知，滤液3中的溶质为NH4Cl、NCl;CoC204→Co203,需发生氧化反应，故在空气中煅烧（或高温加热）得到：4CoC204+3022C0203+8C02o

第二节书面表达Dear Henry,Im Li Hua from your English speaking class lastterm. I m writing to ask for your helpNow I am preparing to attend MUN next weckHowever,have some difficulty with concepts andexpressions of the current affairs and speech skills. Iwonder if I can borrow your Dictionary of englishMedia. which you presented in class. Furthermore.I amnot quite sure about how to make a convincing andpersuasive speech at MUN. Could you be so kind as to giveme some advice?I know you have a very busy schedule, but I'd be verygrateful if you could spare some time to give me face-to-acc instructions. Thank you for your kindnessI'm looking forward to your replyYoursLi Hua

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