ÌâÄ¿ÄÚÈÝ
(16·Ö)ijÑо¿Ð¡×éÄâÓôÖʳÑÎ(º¬Ca2+¡¢Mg2+¡¢SOµÈ)¡¢´Ö¹è(º¬C¼°²»ÓëCl2·´Ó¦µÄ¹ÌÌåÔÓÖÊ)ÖÆÈ¡´¿¹è£¬Éè¼ÆÈçÏµĹ¤ÒÕÁ÷³Ì£º
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©¹¤ÒµÉÏÒ»°ãÊÇÀûÓùýÁ¿½¹Ì¿ÔÚ¸ßÎÂÏ»¹ÔʯӢɰÀ´ÖÆÈ¡´Ö¹è£¬Ð´³ö¸Ã¹ý³ÌµÄ»¯Ñ§·½³Ìʽ£º_______________________________________________________________________¡£
£¨2£©¾«ÖÆ´ÖÑÎË®ËùÐèÊÔ¼ÁΪ¢ÙBaC12£»¢ÚNa2CO3£»¢ÛHC1£»¢ÜNaOH£¬ÆäµÎ¼ÓµÄÏȺó˳ÐòÊÇÏÂÁеÄ________(ÌîÏÂÁи÷ÏîµÄÐòºÅ)¡£
a£®¢Ù¢Ú¢Ü¢Û b£®¢Ü¢Ú¢Ù¢Û c£®¢Ü¢Ù¢Û¢Ú d£®¢Ú¢Ü¢Ù¢Û
ÒÑÖª£¬£¬¼ÙÉè¸Ã´ÖÑÎË®ÖеÄŨ¶È¾ùΪ0.01 mol¡¤L-1£¬ÈôÏò1 L¸Ã´ÖÑÎË®ÖÐÖð½¥µÎÈëÒ»¶¨Á¿Na2CO3ÈÜÒº£¬Ê×ÏȳöÏֵijÁµíÊÇ__________¡£
£¨3£©ÒÑÖªSiCl4µÄ·ÐµãÊÇ57.6¡æ£¬CC14µÄ·ÐµãÊÇ76.8¡æ¡£ÔÚ·´Ó¦Æ÷IÖеõ½µÄSiCl4´ÖÆ·Öк¬ÓÐCCl4£¬´ÓÖеõ½´¿¾»SiCl4¿É²ÉÓõķ½·¨ÊÇÏÂÁи÷ÏîÖеÄ________(ÌîÐòºÅ)¡£
a£®ÕôÁó b£®¸ÉÁó c£®·ÖÒº d£®¹ýÂË
·´Ó¦Æ÷¢òÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________________¡£
£¨4£©ÏÂͼÊÇÓÃÀë×Ó½»»»Ä¤·¨µç½â±¥ºÍʳÑÎË®µÄʾÒâͼ£¬µç½â²ÛÖÐÒõ¼«²úÉúµÄÆøÌåÊÇ_____¡£²úÆ·AµÄ»¯Ñ§Ê½Îª____________¡£
Èô²ÉÓÃÎÞĤµç½â²Ûµç½â±¥ºÍʳÑÎË®¿ÉÖÆÈ¡´ÎÂÈËáÄÆ£¬ÊÔд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ__ ___¡£
£¨16·Ö£©
£¨1£©SiO2+2CSi+2CO¡ü£¨2·Ö£©
£¨2£©a£¨2·Ö£©£»Ì¼Ëá¸Æ£¨»òCaCO3£©£¨2·Ö£©
£¨3£©a£¨2·Ö£©£»SiCl4+2H2Si+4HCl£¨2·Ö£©
£¨4£©ÇâÆø£¨»òH2£©£¨2·Ö£©£»NaOH£¨2·Ö£©£»NaCl+H2O NaClO+H2¡ü£¨2·Ö£©
½âÎöÊÔÌâ·ÖÎö£º£¨1£©CÔÚ¸ßÎÂÌõ¼þÏ»¹ÔSiO2£¬Éú³ÉSiºÍCO£¬»¯Ñ§·½³ÌʽΪ£ºSiO2+2CSi+2CO¡ü
£¨2£©Na2CO3ÈÜÒºµÄ×÷ÓÃΪ³ýÈ¥Ca2+¡¢³ýÈ¥¹ýÁ¿µÄBaCl2ÈÜÒº£¬ËùÒÔNa2CO3µÄ˳ÐòÔÚBaCl2µÄºóÃ棬HClµÄ×÷ÓÃÊdzýÈ¥¹ýÁ¿µÄNa2CO3ºÍNaOH£¬ËùÒÔÔÚ×îºó£¬Òò´ËaÏîÕýÈ·£»ÒòΪKsp(MgCO3) < Ksp(CaCO3)£¬CaCO3¸üÄÑÈÜ£¬ËùÒÔÊ×ÏÈÎö³öµÄ³ÁµíÊÇCaCO3¡£
£¨3£©SiCl4ÓëCCl4ÔÚ³£ÎÂÏÂΪҺÌ壬Ï໥Èܽ⣬µ«·Ðµã²»Í¬£¬ËùÒÔÓÃÕôÁóµÄ·½·¨µÃµ½´¿¾»µÄSiCl4£»·´Ó¦Æ÷IIÖÐH2»¹ÔSiCl4£¬»¯Ñ§·½³ÌʽΪ£ºSiCl4+2H2Si+4HCl¡£
£¨4£©¸ù¾Ý·Åµç˳Ðò£¬Òõ¼«ÉÏH2OµçÀë³öµÄH+·Åµç£¬ËùÒÔµç½â²ÛÖÐÒõ¼«²úÉúµÄÆøÌåÊÇÇâÆø£»Ë®µçÀë³öµÄH+·Åµç£¬´Ù½øH2OµÄµçÀëƽºâÏòÓÒÒƶ¯£¬OH?Ũ¶ÈÔö´ó£¬ËùÒÔ²úÆ·AΪNaOH£»ÎÞĤµç½â²Ûµç½â±¥ºÍʳÑÎË®¿ÉÖÆÈ¡´ÎÂÈËáÄÆ£¬µç½â²úÉúµÄCl2ÓëNaOH·´Ó¦Éú³É´ÎÂÈËáÄÆ£¬ËùÒÔ»¯Ñ§·½³ÌʽΪ£ºNaCl+H2O NaClO+H2¡ü
¿¼µã£º±¾Ì⿼²é»¯Ñ§·½³ÌʽµÄÊéд¡¢³ýÔÓ¡¢µç½âµÄÓ¦Óá£
ij»¯Ñ§ÐËȤС×éÄâ²ÉÓÃÏÂͼËùʾװÖõç½â±¥ºÍÂÈ»¯ÄÆÈÜÒºÖƱ¸H2£¬Í¨¹ýH2»¹ÔÑõ»¯Í²â¶¨CuµÄÏà¶ÔÔ×ÓÖÊÁ¿Ar£¨Cu£©£¬Í¬Ê±¼ìÑéCl2µÄÑõ»¯ÐÔ£¨Í¼ÖмгֺͼÓÈÈÒÇÆ÷ÒÑÂÔÈ¥£©¡£
£¨1£©Ö±Á÷µçÔ´ÖеÄX¼«Îª ¼«£¨Ìî¡°Õý¡±¡¢¡°¸º¡±¡¢¡°Òõ¡±»ò¡°Ñô¡±£©£»Ð´³ö¼××°ÖÃUÐιÜÖз´Ó¦µÄÀë×Ó·½³Ìʽ£º £»ÊµÑ鿪ʼºó£¬ÓÃÌú°ô×÷µç¼«µÄÒ»²àµÄʵÑéÏÖÏóÊÇ ¡£
£¨2£©ÎªÍê³ÉÉÏÊöʵÑ飬ÕýÈ·µÄÁ´½Ó˳ÐòΪ£ºaÁ¬ £¬bÁ¬ £¨ÌîдÁ¬½ÓµÄ×Öĸ£©¡£
£¨3£©×°ÖÃÒÒÖеÄGÆ¿ÄÚÈÜÒº¿ÉÄÜΪ £¨Ìî×Öĸ£©¡£
A£®µí·ÛKIÈÜÒº | B£®NaOHÈÜÒº | C£®Na2SÈÜÒº | D£®Na2SO3ÈÜÒº |
£¨4£©ÔÚ¶ÔÓ²Öʲ£Á§ÊÔ¹ÜÀïµÄÑõ»¯Í·ÛÄ©¼ÓÈÈÇ°ÐèÒª½øÐеIJÙ×÷Ϊ£º ¡£
£¨5£©×°ÖñûÖÐNÆ¿ÄÚÊ¢·ÅµÄÊÔ¼ÁΪ £¬×÷ÓÃÊÇ ¡£
£¨6£©ÎªÁ˲ⶨCuµÄÏà¶ÔÔ×ÓÖÊÁ¿£¬Ä³Í¬Ñ§Í¨¹ýʵÑé²âµÃÈçÒ»ÏÂÊý¾Ý£º
I£®Ñõ»¯ÍÑùÆ·ÖÊÁ¿Îªm1g
¢ò£®·´Ó¦ºóÓ²Öʲ£Á§¹ÜÖÐÊ£Óà¹ÌÌåÖÊÁ¿Îªm2g
¢ó£®·´Ó¦Ç°ºóUÐιܼ°Æä¹ÌÌåÖÊÁ¿²îΪm3g
¢ô£®·´Ó¦Ç°ºóÆ¿¼°ÆäÒºÌåÖÊÁ¿²îΪm4g
¢ÙÇëÑ¡ÔñÀíÂÛÉÏÎó²î×îСµÄÒ»×éÊý¾Ý¼ÆËãAr£¨Cu£©£¬Ar£¨Cu£©= ¡£
¢ÚÈç¹ûÑ¡ÓÃÆäËüÊý¾Ý½øÐмÆË㣬»áµ¼ÖÂAr£¨Cu£© £¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©£¬ÀíÓÉÊÇ ¡£
ÏÂÁÐÓйØͬ·ÖÒì¹¹ÌåÊýÄ¿µÄÐðÊöÖУ¬²»ÕýÈ·µÄÊÇ( )
A£®¼×±½±½»·ÉϵÄÒ»¸öÇâÔ×Ó±»º¬3¸ö̼Ô×ÓµÄÍé»ùÈ¡´ú£¬ËùµÃ²úÎïÓÐ6ÖÖ |
B£®·Ö×Óʽ·ûºÏC5H11ClµÄ»¯ºÏÎïÓÐ6ÖÖ |
C£®ÒÑÖª¶þÂȱ½ÓÐ3ÖÖͬ·ÖÒì¹¹Ì壬ÔòËÄÂȱ½µÄͬ·ÖÒì¹¹ÌåµÄÊýĿΪ3ÖÖ |
D£®·ÆµÄ½á¹¹¼òʽΪ£¬ËüÓëÏõËá·´Ó¦£¬¿ÉÉú³É5ÖÖÒ»Ïõ»ùÈ¡´úÎï |
ϱíÖÐʵÑé²Ù×÷ÄܴﵽʵÑéÄ¿µÄµÄÊÇ
| ʵÑé²Ù×÷ | ʵÑéÄ¿µÄ |
A | Ïò±½·ÓµÄ±¥ºÍÈÜÒºÖеμÓÏ¡äåË® | ÑéÖ¤Èýäå±½·ÓΪ°×É«³Áµí |
B | Ïò¼×ËáÄÆÈÜÒºÖмÓÐÂÖƵÄCu(OH)2Ðü×ÇÒº²¢¼ÓÈÈ | È·¶¨¼×ËáÄÆÖк¬ÓÐÈ©»ù |
C | Ïò¾Æ¾«ºÍÒÒËáµÄ»ìºÏÒºÖмÓÈë½ðÊôÄÆ | È·¶¨¾Æ¾«ÖлìÓд×Ëá |
D | ½«äåÒÒÍéÓëÇâÑõ»¯ÄÆÈÜÒº¹²ÈÈÒ»¶Îʱ¼ä£¬ÔÙÏòÀäÈ´ºóµÄ»ìºÏÒºÖеμÓÏõËáÒøÈÜÒº | ¼ìÑéË®½â²úÎïÖеÄäåÀë×Ó |
ϱíÖÐËùÁеĶ¼ÊÇÍéÌþ£¬ËüÃǵÄһ±ȡ´úÎï¾ùÖ»ÓÐÒ»ÖÖ£¬·ÖÎöϱíÖи÷ÏîµÄÅŲ¼¹æÂÉ£¬°´´Ë¹æÂÉÅŲ¼µÚ6ÏîӦΪ£¨ £©
1 | 2 | 3 | 4 | 5 | ¡¡ |
CH4 | C2H6 | C5H12 | C8H18 | ¡¡ | ¡¡ |