ÌâÄ¿ÄÚÈÝ

Ëá¡¢¼î¡¢ÑÎÊÇÈýÀàÖØÒªÎïÖÊ£®
£¨1£©ÊµÑéÊÒÖÐÓÐһƿ³¨¿Ú·ÅÖõÄŨÑÎËᣮËüµÄÖÊÁ¿·ÖÊýºÍ·ÅÖÃÌìÊýµÄ¹ØϵÈçͼ£®¼ÙÉèË®²»Õô·¢£¬·ÖÎöŨÑÎËáµÄÈÜÖÊÖÊÁ¿·ÖÊý±ä»¯µÄÔ­Òò______£®

£¨2£©¹Û²ìͼƬ£¬ÏòСÊÔ¹ÜÄڵμÓË®ºó£¬³ý¿´µ½Ñõ»¯¸Æ¹ÌÌåÈܽâÍ⣬»¹¿É¹Û²ìµ½µÄÏÖÏóÊÇ______£®½âÊͲúÉú´ËÏÖÏóµÄÔ­Òò______£»______£®

£¨3£©Èçͼ·Ö±ðÊÇä廯Ç⣨HBr£©ºÍÒÒ´¼£¨C2H5OH£©ÔÚË®ÖеÄ΢¹ÛʾÒâͼ£®ÇëÄã½áºÏÒÑÓеÄËá¡¢¼î֪ʶ£¬ÅжÏä廯ÇâµÄË®ÈÜÒºÏÔ______£¬ÒÒ´¼µÄË®ÈÜÒºÏÔ______£¨Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£©£®

£¨4£©ÏòÊ¢ÓÐ10mLÏ¡ÑÎËᣨÆäÖеÎÓÐÉÙÁ¿Ö¸Ê¾¼Á£©µÄÉÕ±­ÖмÓÈëÇâÑõ»¯ÄÆÈÜÒº£¬ÓÃpH¼Æ²â¶¨ÈÜÒºµÄpH£¬ËùµÃÊý¾ÝÈçÏ£®Çë·ÖÎö²¢»Ø´ðÏÂÁÐÎÊÌ⣺
¼ÓÈëNaOHÈÜÒºµÄÌå»ý/mL02468101214
ÉÕ±­ÖÐÈÜÒºµÄpH1.11.21.41.62.57.011.012.0
¢ÙÈôµÎ¼ÓµÄָʾ¼ÁÊÇ×ÏɫʯÈïÈÜÒº£¬µ±¼ÓÈëÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ýΪ13mLʱ£¬ÈÜÒºÏÔ______É«£®
¢ÚÈôµÎ¼Óָʾ¼ÁÊÇÎÞÉ«·Ó̪ÈÜÒº£¬ÈëÇâÑõ»¯ÄÆÈÜÒº³ä·ÖÕñµ´ºó£¬ÈÜÒºÑÕÉ«Îޱ仯£®´ËʱÈÜÒºÖеÄÈÜÖÊÒ»¶¨ÓУ¨Ìѧʽ£©______£»¿ÉÄÜÓÐ______£®

½â£º£¨1£©Í¼Ê¾±íÃ÷£¬Ëæʱ¼äÑÓ³¤ÑÎËáµÄÖÊÁ¿·ÖÊýÖð½¥¼õС£¬Õâ¿ÉÓÃŨÑÎËá¾ßÓÐÒ×»Ó·¢µÄÐÔÖʼÓÒÔ½âÊÍ£ºËæHCl²»¶Ï»Ó·¢£¬ÈÜÒºÖÐÈÜÖÊÖÊÁ¿¼õС£»
£¨2£©Ñõ»¯¸ÆÓöË®·Å³ö´óÁ¿ÈÈ£¬ÊÔ¹ÜÄÚ¿ÕÆøÊÜÈÈÅòÕÍ£¬µ¼¹ÜÓëÖ®ÏàͨµÄUÐ͹ÜÄÚÁ½²àÒºÃæ³öÏָı䣬ºìÄ«Ë®µÄÒºÃæÒÒ¸ßÓÚ¼×£»·Å³öµÄÈÈÁ¿Ê¹´óÊÔ¹ÜÄÚÊ¢·ÅµÄ±¥ºÍÈÜҺζÈÉý¸ß£¬ÓÉÓÚÇâÑõ»¯¸ÆÈܽâ¶ÈËæζÈÉý¸ß¶ø¼õС£¬±¥ºÍÈÜÒº»áÒòÉýζøÎö³ö¹ÌÌ壬¿É¹Û²ìµ½Ê¯»ÒË®±ä»ë×Ç£»
£¨3£©ÓÉͼʾ¿ÉÖª£¬ä廯ÇâµÄË®ÈÜÒºÖк¬ÓÐH+£¬Òò´ËÈÜÒº³ÊËáÐÔ£»¶øÒÒ´¼ÈÜÒºÖмȲ»º¬H+Ò²²»º¬OH-£¬Òò´ËÈÜÒº³ÊÖÐÐÔ£»
£¨4£©¢Ù×ÏɫʯÈïÓö¼îÐÔÈÜÒº±ä³ÉÀ¶É«£¬¸ù¾Ý±íÖÐÊý¾Ý¿ÉÖª£¬µ±µÎÈë10mLÇâÑõ»¯ÄÆÈÜÒºÖÐÇ¡ºÃÍêÈ«·´Ó¦£¬ÈÜÒº³ÊÖÐÐÔ£»Òò´ËµÎÈë13mLÇâÑõ»¯ÄÆÈÜҺʱ£¬ÒòÇâÑõ»¯ÄƹýÁ¿¶øʹÈÜÒº³Ê¼îÐÔ£¬Òò´Ë×ÏɫʯÈï³ÊÏÖÀ¶É«£»
¢ÚµÎÈëÒ»¶¨Á¿ÇâÑõ»¯ÄÆÈÜÒººó£¬·Ó̪ÈÔΪÎÞÉ«£¬¿ÉÍƶϴËʱµÄÈÜÒº¿ÉÄܺ¬ÓÐδ·´Ó¦ÍêµÄÑÎËá¶ø³ÊËáÐÔ£¬»òÇ¡ºÃÍêÈ«·´Ó¦¶ø³ÊÖÐÐÔ£»Òò´Ë£¬´ËʱÈÜÒºÖÐÒ»¶¨º¬·´Ó¦Éú³ÉµÄÂÈ»¯ÄÆ£¬¿ÉÄܺ¬ÓÐδÍêÈ«·´Ó¦µÄÑÎË᣻

¹Ê´ð°¸Îª£º
£¨1£©Å¨ÑÎËáÒ×»Ó·¢£»
£¨2£©ºìÄ«Ë®µÄÒºÃæÒÒ¸ßÓڼס¢³ÎÇåʯ»ÒË®±ä»ë×Ç£»Ñõ»¯¸ÆÓëË®·´Ó¦·ÅÈÈʹÊÔ¹ÜÄÚµÄÆøÌåÊÜÈÈÅòÕÍ£»ÇâÑõ»¯¸ÆµÄÈܽâ¶ÈËæζȵÄÉý¸ß¶ø½µµÍ£¬Ò»²¿·ÖÇâÑõ»¯¸Æ¹ÌÌåÎö³ö£»
£¨3£©ËáÐÔ£»ÖÐÐÔ£»
£¨4£©¢ÙÀ¶£»¢ÚNaCl£»HCl£®
·ÖÎö£º£¨1£©·ÖÎöͼÖÐÑÎËáµÄÈÜÖÊÖÊÁ¿·ÖÊýËæʱ¼äÑÓ³¤µÄ±ä»¯Çé¿ö£¬½áºÏŨÑÎËáµÄ»Ó·¢ÐÔ£¬¶ÔÕâÒ»ÏÖÏó½øÐнâÊÍ£»
£¨2£©¸ù¾ÝÑõ»¯¸ÆÓöË®·Å³ö´óÁ¿ÈÈ£¬·Å³öµÄÈÈÁ¿¿ÉʹװÖÃÄÚ¿ÕÆøÊÜÈÈÅòÕÍ£¬Ò²¿Éʹ±¥ºÍÈÜҺʯ»ÒË®ÈÜҺζÈÉý¸ß£¬½áºÏÇâÑõ»¯¸ÆµÄÈܽâÐÔ£¬Ô¤¼Æ»á³öÏÖµÄÏÖÏó²¢ËµÃ÷Ô­Òò£»
£¨3£©ÈÜÒºÖк¬ÓÐH+£¬ÈÜÒº³ÊËáÐÔ£»ÈÜÒºÖк¬OH-£¬ÈÜÒºÔò³Ê¼îÐÔ£»ÓÉͼËùʾ£¬¸ù¾ÝÈÜÒºÖÐÁ£×Ó£¬ÅжÏÁ½ÈÜÒºµÄËá¼îÐÔ£»
£¨4£©¢ÙÓÉÊý¾Ý±í£¬ÅжÏËæ×ÅÏòÏ¡ÑÎËáÖеμÓÇâÑõ»¯ÄÆÈÜÒºpHµÄ±ä»¯£¬¸ù¾Ý±ä»¯¹æÂÉ£¬Åжϵ±¼ÓÈëÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ýΪ13mLʱ£¬ÈÜÒºµÄËá¼îÐÔ¼°×ÏɫʯÈïµÄ±äÉ«Çé¿ö£»
¢Ú·Ó̪ÓöËá»òÖÐÐÔÈÜÒº¾ùΪÎÞÉ«£¬Ö»ÓÐÓöµ½¼îÐÔÈÜÒº²Å³ÊºìÉ«£»¸ù¾Ý·Ó̪µÄÑÕÉ«£¬ÍƶÏÈÜÒºÖз´Ó¦Çé¿ö£¬ÅжÏÈÜÒºÖÐÈÜÖÊ£®
µãÆÀ£º¸ù¾Ý¶Ôͼʾ»òʵÑéÊý¾ÝµÄ·ÖÎö£¬¿ÆѧµØÍƶÏËù·¢Éú±ä»¯µÄÇé¿ö£¬ÕâÊǽâ¾ö±¾ÌâËùÒªÈÏÕæ¹Ø×¢µÄ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
29¡¢Ëá¡¢¼î¡¢ÑÎÊÇÈýÀàÖØÒªÎïÖÊ£®
£¨1£©ÊµÑéÊÒÖÐÓÐһƿ³¨¿Ú·ÅÖõÄŨÑÎËᣮËüµÄÖÊÁ¿·ÖÊýºÍ·ÅÖÃÌìÊýµÄ¹ØϵÈçͼ£®¼ÙÉèË®²»Õô·¢£¬·ÖÎöŨÑÎËáµÄÈÜÖÊÖÊÁ¿·ÖÊý±ä»¯µÄÔ­Òò
ŨÑÎËáÒ×»Ó·¢
£®

£¨2£©¹Û²ìͼƬ£¬ÏòСÊÔ¹ÜÄڵμÓË®ºó£¬³ý¿´µ½Ñõ»¯¸Æ¹ÌÌåÈܽâÍ⣬»¹¿É¹Û²ìµ½µÄÏÖÏóÊÇ
ºìÄ«Ë®µÄÒºÃæÒÒ¸ßÓڼס¢³ÎÇåʯ»ÒË®±ä»ë×Ç
£®½âÊͲúÉú´ËÏÖÏóµÄÔ­Òò
Ñõ»¯¸ÆÓëË®·´Ó¦·ÅÈÈʹÊÔ¹ÜÄÚµÄÆøÌåÊÜÈÈÅòÕÍ
£»
ÇâÑõ»¯¸ÆµÄÈܽâ¶ÈËæζȵÄÉý¸ß¶ø½µµÍ£¬Ò»²¿·ÖÇâÑõ»¯¸Æ¹ÌÌåÎö³ö
£®

£¨3£©Èçͼ·Ö±ðÊÇä廯Ç⣨HBr£©ºÍÒÒ´¼£¨C2H5OH£©ÔÚË®ÖеÄ΢¹ÛʾÒâͼ£®ÇëÄã½áºÏÒÑÓеÄËá¡¢¼î֪ʶ£¬ÅжÏä廯ÇâµÄË®ÈÜÒºÏÔ
ËáÐÔ
£¬ÒÒ´¼µÄË®ÈÜÒºÏÔ
ÖÐÐÔ
£¨Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£©£®

£¨4£©ÏòÊ¢ÓÐ10mLÏ¡ÑÎËᣨÆäÖеÎÓÐÉÙÁ¿Ö¸Ê¾¼Á£©µÄÉÕ±­ÖмÓÈëÇâÑõ»¯ÄÆÈÜÒº£¬ÓÃpH¼Æ²â¶¨ÈÜÒºµÄpH£¬ËùµÃÊý¾ÝÈçÏ£®Çë·ÖÎö²¢»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙÈôµÎ¼ÓµÄָʾ¼ÁÊÇ×ÏɫʯÈïÈÜÒº£¬µ±¼ÓÈëÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ýΪ13mLʱ£¬ÈÜÒºÏÔ
˦
É«£®
¢ÚÈôµÎ¼Óָʾ¼ÁÊÇÎÞÉ«·Ó̪ÈÜÒº£¬ÈëÇâÑõ»¯ÄÆÈÜÒº³ä·ÖÕñµ´ºó£¬ÈÜÒºÑÕÉ«Îޱ仯£®´ËʱÈÜÒºÖеÄÈÜÖÊÒ»¶¨ÓУ¨Ìѧʽ£©
NaCl
£»¿ÉÄÜÓÐ
HCl
£®
Ëá¡¢¼î¡¢ÑÎÊÇÈýÀàÖØÒªµÄ»¯ºÏÎÉÕ¼îÊÇÇâÑõ»¯ÄƵÄË׳ƣ¬ËüÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬¹ã·ºÓ¦ÓÃÓÚ·ÊÔí¡¢ÔìÖ½µÈ¹¤Òµ£®
£¨1£©ÉÕ¼îÔÚ¹¤ÒµÉÏͨ³£Óõç½â±¥ºÍʳÑÎË®µÄ·½·¨ÖÆÈ¡£¬²úÎï³ýÉÕ¼îÍ⣬»¹ÓÐH2ºÍCl2£¬Çëд³öµç½â±¥ºÍʳÑÎË®µÄ»¯Ñ§·½³Ìʽ
2NaCl+2H2O
 Í¨µç 
.
 
2NaOH+Cl2¡ü+H2¡ü
2NaCl+2H2O
 Í¨µç 
.
 
2NaOH+Cl2¡ü+H2¡ü
£®
£¨2£©ÉÕ¼îÔÚ¿ÕÆøÖзÅÖûá±äÖÊ£¬ÇëÄãÉè¼ÆÒ»¸öʵÑ飬¼ÈÄܼìÑéÉÕ¼î±äÖʵÄͬʱ£¬ÓÖÄܳýÈ¥ÔÓÖÊ£¬Óû¯Ñ§·½³Ìʽ±íʾ
Na2CO3+Ca£¨OH£©2=CaCO3¡ý+2NaOH
Na2CO3+Ca£¨OH£©2=CaCO3¡ý+2NaOH
£®
£¨3£©¼îºÍËáÓкܶ๲ͬµÄÌص㣬Èç¼îÓëËᶼÄܹ»·¢ÉúÖкͷ´Ó¦£¬¼îÓëËᶼÄÜʹËá¼îָʾ¼Á±äÉ«µÈ£¬ÇëÄãÔÙд³öËáºÍ¼îÔÚ×é³ÉºÍÐÔÖÊ·½ÃæµÄ¹²Í¬µã£¨¸÷дһÌõ£©
£¨4£©¼×ͬѧÓùÌÌåÉÕ¼îºÍË®ÅäÖÃ100g 18.5%µÄNaOHÈÜÒº£¬ÒÔÏÂÊÇËûÔÚÅäÖÆÉÕ¼îÈÜҺʱµÄ²Ù×÷£º
¢ÙÓÃͼËùʾµÄÐòºÅ±íʾÕýÈ·ÅäÖƸÃÈÜÒºµÄ²Ù×÷˳ÐòΪ
ECADB
ECADB
£®
¢ÚÈôͼCÖÐíÀÂëµÄÖÊÁ¿Îª15g£¬ÓÎÂëµÄ¶ÁÊýΪ3.5g£¬ÔòСÃ÷³ÆµÃµÄÇâÑõ»¯ÄÆÖÊÁ¿Êµ¼ÊΪ
11.5
11.5
g£®
£¨5£©Í¬Ñ§ÃÇÓÃpHÊÔÖ½²â¶¨ËùÅäÖƵÄÇâÑõ»¯ÄÆÈÜÒºµÄËá¼î¶È£¬ÕýÈ·µÄ·½·¨ÊÇ
Óò£Á§°ô½«ËùÅäÖƵÄÇâÑõ»¯ÄÆÈÜÒºµÎÔÚpHÊÔÖ½ÉÏ£¬ÉÔÍ£»áºÍ±ê×¼±ÈÉ«±È½Ï£¬¶Á³öÈÜÒºµÄpH
Óò£Á§°ô½«ËùÅäÖƵÄÇâÑõ»¯ÄÆÈÜÒºµÎÔÚpHÊÔÖ½ÉÏ£¬ÉÔÍ£»áºÍ±ê×¼±ÈÉ«±È½Ï£¬¶Á³öÈÜÒºµÄpH
£®
£¨6£©Èô8gÖÊÁ¿·ÖÊýΪ20%µÄÇâÑõ»¯ÄÆÈÜÒºÓë22gijÑÎËáÇ¡ºÃÍêÈ«Öкͣ¬ÊÔ¼ÆËã·´Ó¦ºóËùµÃÈÜÒºÈÜÖʵÄÖÊÁ¿·ÖÊý£®
£¨2009?ÐûÎäÇø¶þÄ££©Ëá¡¢¼î¡¢ÑÎÊÇÈýÀàÖØÒªÎïÖÊ£®
£¨1£©ÊµÑéÊÒÖÐÓÐһƿ³¨¿Ú·ÅÖõÄŨÑÎËᣮËüµÄÖÊÁ¿·ÖÊýºÍ·ÅÖÃÌìÊýµÄ¹ØϵÈçͼ£®¼ÙÉèË®²»Õô·¢£¬·ÖÎöŨÑÎËáµÄÈÜÖÊÖÊÁ¿·ÖÊý±ä»¯µÄÔ­Òò£®

£¨2£©¹Û²ìͼƬ£¬ÏòСÊÔ¹ÜÄڵμÓË®ºó£¬³ý¿´µ½Ñõ»¯¸Æ¹ÌÌåÈܽâÍ⣬»¹¿É¹Û²ìµ½µÄÏÖÏóÊÇ£®½âÊͲúÉú´ËÏÖÏóµÄÔ­Òò£»£®

£¨3£©Èçͼ·Ö±ðÊÇä廯Ç⣨HBr£©ºÍÒÒ´¼£¨C2H5OH£©ÔÚË®ÖеÄ΢¹ÛʾÒâͼ£®ÇëÄã½áºÏÒÑÓеÄËá¡¢¼î֪ʶ£¬ÅжÏä廯ÇâµÄË®ÈÜÒºÏÔ£¬ÒÒ´¼µÄË®ÈÜÒºÏÔ£¨Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£©£®

£¨4£©ÏòÊ¢ÓÐ10mLÏ¡ÑÎËᣨÆäÖеÎÓÐÉÙÁ¿Ö¸Ê¾¼Á£©µÄÉÕ±­ÖмÓÈëÇâÑõ»¯ÄÆÈÜÒº£¬ÓÃpH¼Æ²â¶¨ÈÜÒºµÄpH£¬ËùµÃÊý¾ÝÈçÏ£®Çë·ÖÎö²¢»Ø´ðÏÂÁÐÎÊÌ⣺
¼ÓÈëNaOHÈÜÒºµÄÌå»ý/mL2468101214
ÉÕ±­ÖÐÈÜÒºµÄpH1.11.21.41.62.57.011.012.0
¢ÙÈôµÎ¼ÓµÄָʾ¼ÁÊÇ×ÏɫʯÈïÈÜÒº£¬µ±¼ÓÈëÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ýΪ13mLʱ£¬ÈÜÒºÏÔÉ«£®
¢ÚÈôµÎ¼Óָʾ¼ÁÊÇÎÞÉ«·Ó̪ÈÜÒº£¬ÈëÇâÑõ»¯ÄÆÈÜÒº³ä·ÖÕñµ´ºó£¬ÈÜÒºÑÕÉ«Îޱ仯£®´ËʱÈÜÒºÖеÄÈÜÖÊÒ»¶¨ÓУ¨Ìѧʽ£©£»¿ÉÄÜÓУ®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø