ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿»¯Ñ§ÐËȤС×é»î¶¯ÖУ¬¼×¡¢ÒÒ¡¢±ûÈýλͬѧ¶Ôij»¯¹¤³§µÄÎÛˮȡÑùºó½øÐÐÁËÏà¹ØʵÑ飬Çë»Ø´ðÏÂÁÐÎÊÌâ¡£

I.ÓÃpHÊÔÖ½¼ì²âÎÛË®µÄËá¼îÐÔ¡£Èýλͬѧ·Ö±ð½øÐÐÈçÏÂʵÑé:

¼×ͬѧ:È¡pHÊÔÖ½ÓÚ±íÃæÃóÉÏ,Óò£Á§°ôպȡ´ý²âÒºµÎÔÚpHÊÔÖ½ÉÏ£¬²âµÃpH<7.

ÒÒͬѧ:È¡pHÊÔÖ½ÓÚ±íÃæÃóÉÏ£¬ÏÈÓÃÕôÁóˮʪÈó£¬ÔÙÓò£Á§°ôպȡ´ý²âÒºµÎÔÚpHÊÔÖ½ÉÏ£¬²âµÃpH<7¡£

±ûͬѧ:È¡pHÊÔÖ½Ö±½Ó½þÈë´ý²âÒºÖУ¬²âµÃpH <7.

(1)ÒÔÉÏÈýλͬѧÖвÙ×÷¹æ·¶µÄÊÇ___________ͬѧ¡£

¢ò.ÀûÓÃÖкͷ´Ó¦Ô­Àí²â¶¨ÎÛË®ÖÐÎÛȾÎï(ÉèΪÑÎËá)µÄÖÊÁ¿·ÖÊý¡£¼×¡¢ÒÒÁ½Í¬Ñ§Éè¼ÆµÄ·½°¸·Ö±ðÈçͼ¼×¡¢Í¼ÒÒËùʾ:

(2)¼×ͬѧȡÉÙÁ¿Í¼¼×Ëùʾ·´Ó¦ºóµÄÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó3µÎ×ÏɫʯÈïÊÔÒº£¬ÈÜÒº³ÊÀ¶É«£¬ÓÚÊǵóöÁË¡°Á½ÖÖÎïÖÊÒÑÇ¡ºÃÖк͡±µÄ½áÂÛ£¬¼×ͬѧµÄ½áÂÛÊÇ·ñÕýÈ·?_____________ (Ñ¡Ìî¡°ÕýÈ·¡±»ò¡°²»ÕýÈ·¡±)£¬ÀíÓÉÊÇ______________¡£

(3)ÒÒͬѧ°´Í¼ÒÒ·½°¸½øÐÐʵÑ飬ʵÑé¹ý³ÌÖУ¬NaOHÈÜÒº±ØÐëÓýºÍ·µÎ¹ÜÖðµÎ¼ÓÈ룬ÇÒÓò£Á§°ô²»¶Ï½Á°è¡£µ±¹Û²ìµ½____________ÏÖÏóʱ£¬ Í£Ö¹µÎ¼ÓNaOHÈÜÒº£¬´Ëʱ¿ÉÈÏΪËá¼îÇ¡ºÃÍêÈ«·´Ó¦¡£

(4)ÈôÒª³ýÈ¥ÎÛË®ÖеÄÑÎËᣬ´Ó»·±£¡¢Ô­²ÄÁϳɱ¾µÈ½Ç¶È¿¼ÂÇ£¬×îºÃÑ¡ÓÃ________________¡£

A NaOH

B Ca(OH)2

C Fe2O3

D CaCO3

(5)ÈôͼÒÒ·½°¸ÊµÑéÖй²ÏûºÄÁËNaOHÈÜÒº20g,Çë¼ÆËã:

¢Ù20g NaOHÈÜÒºÖÐÈÜÖÊÖÊÁ¿=______________g.

¢ÚÎÛË®ÖÐHClµÄÖÊÁ¿·ÖÊýÊǶàÉÙ? ____ (ÁÐʽ¼ÆËã)

¡¾´ð°¸¡¿¼× ²»ÕýÈ· ÈÜÒº³ÊÀ¶É«£¬ËµÃ÷ÇâÑõ»¯ÄÆÈÜÒºÒѹýÁ¿ ÎÞÉ«ÈÜÒº¸Õ¸Õ±äΪºìɫʱ D 0.008 0.0365%

¡¾½âÎö¡¿

£¨1£©ÓÃpHÊÔÖ½²â¶¨pHÖµµÄ·½·¨£ºÓýྻ¡¢¸ÉÔïµÄ²£Á§°ôպȡ´ý²âÒº£¬µãÔÚpHÊÔÖ½ÉÏ£¬¹Û²ìÑÕÉ«µÄ±ä»¯£¬È»ºóÓë±ê×¼±ÈÉ«¿¨¶ÔÕÕ£¬²»ÄÜÖ±½Ó½«pHÊÔÖ½½þÈë´ý²âÒºÖУ¬»áÎÛȾ´ý²âÒº£¬Ó°Ïì²â¶¨½á¹û£»Ò²²»Äܽ«pHÊÔÖ½Èóʪ£¬»áÓ°Ïì²â¶¨½á¹û£¬¹Ê²Ù×÷¹æ·¶µÄÊÇ£º¼×ͬѧ£»

£¨2£©µÎ¼Ó×ÏɫʯÈïÊÔÒº£¬ÈÜÒºÏÔÀ¶É«£¬×ÏɫʯÈïÊÔÒºÓö¼î±äÀ¶£¬ËµÃ÷ÇâÑõ»¯ÄÆÈÜÒº¹ýÁ¿£¬¹Ê¼×ͬѧµÄ½áÂÛ²»ÕýÈ·£¬ÀíÓÉÊÇ£ºÈÜÒº³ÊÀ¶É«£¬ËµÃ÷ÇâÑõ»¯ÄÆÈÜÒºÒѹýÁ¿£»

£¨3£©ÎÞÉ«·Ó̪ÊÔÒºÔÚÖÐÐÔºÍËáÐÔÈÜÒºÖв»±äÉ«£¬ÔÚ¼îÐÔÈÜÒºÖÐΪºìÉ«£¬¹ÊÏòÎÛË®ÖеμÓÇâÑõ»¯ÄÆÈÜÒº£¬ÇâÑõ»¯ÄÆÓëÑÎËá·´Ó¦Éú³ÉÂÈ»¯ÄƺÍË®£¬µ±ÎÞÉ«ÈÜҺǡºÃ±äΪºìɫʱ£¬ËµÃ÷Ëá¼îÇ¡ºÃÍêÈ«·´Ó¦£¬¹ÊÌÎÞÉ«ÈÜÒº¸Õ¸Õ±äΪºìɫʱ¡£

£¨4£©ÇâÑõ»¯ÄÆ¡¢ÇâÑõ»¯¸Æ¡¢ÈýÑõ»¯¶þÌú¡¢Ì¼Ëá¸Æ¾ùÄÜÓëÑÎËá·´Ó¦£¬ÆäÖÐ̼Ëá¸Æ¼Û¸ñ×îµÍ£¬¶øÇÒ²»ÏñÇâÑõ»¯ÄÆ¡¢ÇâÑõ»¯¸ÆÒ»Ñù¾ßÓи¯Ê´ÐÔ£¬Ì¼Ëá¸Æ²»ÈÜÓÚË®£¬¶àÓàµÄ̼Ëá¸Æ²»»áÎÛȾ¿ÕÆø£¬¹ÊÑ¡Ôñ̼Ëá¸Æ×îºÏÀí¡£

¹ÊÑ¡D¡£

£¨5£©¢Ù20g NaOHÈÜÒºÖÐÈÜÖÊÖÊÁ¿Îª£º20g¡Á0.04%=0.008g£»

¢Ú½â£ºÉèÎÛË®ÖÐHClµÄÖÊÁ¿·ÖÊýÊÇx

x=0.0365%

´ð£ºÎÛË®ÖÐHClµÄÖÊÁ¿·ÖÊýÊÇ0.0365%¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Na2CO3ºÍNaHCO3ÓÐÏàËƵĻ¯Ñ§ÐÔÖÊ£¬¶¼ÄÜÓëËᡢijЩ¼î·´Ó¦¡£µ«Ì¼ËáÄÆÎȶ¨£¬¶ø NaHCO3ÊÜÈȷֽ⣺2NaHCO3X+H2O+CO2¡ü¡£

£¨1£©Éú³ÉÎïXµÄ»¯Ñ§Ê½Îª______¡£

£¨2£©¹¤ÒµÉÏÓÃ̼ËáÄÆÖÆÉÕ¼îµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ______¡£

£¨3£©Í¬Ñ§ÃÇÔÚ̽¾¿Æ仯ѧÐÔÖÊʵÑéºó£¬½«Ê£Óà¹ÌÌåÒ©Æ·¼¯ÖзÅÔÚÉÕ±­ÖС£Ð¡Ã÷ΪÁËŪÇåÊ£Óà¹ÌÌåµÄ³É·Ý£¬½øÐÐÁËÈçϵÄ̽¾¿£º

£¨Ìá³ö²ÂÏ룩ʣÓà¹ÌÌåÊÇ£ºI£®Na2CO3£»¢ò£®NaHCO3£»¢ó£®______¡£

£¨½øÐÐʵÑ飩

²Ù×÷

ÏÖÏó

½áÂÛ

¢ÙÈ¡ÉÙÁ¿Ê£Óà¹ÌÌåÓÚÊÔ¹ÜÖУ¬¼ÓÈë×ãÁ¿Ë®£¬³ä·ÖÈܽâºó£¬µÎ¼Ó¼¸µÎ·Ó̪ÈÜÒº¡£

ÈÜÒºÓÉÎÞÉ«±äºìÉ«¡£

Ê£Óà¹ÌÌåÒ»¶¨ÓУº

______ ¡£

¢ÚÈ¡ÊÊÁ¿Ê£Óà¹ÌÌåÓÚÊÔ¹ÜÖУ¬¹Ì¶¨ÔÚÌú¼Ų̈ÉÏ£¬³ä·Ö¼ÓÈÈ¡£

°×É«¹ÌÌåÖÊÁ¿¼õÉÙ£¬

ÊÔ¹ÜÄÚ±ÚÓн϶àË®Öé³öÏÖ¡£

£¨½áÂÛ·ÖÎö£©

¢Ù¾ÝÉÏÊöʵÑéÏÖÏó£¬Ð¡Ã÷ÈÏΪ²ÂÏë¢óÕýÈ·¡£µ«ÓÐͬѧÈÏΪСÃ÷µÄÅжϲ»ÑϽ÷£¬ÒòΪ²ÂÏë______Ò²ÓÐÏàͬÏÖÏó¡£

¢ÚͬѧÃÇÌÖÂÛºóÈÏΪ£¬ÈçҪȷ¶¨Ê£Óà¹ÌÌå³É·Ö£¬Ð¡Ã÷Ö»Òª×öʵÑé²Ù×÷¢Ú£¬²¢²¹³ä²Ù×÷______£¬ÔÙͨ¹ý______¼´¿ÉµÃ³ö½áÂÛ¡£

£¨ÍØÕ¹·´Ë¼£©

ÓÐͬѧÌá³ö£¬Ì¼Ëá¸ÆºÍ̼ËáÇâ¸ÆÄܲ»ÄÜ»¥Ïàת»¯ÄØ£¿Ð¡Ã÷ÈÏΪ¿ÉÒÔ£¬Òò¿Î±¾¡°×ÊÁÏ¿¨Æ¬¡±ÖнéÉܵġ¶Ê¯ËñºÍÖÓÈéʯµÄÐγɡ·¾ÍÊÇ¡°CaCO3¡úCa(HCO3)2¡úCaCO3¡±µÄ¹ý³Ì¡£Ôò¡°CaCO3¡úCa(HCO3)2¡±»¯Ñ§·´Ó¦·½³ÌʽΪ______¡£

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

¾«Ó¢¼Ò½ÌÍø