ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿¸ù¾ÝϱíÖÐÈýÖÖ»¯·ÊµÄÐÅÏ¢£¬»Ø´ðÓйØÎÊÌâ¡£

Ãû³Æ

ÄòËØ

ÏõËáï§

̼ËáÇâï§

»¯Ñ§Ê½

CO£¨NH2£©2

NH4NO3

NH4HCO3

Êг¡¼Û¸ñ

1080Ôª/¶Ö

810Ôª/¶Ö

330Ôª/¶Ö

£¨1£©Ì¼ËáÇâ淋ÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª_____¡£

£¨2£©ÏõËáï§ÖеªÔªËصÄÖÊÁ¿·ÖÊýΪ_____¡£

£¨3£©´¿¾»µÄÄòËØÖÐ̼ԪËØÓëÇâÔªËصÄÖÊÁ¿±ÈΪ_____¡£

£¨4£©·Ö±ðÓÃ10000Ôª²É¹ºÄòËØ¡¢ÏõËá李¢Ì¼ËáÇâï§ÈýÖÖ»¯·Ê£¬Ëù¹ºµÃµÄ»¯·ÊÖеªÔªËصÄÖÊÁ¿·Ö±ðΪx¡¢y¡¢z£¬Ôòx¡¢y¡¢zÖ®¼äµÄ¹ØϵÊÇx_____y_____z£¨Ó㾡¢£½»ò£¼±íʾ£©¡£

¡¾´ð°¸¡¿79 35% 3£º1 £½ £¼

¡¾½âÎö¡¿

£¨1£©¸ù¾ÝÏà¶Ô·Ö×ÓµÄÖÊÁ¿Îª×é³É·Ö×ӵĸ÷Ô­×ÓµÄÏà¶ÔÔ­×ÓÖÊÁ¿Ö®ºÍ£¬¿ÉµÃ̼ËáÇâ淋ÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª£º14+5+12+16¡Á3£½79£»

£¨2£©ÏõËáï§ÖеªÔªËصÄÖÊÁ¿·ÖÊýΪ=£»

£¨3£©´¿¾»µÄÄòËØÖÐ̼ԪËØÓëÇâÔªËصÄÖÊÁ¿±ÈΪ12£º£¨1¡Á4£©£½3£º1£»

£¨4£©1ÍòÔª¹ºÂòµÄ»¯·ÊËùº¬µªÔªËصÄÖÊÁ¿·Ö±ðΪ£º

CO£¨NH2£©2ÖеªÔªËصÄÖÊÁ¿£¨x£©=£¬

NH4NO3ÖеªÔªËصÄÖÊÁ¿£¨y£©=£¬

NH4HCO3 ÖеªÔªËصÄÖÊÁ¿£¨z£©=£¬

ËùÒÔx£½y£¼z¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ä³Ð£Í¬Ñ§ÃÇ¿ªÕ¹½ðÊô»¯Ñ§ÐÔÖʵÄʵÑé̽¾¿»î¶¯¡£

£¨1£©Ä³Í¬Ñ§Éè¼Æ²»Í¬ÊµÑé·½°¸£¬ÑéÖ¤ÌúºÍÍ­Á½ÖÖ½ðÊô»î¶¯ÐÔ

ʵÑé²½Öè

ʵÑéÏÖÏó

ʵÑé½áÂÛ

¢Ù·Ö±ðÈ¡µÈÁ¿µÄÌúƬºÍͭƬÓÚÁ½ÊÔ¹ÜÖУ¬¼ÓÈëµÈÁ¿µÄÏ¡ÁòËá

________

Ìú±ÈÍ­»î¶¯ÐÔÇ¿

¢Ú________

________

£¨2£©Ä³Ð¡×é×öþÓëÏ¡ÁòËᷴӦʵÑéʱ£¬·¢Ïָ÷´Ó¦ºÜ¾çÁÒ£¬»¹¹Û²ìµ½ÊÔ¹ÜÄÚ²úÉú¡°°×Îí¡±£¬¼´¶Ô¡°°×Îí¡±½øÐÐÈçÏÂ̽¾¿:·´Ó¦¹ý³ÌÖУ¬¡°°×Îí¡± ²úÉúµÄÔ­ÒòÊÇ________________________¡£Ð´³öþÓëÏ¡ÁòËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ____________________¡£ÓÃȼ×ŵÄľÌõ·ÅÔÚÊԹܿڣ¬¿ÉÌýµ½±¬ÃùÉù,µ«ÊÔ¹ÜûÓб¬Õ¨µÄÔ­ÒòÊÇ____________________¡£

[Ìá³öÎÊÌâ]ʵÑé¹ý³ÌÖУ¬´ó¼ÒÖ»Ìý¼ûÇâÆøȼÉյı¬ÃùÉù¶øûÓп´¼û»ðÑ棬С×éÌÖÂÛÈçºÎʵÏÖÇâÆø³ÖÐøµØȼÉÕ²¢¿´¼ûµ­À¶É«»ðÑæÄØ?

[²éÔÄ×ÊÁÏ]Ë®ÕôµÄ´æÔÚ£¬¶ÔÇâÆøȼÉÕ»ðÑæµÄ´«²¥ÓкÜÃ÷ÏÔµÄ×èÖÍ×÷Ó㬼´Ë®ÕôÆøŨ¶ÈÔ½¸ß£¬»ðÑæ´«²¥µÄËÙ¶ÈÔ½Âý¡£

[Éè¼ÆʵÑé]¾­ÀÏʦָµ¼£¬Í¬Ñ§Ãǽ«ÊµÑé½øÐÐÁ˸Ľø£¬ÈçͼËùʾ¡£


[ʵÑé²Ù×÷]È¡0.3 gþÌõ·ÅÈë×°ÓÐ5.0 gÏ¡ÁòËáµÄÊÔ¹ÜÖУ¬½«ÊԹܷÅÈëÊ¢ÓÐÀäË®µÄ׶ÐÎÆ¿ÖУ¬·´Ó¦¿ªÊ¼ºó£¬ÓÃȼ×ŵÄľÌõµãȼÊԹܿڵÄÇâÆø£¬¹Û²ìµ½ÇâÆø³ÖÐøȼÉÕ£¬»ðÑæ³ÊÏÖµ­À¶É«¡£

[½âÊÍÓ뷴˼]¢Ù׶ÐÎÆ¿ÖÐÀäË®µÄ×÷ÓÃÊÇ__________¡£

¢ÚÔÚʵÑéÀäÈ´»Ö¸´ÊÒεĹý³ÌÖУ¬Í¬Ñ§ÃÇ»¹·¢ÏÖþÌõÏûʧһ¶Îʱ¼äºóÊÔ¹ÜÄÚ²¿²úÉúÁË°×É«¾§Ìå¡£¶Ô´Ë¡°ÒâÍ⡱µÄÏÖÏó£¬ÄãµÄ½âÊÍÊÇ________¡£Çë¼ÆËã´ËʱÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ________£¨Áгö¼ÆËãʽ¼´¿É£©¡££¨ÊÒÎÂÏÂÁòËáþµÄÈܽâ¶ÈΪ33.5 g£©

¡¾ÌâÄ¿¡¿Ä³¿ÎÍâÐËȤС×éµÄͬѧ¶Ô¶þÑõ»¯Ì¼µÄÖÆÈ¡ºÍÐÔÖʽøÐÐÏà¹Ø̽¾¿¡£

£¨ÊµÑé»Ø¹Ë£©£¨1£©ÊµÑéÊÒÓôóÀíʯÓëÏ¡ÑÎËáÖÆÈ¡¶þÑõ»¯Ì¼µÄ»¯Ñ§·½³ÌʽΪ_______________¡£

£¨2£©ÏÂͼ¿ÉÓÃÓÚʵÑéÊÒÖÆÈ¡CO2·¢Éú×°ÖõÄÊÇ__________£¨Ìî×Öĸ£©¡£


£¨ÊµÑé̽¾¿£©CO2ÓëNaOHÈÜÒº·´Ó¦

£¨²éÔÄ×ÊÁÏ£©

£¨1£©¹ýÁ¿CO2ͨÈëNaOHÈÜÒº£¬·¢ÉúÁ½²½·´Ó¦¡£

µÚÒ»²½£º____________________________£¨Óû¯Ñ§·½³Ìʽ±íʾ£©

µÚ¶þ²½£ºNa2CO3+H2O+CO2=2NaHCO3

£¨2£©Na2CO3ºÍNaHCO3²¿·ÖÈܽâ¶È±í

ζÈ/¡æ

0

15

20

30

40

50

60

NaHCO3/g

6.9

8.72

9.6

11.1

12.7

14.45

16.4

Na2CO3/g

7.1

13.25

21.8

39.7

48.8

47.3

46.4

£¨ÊµÑéÉè¼Æ£©ÔÚÊÒÎÂΪ15¡æʱ£¬½«10gNaOH¹ÌÌåÍêÈ«ÈܽâÓÚ80gË®ÖУ¬¶øºóÔÈËÙͨÈëCO2£¬Í¬Ê±Óô«¸ÐÆ÷²â¶¨ÈÜÒºµÄpH±ä»¯£¬½á¹ûÈçͼ1Ëùʾ¡£¶à´ÎÖظ´ÊµÑ飬ËùµÃʵÑé½á¹û»ù±¾Ò»Ö¡£


£¨1£©Í¨¹ýͼÏñ·ÖÎö£¬NaHCO3ÈÜÒºÏÔ_____£¨Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£©¡£

£¨2£©ÎªÁËÈ·¶¨MµãÈÜÒºµÄ³É·Ö£¬Í¬Ñ§ÃǽøÐÐÈçÏÂʵÑ飺

ʵÑé²½Öè

ʵÑéÏÖÏó

³õ²½ÊµÑé½áÂÛ

¢ÙÈ¡Ñù£¬µÎ¼Ó¹ýÁ¿µÄBaCl2ÈÜÒº

²úÉú°×É«³Áµí

º¬ÓÐNa2CO3¡¢NaOH

¢ÚÔÚ¢Ù·´Ó¦ºóµÄÉϲãÇåÒºÖеμÓ____

_____

£¨3£©ÊµÑé¹ý³ÌÖУ¬Í¬Ñ§ÃÇ·¢ÏÖ18minºóÈÜÒºµÄpH»ù±¾²»Ôٱ仯£¬Ôò´ËʱÈÜÒºÖпɹ۲쵽µÄÏÖÏóÊÇ______________________________¡£

£¨ÍØÕ¹ÑÓÉ죩ijͬѧÏòµÈÌå»ý¡¢º¬µÈ̼ԭ×ÓÊýµÄ̼ËáÇâÄƺÍ̼ËáÄÆÈÜÒºÖУ¬·Ö±ðÖðµÎ¼ÓÈëÏàͬŨ¶ÈµÄÏ¡ÑÎËᣬ²âµÃÏûºÄÑÎËáÓëÉú³ÉCO2µÄÖÊÁ¿Ö®¼ä¹ØϵÈçͼ2¡¢Í¼3Ëùʾ¡£(ºöÂÔCO2ÔÚË®ÖеÄÈܽâ)


£¨1£©Í¼2ÖÐAµãÈÜÒºÖеÄÈÜÖÊÊÇ_________£¨Ð´»¯Ñ§Ê½ £©£¬Í¼3ÖдÓBµã¿ªÊ¼µÄ±ä»¯¹ØϵÓëͼ2ÍêÈ«ÖغÏ,ͼ3ÖÐBµãÈÜÒºÖеÄÈÜÖÊÊÇ__________£¨Ð´»¯Ñ§Ê½£©¡£

£¨2£©Ð´³öOB¶Î·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ______________________________¡£

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

¾«Ó¢¼Ò½ÌÍø