µç¯¼ÓÈÈʱÓô¿ÑõÆøÑõ»¯¹ÜÄÚÑùÆ·£¬¸ù¾Ý²úÎïµÄÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É£¬Í¼ÖÐËùÁÐ×°ÖÃÊÇÓÃȼÉÕ·¨È·¶¨ÓлúÎï·Ö×Óʽ³£ÓõÄ×°Öá£

(1)²úÉúµÄÑõÆø°´´Ó×óµ½ÓÒÁ÷Ïò£¬ËùѡװÖø÷µ¼¹ÜµÄÁ¬½Ó˳ÐòÊÇ________________________¡£

(2)C×°ÖÃÖÐŨH2SO4µÄ×÷ÓÃÊÇ___________________________________________¡£

(3)D×°ÖÃÖÐMnO2µÄ×÷ÓÃÊÇ_____________________________________________¡£

(4)ȼÉÕ¹ÜÖÐCuOµÄ×÷ÓÃÊÇ______________________________________________¡£

(5)Èô׼ȷ³ÆÈ¡0.90 gÑùÆ·(Ö»º¬C¡¢H¡¢OÈýÖÖÔªËØÖеÄÁ½ÖÖ»òÈýÖÖ)£¬¾­³ä·ÖȼÉÕºó£¬A¹ÜÖÊÁ¿Ôö¼Ó1.32 g£¬B¹ÜÖÊÁ¿Ôö¼Ó0.54 g£¬Ôò¸ÃÓлúÎïµÄʵÑéʽΪ_______________________¡£

(6)ҪȷÁ¢¸ÃÓлúÎïµÄ·Ö×Óʽ£¬»¹ÒªÖªµÀ_________________________________________¡£

½âÎö£º(5)n(CO2)=,n(H2O)=

m(O)=0.90 g-m(C)-m(H)=0.90 g-0.03 mol¡Á12 g¡¤mol-1-0.03 mol¡Á2¡Á1 g¡¤mol-1=0.48 g,n(O)= £¬¹ÊʵÑéʽΪCH2O¡£

´ð°¸£º(1)g¡¢f¡¢e¡¢h¡¢i¡¢c(»òd)¡¢d(»òc)¡¢a(»òb)¡¢b(»òa)

(2)ÎüÊÕË®·Ö£¬µÃµ½¸ÉÔïµÄÑõÆø

(3)´ß»¯¼Á£¬¼Ó¿ìO2µÄÉú³ÉËÙÂÊ

(4)ʹÓлúÎï³ä·ÖÑõ»¯Éú³ÉCO2ºÍH2O

(5)CH2O

(6)²â³öÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§Éϳ£ÓÃȼÉÕ·¨È·¶¨ÓлúÎïµÄ×é³É£®ÕâÖÖ·½·¨ÊÇÔڵ篼ÓÈÈʱÓô¿ÑõÑõ»¯¹ÜÄÚÑùÆ·£¬¸ù¾Ý²úÎïÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É£¬×°ÖÃÈçÏÂͼËùʾ£¬ÊÇÓÃȼÉÕ·¨È·¶¨ÓлúÎﻯѧʽ³£ÓõÄ×°Öã®

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²úÉúµÄÑõÆø°´´Ó×óµ½ÓÒÁ÷Ïò£¬ËùÑ¡Ôñ×°Öø÷µ¼¹ÜµÄÁ¬½Ó˳ÐòÊÇ
g f e h i c d£¨d£¬c£©ab£¨b£¬a£©
g f e h i c d£¨d£¬c£©ab£¨b£¬a£©
£®
£¨2£©C×°ÖÃÖÐŨH2SO4µÄ×÷ÓÃΪ
ÎüÊÕË®·Ö¡¢¸ÉÔïÑõÆø
ÎüÊÕË®·Ö¡¢¸ÉÔïÑõÆø
£®
£¨3£©D×°ÖÃÖÐMnO2µÄ×÷ÓÃΪ
´ß»¯¼Á¡¢¼Ó¿ì²úÉúO2µÄËÙÂÊ
´ß»¯¼Á¡¢¼Ó¿ì²úÉúO2µÄËÙÂÊ
£®
£¨4£©EÖÐCuOµÄ×÷ÓÃΪ
ʹÓлúÎï¸ü³ä·ÖÑõ»¯ÎªCO2¡¢H2O
ʹÓлúÎï¸ü³ä·ÖÑõ»¯ÎªCO2¡¢H2O
£®
£¨5£©Èô׼ȷ³ÆÈ¡0.90gÑùÆ·£¨Ö»º¬C¡¢H¡¢OÈýÖÖÔªËØÖеÄÁ½ÖÖ»òÈýÖÖ£©¾­³ä·ÖȼÉÕºó£¬A¹ÜÖÊÁ¿Ôö¼Ó1.32g£¬B¹ÜÖÊÁ¿Ôö¼Ó0.54g£¬Ôò¸ÃÓлúÎï×î¼òʽΪ
CH2O
CH2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

µç¯¼ÓÈÈʱÓô¿O2Ñõ»¯¹ÜÄÚÑùÆ·£¬¸ù¾Ý²úÎïµÄÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É£®ÏÂÁÐ×°ÖÃÊÇÓÃȼÉÕ·¨È·¶¨ÓлúÎï·Ö×Óʽ³£ÓõÄ×°Öã®

£¨1£©²úÉúµÄO2°´´Ó×óµ½ÓÒµÄÁ÷Ïò£¬ËùѡװÖø÷µ¼¹ÜµÄÕýÈ·Á¬½Ó˳ÐòÊÇ
g¡úf¡úe¡úh¡úi¡úc£¨»òd£©¡úd£¨»òc£©¡úa£¨»òb£©¡úb£¨»òa£©
g¡úf¡úe¡úh¡úi¡úc£¨»òd£©¡úd£¨»òc£©¡úa£¨»òb£©¡úb£¨»òa£©
£®
£¨2£©C×°ÖÃÖÐŨÁòËáµÄ×÷ÓÃÊÇ
ÎüÊÕË®·Ö¡¢¸ÉÔïÑõÆø
ÎüÊÕË®·Ö¡¢¸ÉÔïÑõÆø
£®
£¨3£©D×°ÖÃÖÐMnO2µÄ×÷ÓÃÊÇ
×÷´ß»¯¼Á¡¢¼Ó¿ì²úÉúO2µÄËÙÂÊ
×÷´ß»¯¼Á¡¢¼Ó¿ì²úÉúO2µÄËÙÂÊ
£®
£¨4£©È¼ÉÕ¹ÜÖÐCuOµÄ×÷ÓÃÊÇ
ʹÓлúÎï¸ü³ä·ÖÑõ»¯ÎªCO2¡¢H2O£»
ʹÓлúÎï¸ü³ä·ÖÑõ»¯ÎªCO2¡¢H2O£»
£®
£¨5£©ÈôʵÑéÖÐËùÈ¡ÑùÆ·Ö»º¬C¡¢H¡¢OÈýÖÖÔªËØÖеÄÁ½ÖÖ»òÈýÖÖ£¬×¼È·³ÆÈ¡0.92gÑùÆ·£¬¾­³ä·Ö·´Ó¦ºó£¬A¹ÜÖÊÁ¿Ôö¼Ó1.76g£¬B¹ÜÖÊÁ¿Ôö¼Ó1.08g£¬Ôò¸ÃÑùÆ·µÄʵÑéʽΪ
C2H6O
C2H6O
£®
£¨6£©ÓÃÖÊÆ×ÒDzⶨÆäÏà¶Ô·Ö×ÓÖÊÁ¿£¬µÃÈçͼһËùʾµÄÖÊÆ×ͼ£¬Ôò¸ÃÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª
46
46
£®
£¨7£©ÄÜ·ñ¸ù¾ÝAµÄʵÑéʽȷ¶¨AµÄ·Ö×Óʽ
ÄÜ
ÄÜ
£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©£¬ÈôÄÜ£¬ÔòAµÄ·Ö×ÓʽÊÇ
C2H6O
C2H6O
£¨Èô²»ÄÜ£¬Ôò´Ë¿Õ²»Ì£®
£¨8£©¸ÃÎïÖʵĺ˴ʲÕñÇâÆ×Èçͼ¶þËùʾ£¬ÔòÆä½á¹¹¼òʽΪ
CH3CH2OH
CH3CH2OH
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§Éϳ£ÓÃȼÉÕ·¨È·¶¨ÓлúÎïµÄ×é³É£¬ÈçÏÂͼËùʾװÖÃÊÇÓÃȼÉÕ·¨È·¶¨Ìþ»òÌþµÄº¬ÑõÑÜÉúÎï·Ö×ÓʽµÄ³£ÓÃ×°Öã¬ÕâÖÖ·½·¨ÊÇÔڵ篼ÓÈÈʱÓô¿ÑõÑõ»¯¹ÜÄÚÑùÆ·£¬¸ù¾Ý²úÎïµÄÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É¡£

Íê³ÉÏÂÁÐÎÊÌ⣺

(1)AÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________________________________¡£

(2)B×°ÖõÄ×÷ÓÃÊÇ_______________________________________________£¬È¼ÉÕ¹ÜCÖÐCuOµÄ×÷ÓÃÊÇ______________________________________________________¡£

(3)²úÉúÑõÆø°´´Ó×óÏòÓÒÁ÷Ïò£¬È¼ÉÕ¹ÜCÓë×°ÖÃD¡¢EµÄÁ¬½Ó˳ÐòÊÇ£ºC¡ú________¡ú__________¡£

(4)·´Ó¦½áÊøºó£¬´ÓA×°ÖÃ×¶ÐÎÆ¿ÄÚµÄÊ£ÓàÎïÖÊÖзÖÀë³ö¹ÌÌ壬ÐèÒª½øÐеÄʵÑé²Ù×÷ÊÇ____________________________________________________________________¡£

(5)׼ȷ³ÆÈ¡1.8 gÌþµÄº¬ÑõÑÜÉúÎïXµÄÑùÆ·£¬¾­³ä·ÖȼÉÕºó£¬D¹ÜÖÊÁ¿Ôö¼Ó2.64 g£¬E¹ÜÖÊÁ¿Ôö¼Ó1.08 g£¬Ôò¸ÃÓлúÎïµÄʵÑéʽÊÇ___________________________¡£ÊµÑé²âµÃXµÄÕôÆøÃܶÈÊÇͬÎÂͬѹÏÂÇâÆøÃܶȵÄ45±¶£¬ÔòXµÄ·Ö×ÓʽΪ_________________________£¬1 mol X·Ö±ðÓë×ãÁ¿Na¡¢NaHCO3·´Ó¦·Å³öµÄÆøÌåÔÚÏàͬÌõ¼þϵÄÌå»ý±ÈΪ1¡Ã1£¬X¿ÉÄܵĽṹ¼òʽΪ____________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º×¨ÏîÌâ ÌâÐÍ£ºÊµÑéÌâ

»¯Ñ§Éϳ£ÓÃȼÉÕ·¨È·¶¨ÓлúÎïµÄ×é³É¡£ÕâÖÖ·½·¨ÊÇÔڵ篼ÓÈÈʱÓô¿ÑõÑõ»¯¹ÜÄÚÑùÆ·£¬¸ù¾Ý²úÎïÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É£¬×°ÖÃÈçÏÂͼËùʾ£¬ÊÇÓÃȼÉÕ·¨È·¶¨ÓлúÎﻯѧʽ³£ÓõÄ×°Öá£
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)²úÉúµÄÑõÆø°´´Ó×óµ½ÓÒÁ÷Ïò£¬ËùÑ¡Ôñ×°Öø÷µ¼¹ÜµÄÁ¬½Ó˳ÐòÊÇ____________¡£
(2)C×°ÖÃÖÐŨH2SO4µÄ×÷ÓÃΪ____________¡£
(3)D×°ÖÃÖÐMnO2µÄ×÷ÓÃΪ____________¡£
(4)EÖÐCuOµÄ×÷ÓÃΪ______________¡£
(5)Èô׼ȷ³ÆÈ¡0.90 gÑùÆ·(Ö»º¬C¡¢H¡¢OÈýÖÖÔªËØÖеÄÁ½ÖÖ»òÈýÖÖ)¾­³ä·ÖȼÉÕºó£¬A¹ÜÖÊÁ¿Ôö¼Ó1.32 g£¬B¹ÜÖÊÁ¿Ôö¼Ó0.54 g£¬Ôò¸ÃÓлúÎï×î¼òʽΪ______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

  »¯Ñ§Éϳ£ÓÃȼÉÕ·¨È·¶¨ÓлúÎïµÄ×é³É¡£ÈçͼËùʾװÖÃÊÇÓÃȼÉÕ·¨È·¶¨Ìþ»òÌþµÄº¬ÑõÑÜÉúÎï·Ö×ÓʽµÄ³£ÓÃ×°Öã¬ÕâÖÖ·½·¨ÊÇÔڵ篼ÓÈÈʱÓô¿ÑõÑõ»¯¹ÜÄÚÑùÆ·£¬¸ù¾Ý²úÎïµÄÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©AÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________________________________¡£

£¨2£©B×°ÖõÄ×÷ÓÃÊÇ_____________£¬È¼ÉÕ¹ÜCÖÐCuOµÄ×÷ÓÃÊÇ______________¡£

£¨3£©²úÉúÑõÆø°´´Ó×óÏòÓÒÁ÷Ïò£¬È¼ÉÕ¹ÜCÓë×°ÖÃD¡¢EµÄÁ¬½Ó˳ÐòÊÇ£ºC______________¡£

£¨4£©·´Ó¦½áÊøºó£¬´ÓA×°ÖÃ×¶ÐÎÆ¿ÄÚµÄÊ£ÓàÎïÖÊÖзÖÀë³ö¹ÌÌ壬ÐèÒª½øÐеÄʵÑé²Ù×÷ÊÇ

________________________¡£

£¨5£©×¼È·³ÆÈ¡1.8gÌþµÄº¬ÑõÑÜÉúÎïXµÄÑùÆ·£¬¾­³ä·ÖȼÉÕºó£¬D¹ÜÖÊÁ¿Ôö¼Ó2.64g£¬E¹ÜÖÊÁ¿Ôö¼Ó1.08g£¬Ôò¸ÃÓлúÎïµÄʵÑéʽÊÇ____________________¡£ÊµÑé²âµÃXµÄÕôÆøÃܶÈÊÇͬÎÂͬѹÏÂÇâÆøÃܶȵÄ45±¶£¬ÔòXµÄ·Ö×ÓʽΪ______________£¬1molX·Ö±ðÓë×ãÁ¿Na¡¢NaHCO3·´Ó¦·Å³öµÄÆøÌåÔÚÏàͬÌõ¼þϵÄÌå»ý±ÈΪ1¡Ã1£¬XËùÓпÉÄܵĽṹ¼òʽΪ_____________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸