Çë»Ø´ðÏÂÁÐÎÊÌ⣮
£¨1£©Ë®ÖдæÔÚƽºâ£ºH2O?H++OH-¡÷H£¾0£¬ÓûʹƽºâÓÒÒÆ£¬ÇÒÈÜÒºÏÔËáÐÔ£¬¿ÉÑ¡µÄ·½·¨ÊÇ______£®£¨Ñ¡ÌîÐòºÅ£©
a£®ÏòË®ÖмÓÈëNaHSO4¹ÌÌå           b£®ÏòË®ÖмÓNa2CO3¹ÌÌå
c£®¼ÓÈÈÖÁ100¡æd£®ÏòË®ÖмÓÈë £¨NH4£©2SO4¹ÌÌå
£¨2£©A¡¢B¡¢C¡¢D 4ÖÖÒ×ÈÜÓÚË®µÄÇ¿µç½âÖÊÔÚË®ÖпɵçÀë²úÉúÏÂÁÐÀë×Ó£¨¸÷ÖÖÀë×Ó²»Öظ´£©£®
ÑôÀë×Ó Ba2+¡¢H+¡¢Na+¡¢Mg2+
ÒõÀë×Ó Cl-¡¢SO42-¡¢CO32-¡¢OH-
¢Ù²»½øÐÐÈκÎʵÑé¼´¿ÉÅжϣ¬¸Ã4ÖÖ»¯ºÏÖÐÓÐ2ÖÖÊÇ______¡¢______£¨Ìѧʽ£©£¬¸Ã2ÖÖ»¯ºÏÎïµÄÈÜÒº»ìºÏʱ·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£®
¢ÚΪȷ¶¨A¡¢B¡¢C¡¢DµÄ³É·Ö£¬Ä³Ì½¾¿ÐÔѧϰС×é½øÐÐÏÂÁÐʵÑ飬ÇëÍê³ÉʵÑé½áÂÛ£º
ʵÑéÐòºÅ ʵÑéÄÚÈÝ ÊµÑé½áÂÛ
1 ÓÃpHÊÔÖ½²âµÃA¡¢CÁ½ÈÜÒº³Ê¼îÐÔ£¬B¡¢DÁ½ÈÜÒº³ÊËáÐÔ
2 AÈÜÒºÓëBÈÜÒº»ìºÏʱ£¬ÓÐÎÞÉ«ÆøÅݲúÉú »¯ºÏÎïAΪ______£¨Ìѧʽ£©BÈÜÒºÖп϶¨´óÁ¿º¬ÓÐ______£¨ÌîÀë×Ó·ûºÅ£©£»
3 CÓëDÁ½ÈÜÒº»ìºÏʱ£¬Óа×É«³ÁµíÉú³É£¬¹ýÂË£¬Ïò³ÁµíÖеμӹýÁ¿ÑÎËᣬ³Áµí²¿·ÖÈܽ⠻¯ºÏÎïBΪ______£¬»¯ºÏÎïDΪ______£¨Ìѧʽ£©£®
£¨1£©Ë®ÖдæÔÚƽºâ£ºH2O?H++OH-¡÷H£¾0£¬ÓûʹƽºâÓÒÒÆ£¬ÇÒÈÜÒºÏÔËáÐÔ£¬ËµÃ÷ÊÇÑôÀë×ÓË®½â½áºÏÇâÑõ¸ùÀë×ÓÉú³ÉÈõµç½âÖÊ£¬ÈÜÒº³ÊËáÐÔ£»
a¡¢ÏòË®ÖмÓÈëNaHSO4¹ÌÌ壬ÒÖÖÆË®µÄµçÀ룬ƽºâÄæÏò½øÐУ¬¹Êa²»·ûºÏ£»
b£®ÏòË®ÖмÓNa2CO3¹ÌÌ壬̼Ëá¸ùÀë×ÓË®½â´Ù½øË®µÄµçÀ룬ÈÜÒº³Ê¼îÐÔ£¬¹Êb²»·ûºÏ£»
c£®¼ÓÈÈÖÁ100¡æ£¬´Ù½øË®µÄµçÀ룬ÈÜÒºÊÇÖÐÐÔ£¬¹Êc²»·ûºÏ£»
d£®ÏòË®ÖмÓÈë £¨NH4£©2SO4¹ÌÌ壬笠ùÀë×ÓË®½â´Ù½øË®µÄµçÀ룬ÈÜÒº³ÊËáÐÔ£¬¹Êd·ûºÏ£»
¹ÊÑ¡d£®
£¨2£©¢ÙA¡¢B¡¢C¡¢D 4ÖÖÒ×ÈÜÓÚË®µÄÇ¿µç½âÖÊÔÚË®ÖпɵçÀë²úÉúÏÂÁÐÀë×Ó£¬ÒÀ¾ÝÀë×Ó¹²´æ·ÖÎöÅжϣ¬ÒõÀë×ÓÖÐ̼Ëá¸ùÀë×ÓÖ»ÄܺÍÄÆÀë×Ó×é³ÉÎïÖÊ̼ËáÄÆ£»ÇâÑõ¸ùÀë×ÓÖ»ÄܺͱµÀë×Ó×é³ÉÎïÖÊÇâÑõ»¯±µ£»¸Ã2ÖÖ»¯ºÏÎïµÄÈÜÒº»ìºÏʱ·´Ó¦µÄÀë×Ó·½³Ìʽ£ºBa2++CO32-=BaCO3¡ý£¬¹Ê´ð°¸Îª£ºNa2CO3£»Ba£¨OH£©2 £»Ba2++CO32-=BaCO3¡ý
¢ÚÓÃpHÊÔÖ½²âµÃA¡¢CÁ½ÈÜÒº³Ê¼îÐÔ£¬ËùÒÔΪNa2CO3ºÍBa£¨OH£©2 £»B¡¢DÁ½ÈÜÒº³ÊËáÐÔΪMgCl2¡¢H2SO4»òMgSO4¡¢HCl£»
AÈÜÒºÓëBÈÜÒº»ìºÏʱ£¬ÓÐÎÞÉ«ÆøÅݲúÉú£¬Ö¤Ã÷AΪNa2CO3£»BΪËẬÓÐH+£»CΪBa£¨OH£©2 £»
CÓëDÁ½ÈÜÒº»ìºÏʱ£¬Óа×É«³ÁµíÉú³É£¬¹ýÂË£¬Ïò³ÁµíÖеμӹýÁ¿ÑÎËᣬ³Áµí²¿·ÖÈܽ⣬ÈܽâµÄ³ÁµíÊÇÇâÑõ»¯Ã¾³Áµí£¬²»ÈܽâµÄÊÇÁòËá±µ£¬ÅжÏDÖÐΪMgSO4£»BΪHCl£»
¹Ê´ð°¸Îª£ºNa2CO3 £»H+£»HCl£»MgSO4£»
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¢ñÏÖÓûÓÃ̼Ëá¸Æ¹ÌÌåºÍÏ¡ÑÎËá·´Ó¦ÖÆÈ¡CO2ÆøÌ壬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³ö·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ
CaCO3+2H+=Ca2++H2O+CO2¡ü
CaCO3+2H+=Ca2++H2O+CO2¡ü
£»
£¨2£©ÊµÑé¹ý³ÌÖлæÖƳöÉú³ÉCO2µÄÌå»ý[V£¨CO2£©]Óëʱ¼ä£¨t£©µÄ¹ØϵÈçͼËùʾ£¬ÊÔ·ÖÎöÅжÏOE¶Î¡¢EF¶Î¡¢FG¶Î·´Ó¦ËÙÂÊ£¨·Ö±ðÓæԣ¨OE£©¡¢¦Ô£¨EF£©¡¢¦Ô£¨FG£©±íʾ£©Äĸö×î¿ì
¦Ô£¨EF£©
¦Ô£¨EF£©
£»
±È½ÏOE¶ÎºÍEF¶Î£¬ËµÃ÷EF¶ÎËÙÂʱ仯µÄÖ÷ÒªÔ­Òò¿ÉÄÜÊÇ
ζȽϸߡ¢Å¨¶È½Ï´ó
ζȽϸߡ¢Å¨¶È½Ï´ó

¢ò·´Ó¦A+3B=2C+2D£®ÔÚËÄÖÖ²»Í¬µÄÇé¿öϵķ´Ó¦ËÙÂÊ·Ö±ðΪ£º
¢Ù¦Ô£¨A£©=0.15mol/£¨L?s£©     ¢Ú¦Ô£¨B£©=0.6mol/£¨L?s£©
¢Û¦Ô£¨C£©=0.4mol/£¨L?s£©      ¢Ü¦Ô£¨D£©=0.45mol/£¨L?s£©
¸Ã·´Ó¦½øÐеĿìÂý˳ÐòΪ
¢Ü£¾¢Ú=¢Û£¾¢Ù
¢Ü£¾¢Ú=¢Û£¾¢Ù
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÖÓÐA¡¢BÀöÖÖÌþ£¬ÒÑÖªAµÄ·Ö×ÓʽΪC5Hm£¬¶øBµÄ×î¼òʽΪC5Hn£¨m¡¢n¾ùΪÕýÕûÊý£©£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÏÂÁйØÓÚÌþAºÍÌþBµÄ˵·¨²»ÕýÈ·µÄÊÇ
d
d
£¨ÌîÐòºÅ£©£®
a£®ÌþAºÍÌþB¿ÉÄÜ»¥ÎªÍ¬ÏµÎï
b£®ÌþAºÍÌþB¿ÉÄÜ»¥ÎªÍ¬·ÖÒì¹¹Ìå
c£®µ±m=12ʱ£¬ÌþAÒ»¶¨ÎªÍéÌþ
d£®µ±n=11ʱ£¬ÌþB¿ÉÄܵķÖ×ÓʽÓÐ2ÖÖ
£¨2£©ÈôÌþAΪÁ´Ìþ£¬ÇÒ·Ö×ÓÖÐËùÓÐ̼ԭ×Ó¶¼ÔÚͬһÌõÖ±ÏßÉÏ£¬ÔòAµÄ½á¹¹¼òʽΪ
CH¡ÔC-C¡ÔC-CH3
CH¡ÔC-C¡ÔC-CH3
£®
£¨3£©ÈôÌþAΪÁ´Ìþ£¬ÇÒ·Ö×ÓÖÐËùÓÐ̼ԭ×ÓÒ»¶¨¹²Ã棬ÔÚÒ»¶¨Ìõ¼þÏ£¬1mol A×î¶à¿ÉÓë1mol H2¼Ó³É£¬ÔòAµÄÃû³ÆÊÇ
3-¼×»ù-1-¶¡Ï©
3-¼×»ù-1-¶¡Ï©
£®
£¨4£©ÈôÌþBΪ±½µÄͬϵÎȡһ¶¨Á¿µÄÌþBÍêȫȼÉÕºó£¬Éú³ÉÎïÏÈͨ¹ý×ãÁ¿µÄŨÁòËᣬŨÁòËáµÄÖÊÁ¿Ôö¼Ó1.26g£¬ÔÙͨ¹ý×ãÁ¿µÄ¼îʯ»Ò£¬¼îʯ»ÒµÄÖÊÁ¿Ôö¼Ó4.4g£¬ÔòÌþBµÄ·Ö×ÓʽΪ
C10H14
C10H14
£»ÈôÆä±½»·ÉϵÄÒ»äå´úÎïÖ»ÓÐÒ»ÖÖ£¬Ôò·ûºÏ´ËÌõ¼þµÄÌþBÓÐ
4
4
ÖÖ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¿×ȸʯÖ÷Òªº¬Cu2£¨OH£©2CO3£¬»¹º¬ÉÙÁ¿Fe¡¢SiµÄ»¯ºÏÎʵÑéÊÒÒÔ¿×ȸʯΪԭÁÏÖƱ¸CuSO4?5H2O¼°CaCO3£¬²½ÖèÈçÏ£º
Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈÜÒºAµÄ½ðÊôÀë×ÓÓÐCu2+¡¢Fe2+¡¢Fe3+£®´ÓÏÂÁÐËù¸øÊÔ¼ÁÖÐÑ¡Ôñ£ºÊµÑé²½ÖèÖÐÊÔ¼Á¢ÙΪ
c
c
£¨Ìî´úºÅ£©£¬¼ìÑéÈÜÒºAÖÐFe3+µÄ×î¼ÑÊÔ¼ÁΪ
d
d
£¨Ìî´úºÅ£©£®
a£®KMnO4      b£®£¨NH4£©2S       c£®H2O2      d£®KSCN
£¨2£©ÓÉÈÜÒºC»ñµÃCuSO4?5H2O£¬ÐèÒª¾­¹ý¼ÓÈÈÕô·¢¡¢
ÀäÈ´½á¾§
ÀäÈ´½á¾§
¡¢¹ýÂ˵ȲÙ×÷£®¹ýÂËʱΪÁË·ÀÖ¹ÂËÒº·É½¦£¬Ó¦
ʹ©¶·Ï¶˲£Á§¹ÜÓëÉÕ±­±Ú½ô¿¿
ʹ©¶·Ï¶˲£Á§¹ÜÓëÉÕ±­±Ú½ô¿¿
£®
£¨3£©ÖƱ¸CaCO3ʱ£¬ÈôʵÑé¹ý³ÌÖÐÓа±ÆøÒݳö£¬Ó¦Ñ¡ÓÃÏÂÁÐ
b
b
×°ÖÃÎüÊÕ£¨Ìî´úºÅ£©£®

£¨4£©Óû²â¶¨ÈÜÒºAÖÐFe2+µÄŨ¶È£¬ÐèÒªÓÃÈÝÁ¿Æ¿ÅäÖÆKMnO4±ê×¼ÈÜÒº½øÐÐÑõ»¯»¹Ô­µÎ¶¨£¬ÅäÖÆʱµÈÒºÃæ½Ó½üÈÝÁ¿Æ¿¿Ì¶ÈÏßʱ£¬Ó¦¸Ã¼ÌÐø½øÐеIJÙ×÷ÊÇ
ÓýºÍ·µÎ¹ÜÏòÈÝÁ¿Æ¿ÖеμÓÕôÁóË®£¬ÖÁÈÜÒºµÄ°¼ÒºÃæÕýºÃÓë¿Ì¶ÈÏßÏàƽ
ÓýºÍ·µÎ¹ÜÏòÈÝÁ¿Æ¿ÖеμÓÕôÁóË®£¬ÖÁÈÜÒºµÄ°¼ÒºÃæÕýºÃÓë¿Ì¶ÈÏßÏàƽ
£®µÎ¶¨Ê±ÐèÒªÓÃÓÃKMnO4±ê×¼ÈÜÒºÈóÏ´µÎ¶¨¹Ü£¬ÈóÏ´ºóµÄ·ÏÒºÓ¦´ÓËáʽµÎ¶¨¹ÜµÄ
Ï¿Ú
Ï¿Ú
Åųö£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡¢B¡¢C¡¢D¡¢EÎåÖÖ¶ÌÖÜÆÚÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£®BÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇÆä´ÎÍâ²ãµç×ÓÊýµÄ2±¶£»AµÄÒ»ÖÖÔ­×ÓÖУ¬ÖÊÁ¿ÊýÓëÖÊ×ÓÊýÖ®²îΪÁ㣮DÔªËصÄÔ­×Ó×îÍâ²ãµç×ÓÊýΪm£¬´ÎÍâ²ãµç×ÓÊýΪn£»EÔªËصÄÔ­×ÓL²ãµç×ÓÊýΪm+n£¬M²ãµç×ÓÊýΪ
m2
-n
£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©BÔªËØÊÇ
̼
̼
£¬DÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ
µÚ¶þÖÜÆÚµÚ¢öA×å
µÚ¶þÖÜÆÚµÚ¢öA×å
£»
£¨2£©CÓëEÐγɵĻ¯ºÏÎïE3CÊôÓÚ
Àë×Ó
Àë×Ó
¾§Ì壨Ìî¡°Ô­×Ó¡±¡¢¡°Àë×Ó¡±»ò¡°·Ö×Ó¡±£©£»
£¨3£©ÓÉA¡¢D¡¢EÔªËØ×é³ÉµÄ»¯ºÏÎïÖдæÔÚµÄ×÷ÓÃÁ¦ÊÇ
Àë×Ó¼ü¡¢¹²¼Û¼ü
Àë×Ó¼ü¡¢¹²¼Û¼ü
£»
£¨4£©Ð´³öÒ»¸öEºÍDÐγɵĻ¯ºÏÎïÓëË®·´Ó¦µÄ»¯Ñ§·½³Ìʽ
Na2O+H2O=2NaOH»ò2Na2O2+2H2O=4NaOH+O2¡ü
Na2O+H2O=2NaOH»ò2Na2O2+2H2O=4NaOH+O2¡ü
£»
£¨5£©CµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÓëÆäÇ⻯Îï·´Ó¦Éú³ÉÒ»ÖÖÑÎX£¬XµÄË®ÈÜÒºÏÔ
Ëá
Ëá
ÐÔ£¨Ìî¡°Ëᡱ¡¢¡°¼î¡±»ò¡°ÖС±£©£¬ÆäÔ­ÒòÓÃÀë×Ó·½³Ìʽ±íʾ£º
NH4++H2O?NH3?H2O+H+
NH4++H2O?NH3?H2O+H+
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¹¤ÒµÉÏÓÃÂÁÍÁ¿ó£¨º¬Ñõ»¯ÂÁ¡¢Ñõ»¯Ìú£©ÖÆÈ¡ÂÁµÄ¹ý³ÌÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©³ÁµíCµÄ»¯Ñ§Ê½Îª
Fe2O3
Fe2O3
£¬¸ÃÎïÖÊÓÃ;³ýÁËÓÃÓÚ½ðÊôÒ±Á¶ÒÔÍ⣬»¹¿ÉÓÃ×÷
ÑÕÁÏ
ÑÕÁÏ
£®
£¨2£©Éú²ú¹ý³ÌÖУ¬³ýNaOH¡¢H2O¿ÉÒÔÑ­»·Ê¹ÓÃÍ⣬»¹¿ÉÒÔÑ­»·Ê¹ÓõÄÎïÖÊÓÐ
CaOºÍCO2
CaOºÍCO2
£¨Ìѧʽ£©£®Óô˷¨ÖÆÈ¡ÂÁµÄ¸±²úÆ·ÊÇ
Fe2O3ºÍO2
Fe2O3ºÍO2
£¨Ìѧʽ£©£®
£¨3£©²Ù×÷¢ñ¡¢²Ù×÷¢òºÍ²Ù×÷¢ó¶¼ÊÇ
¹ýÂË
¹ýÂË
£¨Ìî²Ù×÷Ãû³Æ£©£¬ÊµÑéÊÒҪϴµÓAl£¨OH£©3³ÁµíÓ¦¸ÃÔÚ
¹ýÂË
¹ýÂË
×°ÖÃÖнøÐУ¬Ï´µÓ·½·¨ÊÇ
Ïò©¶·ÖмÓÕôÁóË®ÖÁ½þû³Áµí£¬Ê¹Ë®×ÔÈ»Á÷Í꣬Öظ´²Ù×÷2¡«3´Î
Ïò©¶·ÖмÓÕôÁóË®ÖÁ½þû³Áµí£¬Ê¹Ë®×ÔÈ»Á÷Í꣬Öظ´²Ù×÷2¡«3´Î
£®
£¨4£©µç½âÈÛÈÚµÄÑõ»¯ÂÁʱ£¬ÈôµÃµ½±ê×¼×´¿öÏÂ22.4L O2£¬ÔòͬʱÉú³ÉÂÁµÄÖÊÁ¿Îª
36g
36g
£®
£¨5£©Ð´³öNa2CO3ÈÜÒºÓëCaO·´Ó¦µÄÀë×Ó·½³Ìʽ£º
CO32-+CaO+H2O¨TCaCO3¡ý+2OH-
CO32-+CaO+H2O¨TCaCO3¡ý+2OH-
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸