6£®1£¬2£¬3£¬4-ËÄÇ⻯ÝÁµÄ½á¹¹¼òʽÊÇ£¬·Ö×ÓʽÊÇC10H12£®³£ÎÂÏÂΪÎÞÉ«ÒºÌ壬Óд̼¤ÐÔÆø棬·Ðµã207¡æ£¬²»ÈÜÓÚË®£¬ÊÇÒ»ÖÖÓÅÁ¼µÄÈܼÁ£¬ËüÓëÒºäå·¢Éú·´Ó¦£ºC10H12+4Br2?C10H8Br4+4HBr£®Éú³ÉµÄËÄä廯ÝÁ³£ÎÂÏÂΪ¹Ì̬£¬²»ÈÜÓÚË®£®ÓÐÈËÓÃËÄÇ⻯ÝÁ¡¢Òºäå¡¢ÕôÁóË®ºÍ´¿Ìú·ÛΪԭÁÏ£¬ÖƱ¸ÉÙÁ¿±¥ºÍÇâäåËáÈÜÒº£¬ÊµÑé²½ÖèÈçÏ£º
¢Ù°´Ò»¶¨ÖÊÁ¿±È°ÑËÄÇ⻯ÝÁºÍË®¼ÓÈëÊʵ±µÄÈÝÆ÷ÖУ¬¼ÓÈëÉÙÁ¿´¿Ìú·Û£®
¢ÚÂýÂýµÎÈëÒºä壬²»¶Ï½Á°è£¬Ö±µ½·´Ó¦ÍêÈ«£®
¢ÛÈ¡Ï·´Ó¦ÈÝÆ÷£¬²¹³äÉÙÁ¿ËÄÇ⻯ÝÁ£¬Ö±µ½ÈÜÒºÑÕÉ«Ïûʧ£®¹ýÂË£¬½«ÂËÒºµ¹Èë·ÖҺ©¶·£¬¾²Öã®
¢Ü·ÖÒº£¬µÃµ½µÄ¡°Ë®²ã¡±¼´ÇâäåËáÈÜÒº£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÈçͼʾÒâͼÖеÄ×°Öã¬Êʺϲ½Öè¢ÙºÍ¢Ú²Ù×÷µÄÊÇD£®
£¨2£©²½Öè¢ÚÖÐÈçºÎÅжϡ°·´Ó¦ÍêÈ«¡±¼ÓÈëµÄÒºäåÑÕÉ«»ù±¾²»ÍÊ£®
£¨3£©²½Öè¢ÛÖв¹³äÉÙÁ¿ËÄÇ⻯ÝÁµÄÄ¿µÄÊdzý³öÈ¥¹ýÁ¿µÄBr2£®
£¨4£©²½Öè¢ÛÖйýÂ˺óµÃµ½µÄ¹ÌÌåÎïÖÊÊÇËÄä廯ÝÁ¡¢Ìú·Û£®
£¨5£©ÒÑÖªÔÚʵÑéÌõ¼þÏ£¬±¥ºÍÇâäåËáË®ÈÜÒºÖÐÇâäåËáµÄÖÊÁ¿·ÖÊýÊÇ66%£¬Èç¹ûäå
»¯·´Ó¦½øÐÐÍê³É£¬Ôò²½Öè¢ÙÖÐËÄÇ⻯ÝÁºÍË®µÄÖÊÁ¿±ÈÔ¼ÊÇ1£º1.3£¨±£ÁôСÊýµãºó1룩£®

·ÖÎö ±¾ÌâÊÇÀûÓÃËÄÇ⻯ÝÁµÄÐÔÖÊÖƱ¸±¥ºÍÇâäåËáÈÜÒºµÄʵÑé̽¾¿£¬Éæ¼°¸ù¾Ý·´Ó¦ÎïÒºäåºÍÉú³ÉÎïä廯ÇâÒ×»Ó·¢µÄÌصãÑ¡Ôñ·´Ó¦×°ÖÃD½øÐÐʵÑ飬ͨ¹ýµÎ¼ÓµÄÒºäåÑÕÉ«²»ÔÙÍÊÈ¥ÅжϷ´Ó¦ÍêÈ«£¬¿¼²éÁË·´Ó¦ºó»ìºÏÎïµÄ·ÖÀëÓëÌá´¿£¬²¢½áºÏ²úÆ·µÄÖÊÁ¿·ÖÊý¼ÆËãÔ­ÁϵÄÅä±È£¬½¨Òé½èÖúÔ­×ÓÊغã½øÐмÆË㣬¾Ý´ËÅжϽâ´ð£»
£¨1£©¸ù¾Ý¢Ù¢ÚµÄ²Ù×÷¼°Ê¹ÓÃÊÔ¼ÁÒ×»Ó·¢µÄÌصãÅжϺÏÀíµÄ×°Öã»
£¨2£©ÒºäåÊÇÉîºìºÖÉ«£¬×î³õµÎÈëµ½·´Ó¦ÈÝÆ÷ÄÚÁ¢¼´²ÎÓë·´Ó¦¶øÍÊÉ«£¬Èç¹û½øÐÐÍêÈ«£¬µÎÈëµÄÒºä彫ÈÔ±£ÁôÔÚ·´Ó¦ÈÝÆ÷ÄÚ£»
£¨3£©¸ù¾Ý²½Öè¢ÚÐèÒª¼ÓÈëÁ˹ýÁ¿µÄÒºäå½øÐÐÅжϣ»
£¨4£©¸ù¾ÝËÄä廯ÝÁ³£ÎÂÏÂΪ¹Ì̬ºÍ¼ÓÈëÌú·Û×÷´ß»¯¼Á¿¼ÂÇ£»
£¨5£©±¥ºÍÇâäåËáË®ÈÜÒºÖÐÇâäåËáµÄÖÊÁ¿·ÖÊýÊÇ66%£¬¿ÉÉèÈÜÒºµÄÖÊÁ¿Îª100g£¬ÔòÈÜÒºÀïHBrµÄÖÊÁ¿Îª66g£¬¿É¼ÆËã³öÆäÎïÖʵÄÁ¿£¬ÔÙ½áºÏ·´Ó¦Ô­Àí¿ÉÍƳöËÄÇ⻯ÝÁµÄÎïÖʵÄÁ¿£¬´Ó¶ø¿É¼ÆËã³öËÄÇ⻯ÝÁÓëË®µÄÖÊÁ¿±È£®

½â´ð ½â£º£¨1£©°´ÕÕ²½Öè¢Ù£¬±¾ÊµÑéÊÇ°ÑËÄÇ⻯ÝÁ¡¢Ë®ºÍÌú·Û¼ÓÈëͬһÈÝÆ÷Öз´Ó¦£¬ËùÒÔB²»ÕýÈ·£®ÈôÓÃAÖеÄ×°ÖýøÐз´Ó¦£¬äåÕôÆøºÍ´Ì¼¤ÐÔÆøζµÄËÄÇ⻯ÝÁ¶¼»áÎÛȾ»·¾³£¬Í¬ÑùC×°ÖÃÔÚ²½Öè¢Ú·´Ó¦Ê±£¬Ò²»áÔì³É»·¾³ÎÛȾ£¬ËùÒÔA¡¢C¶¼²»¶Ô£¬¹ÊÑ¡D×°Ö㬹ʴð°¸Îª£ºD£»
£¨2£©°éËæ·´Ó¦µÄ½øÐУ¬×î³õµÎÈëµÄºìºÖÉ«ÒºäåÁ¢¼´ÍÊÉ«£¬µ±·´Ó¦ÍêÈ«ºó£¬µÎÈëµÄÒºäåÑÕÉ«²»ÔÙÍÊÈ¥£¬¹Ê´ð°¸Îª£º¼ÓÈëµÄÒºäåÑÕÉ«»ù±¾²»ÍÊ£»
£¨3£©ÓÉÓÚ²½Öè¢Ú¼ÓÈëÁ˹ýÁ¿µÄÒºä壬²½Öè¢ÛÖв¹³äÉÙÁ¿ËÄÇ⻯ÝÁ³ýµôÊ£ÓàµÄäåµ¥ÖÊ£¬¹Ê´ð°¸Îª£º³ý³öÈ¥¹ýÁ¿µÄBr2£»
£¨4£©ÓÉÓÚÉú³ÉµÄËÄä廯ÝÁ³£ÎÂÏÂΪ¹Ì̬£¬ËùÒÔ¹ýÂ˺óµÃµ½µÄ¹ÌÌåÎïÖÊÊÇËÄä廯ÝÁºÍÌú·Û£¬¹Ê´ð°¸Îª£ºËÄä廯ÝÁºÍÌú·Û£»
£¨5£©ÉèÈÜÒºµÄÖÊÁ¿Îª100g£¬ÔòÈÜÒºÀïHBrµÄÖÊÁ¿Îª66g£¬Ë®µÄÖÊÁ¿Îª34g£¬HBrµÄÎïÖʵÄÁ¿$\frac{66g}{81g/mol}$=$\frac{22}{27}$mol£¬¸ù¾ÝC10H12+4Br2?C10H8Br4+4HBr¿ÉÖªËÄÇ⻯ÝÁµÄÎïÖʵÄÁ¿Îª$\frac{22}{27}$mol¡Â4=$\frac{11}{54}$mol£¬ÖÊÁ¿Îª$\frac{11}{54}$mol¡Á134g/mol=$\frac{737}{27}$g£¬Ôò²½Öè¢ÙÖÐËÄÇ⻯ÝÁºÍË®µÄÖÊÁ¿±ÈÔ¼ÊÇ$\frac{737}{27}$g£º36g¡Ö1£º1.3£¬¹Ê´ð°¸Îª£º1.3£®

µãÆÀ ±¾Ì⿼²éÁËÓÃËÄÇ⻯ÝÁ¡¢Òºäå¡¢ÕôÁóË®ºÍ´¿Ìú·ÛΪԭÁÏ£¬ÖƱ¸ÉÙÁ¿±¥ºÍÇâäåËáÈÜÒº£¬Éæ¼°ÁË×°ÖõÄÑ¡Ôñ¡¢»¯Ñ§·½³ÌʽµÄÊéд¼°Åäƽ£¬³ä·Ö¿¼²éÁËѧÉúµÄ·ÖÎö¡¢Àí½âÄÜÁ¦¼°¶ÔËùѧ֪ʶÕÆÎÕµÄÊìÁ·³Ì¶È£¬±¾ÌâÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

5£®ÒÑ֪ij»ìºÏÎïÖк¬ÓÐAµ¥Öʼ°Ó뻯ºÏÎïB£¬ÔÚÒ»¶¨Ìõ¼þÏ¿ɰ´Í¼Ëùʾ·¢Éúת»¯£®

Çë»Ø´ð£º
£¨1£©³ÁµíH±ä³ÁµíIµÄ»¯Ñ§·½³Ìʽ4Fe£¨OH£©2+O2+2H2O¨T4Fe£¨OH£©3£»
£¨2£©·´Ó¦¢ÚµÄ»¯Ñ§·½³Ìʽ2Al+2NaOH+2H2O¨T2NaAlO2+3H2¡ü£»
£¨3£©ÈÜÒºEÖÐͨÈë×ãÁ¿¶þÑõ»¯Ì¼µÄÀë×Ó·½³ÌʽAlO2-+CO2+2H2O=Al£¨OH£©3¡ý+HCO3-£»
£¨4£©ÈçºÎ¼ìÑéÈÜÒºGÖеÄÑôÀë×ÓÈ¡Ñù£¬µÎ¼ÓÉÙÁ¿KSCNÈÜÒº£¬Èô±äѪºìÉ«£¬Ôòº¬ÓÐFe3+£»
£¨5£©ÈôÒª±£´æFÈÜÒº£¬Ó¦²Éȡʲô´ëÊ©£º¼ÓÌú·ÛºÍÏ¡ÑÎËᣮ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

17£®Ð¿Ã̷ϵç³Ø¿É»ØÊÕп¡¢ÃÌÔªËØÉú²úÁòËáп¼°Ì¼ËáÃÌ£¬ÆäÖÐÉú²ú̼ËáÃ̵ÄÖ÷Òª¹¤ÒÕÈçÏ£º

ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÁòËáпˮÈÜÒºÏÔËáÐÔ£¬ÓÃÀë×Ó·½³Ìʽ˵Ã÷Zn2++2H2O?Zn£¨OH£©2+2H+£®
£¨2£©Ô­ÁÏÃÌ·Û´ÖÆ·ÖÐÖ÷Òª³É·ÖΪMnO2ºÍÌ¿£¬ÔÚ¹ý³Ì¢Ù±ºÉÕʱ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪMnO2+C¨TMnO+CO¡ü£¬¸Ã·´Ó¦µÄÑõ»¯²úÎïÊÇCO£¬Ã¿Éú³É44.8L£¨±ê×¼×´¿öÏ£©ÆøÌåʱתÒƵç×ÓÊýΪ4NA£®
£¨3£©ÔÚ50¡«55¡æʱÏò¹ý³Ì¢ÛÖÐMnSO4µÄĸҺÖмÓÈë×ãÁ¿NH4HCO3£¬ÓÐÆøÌåÉú³É£¬Æ仯ѧ·½³ÌʽΪ£ºMnSO4+2NH4HCO3=£¨NH4£©2SO4+MnCO3¡ý+H2O+CO2¡ü£®
£¨4£©ÒÑÖªÈýÖÖÀë×Ó³ÁµíµÄpH·¶Î§ÎªFe3+£º2.7¡«3.7£® Mn2+£º8.6¡«10.1£¬Fe2+£º7.6¡«9.6£®Ï±íÊÇÉÏÊö¹ý³Ì¢ÚÖгýÈ¥Fe2+µÄÄ£Äâ²Ù×÷·½·¨£¬ÇëÍê³ÉϱíÄÚÈÝ£º
ʵÑé²Ù×÷Àë×Ó·½³Ìʽ
²½Öè1£ºÈ¡ÉÙÁ¿ÂËÒºÓÚÊÔ¹ÜÖУ¬¼ÓÈë¹ýÁ¿ËữµÄH2O2ÈÜÒº£¬Õñµ´£®
2Fe2++H2O2+2H+=2Fe3++2H2O
²½Öè2£º½«pHµ÷µ½3.7¡«8.6£¬Ê¹Fe3+³ÁµíÍêÈ«£®Fe3++3H2O?Fe£¨OH£©3+3H+

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

14£®700¡æʱ£¬ÏòÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨Á¿µÄCOºÍH2O£¬·¢Éú·´Ó¦£ºCO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£©£®·´Ó¦¹ý³ÌÖвⶨµÄ²¿·ÖÊý¾Ý¼ûÏÂ±í£¨±íÖÐt1£¾t2£©£¬ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
·´Ó¦Ê±¼ä/minn£¨CO£©/molH2O/mol
01.200.60
t10.80
t20.20
A£®·´Ó¦ÔÚt1minÄÚµÄƽ¾ùËÙÂÊΪv£¨H2£©=0.40/t1mol•L-1•min-1
B£®Î¶ÈÉýÖÁ800¡æ£¬ÉÏÊö·´Ó¦Æ½ºâ³£ÊýΪ0.64£¬ÔòÕý·´Ó¦ÎªÎüÈÈ·´Ó¦
C£®±£³ÖÆäËûÌõ¼þ²»±ä£¬ÏòƽºâÌåϵÖÐÔÙͨÈë0.20molH2O£¬ÓëԭƽºâÏà±È£¬´ïµ½ÐÂƽºâʱCOת»¯ÂʼõС
D£®±£³ÖÆäËûÌõ¼þ²»±ä£¬ÆðʼʱÏòÈÝÆ÷ÖгäÈë0.60molCOºÍ1.20 molH2O£¬µ½´ïƽºâʱ£¬n£¨CO2£©=0.40 mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

1£®ÔÚ10LÈÝÆ÷ÖУ¬¼ÓÈë2molµÄSO2£¨g£©ºÍ2molµÄNO2£¨g£©£¬±£³ÖζȺ㶨£¬·¢Éú·´Ó¦£ºSO2£¨g£©+NO2£¨g£©?SO3£¨g£©+NO£¨g£©£®µ±´ïµ½Æ½ºâ״̬ʱ£¬²âµÃÈÝÆ÷ÖÐSO2£¨g£©µÄת»¯ÂÊΪ50%£®
£¨1£©¸ÃζÈϵĸ÷´Ó¦µÄƽºâ³£Êý=1£¨ÓÃÊýÖµ±íʾ£©
£¨2£©¸ÃζÈÏ¡¢¸ÃÈÝÆ÷ÖУ¬ÔÙ¼ÌÐø¼ÓÈë1molµÄSO2£¨g£©£¬Ôò£º
¢Ù»¯Ñ§Æ½ºâ½«ÏòÕý·´Ó¦·½ÏòÒƶ¯£¬NO2µÄת»¯Âʽ«Ôö´ó£»
¢Ú¾­¼ÆË㣬µ±´ïµ½ÐµÄƽºâ״̬ʱ£¬ÈÝÆ÷ÖÐSO2£¨g£©µÄŨ¶ÈÊÇ0.18mol/L£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

11£®ÓÃÒÑÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÑÎËáµÎ¶¨Î´ÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÇâÑõ»¯ÄÆÈÜÒº£¬ÏÂÁвÙ×÷»áµ¼Ö²ⶨ½á¹ûÆ«¸ßµÄÊÇ£¨¡¡¡¡£©
A£®Óñê×¼µÄÑÎËáÈÜÒºÈóÏ´ËáʽµÎ¶¨¹Ü2-3´Î
B£®Óôý²â¼îÒºÈóÏ´¼îʽµÎ¶¨¹Ü2-3´Î
C£®Óôý²â¼îÒºÈóϴ׶ÐÎÆ¿2-3´Î
D£®ÓÃÕôÁóË®Èóϴ׶ÐÎÆ¿2-3´Î

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

18£®Íê³ÉÒÔÏ»¯Ñ§·½³Ìʽ
£¨1£©3S+6KOH¨T2K2S+1K2SO3+3H2O
£¨2£©2 KMnO4+16HBr=5Br2+2MnBr2+2KBr+8H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

15£®Îª²â¶¨NaOH¡¢Na2C03»ìºÏÎïÖÐNa2C03µÄÖÊÁ¿·ÖÊý£¬¼×ͬѧÄâ²ÉÓÃͼ£¨1£©ËùʾװÖÃÀ´²â¶¨£®ÒÒͬѧÄâ²ÉÓÃͼ£¨2£©ËùʾװÖÃÀ´²â¶¨£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¼×ͬѧÓÃͼ£¨1£©×°ÖÃÀ´²â¶¨£¬ÔÚûÓзÅÑùÆ·Ç°£¬¼ì²é¸Ã×°ÖÃÆøÃÜÐԵķ½·¨Ö®Ò»ÊÇ´ÓËáʽµÎ¶¨¹ÜÏòÉÕÆ¿ÖмÓÈëÒ»¶¨Á¿µÄË®£¬¹Û²ìµ½×¢ÉäÆ÷Ôö´óµÄÌå»ýÓë¼ÓÈëË®µÄÌå»ýÏàµÈÏÖÏó£¬Ö¤Ã÷´Ë×°Á¿²»Â©Æø£®ÊµÑéʱ³ÆµÃÑùÆ·µÄÖÊÁ¿Îªmg£¬ËáʽµÎ¶¨¹ÜµÄÆðʼ¶ÁÊýΪamL£¬ÖÕÁ˶ÁÊýΪbmL£¬×¢ÉäÆ÷²â¶¨ÅųöµÄÆøÌåΪc mL£¨ÒÑÕÛËã³É±ê×¼×´¿ö£©£¬¸ÃÑùÆ·ÖÐNa2C03µÄÖÊÁ¿·ÖÊýΪ£¨Óú¬a¡¢b¡¢c¡¢mµÄ´úÊýʽ±íʾ£©$\frac{53£¨c-b+a£©}{11200m}$£®
£¨2£©±¾ÊµÑéÑ¡ÓÃÏ¡H2S04¶øδѡÓÃÏ¡ÑÎËáµÄÔ­ÒòÊÇÏ¡ÁòËá²»Ò×»Ó·¢£¬Ï¡ÑÎËáÒ×»Ó·¢³öÂÈ»¯ÇâÆøÌ壬²úÉúÎó²î£®
£¨3£©ÒÒͬѧ¹Û²ìÁ˼×ͬѧµÄʵÑéºóºÜÊÜÆô·¢£¬²¢·¢ÏÖ×Ô¼ºµÄͼ£¨2£©Ô­ÊµÑéÉè¼Æ·½°¸»áÔì³ÉÎó²î£®ÆäÔ­ÒòÖ®Ò»¿ÉÄÜÊÇC02ÔÚË®ÖÐÈܽâ¶È½Ï´ó£¬ÒýÆð²âÁ¿ÆøÌåÌå»ý¼õС£®±ûͬѧ¾­Ë¼¿¼ºó£®½¨ÒéÒÒͬѧ½«Í¼£¨2£©ÖÐij²¿·ÖÉÔ×÷¸Ä½ø£¬¾Í¿É±ÜÃâ´ËÖÖÎó²î£¬Ð´³öÄãµÄ¸Ä½ø·½°¸½«Ë®»»³É±¥ºÍNaHCO3ÈÜÒº£¬»òÔÚÖüˮƿÖеÄË®ÃæÉϼÓÒ»²ãÖ²ÎïÓͶ¼¿É·ÀÖ¹CO2ÈÜÓÚË®£®
£¨4£©¶¡Í¬Ñ§Ìá³ö½«Í¼£¨2£©×°ÖÃÖÐÁ¿Æø×°ÖÃÈ¥µô£¬Ö»Ó÷´Ó¦×°ÖúÍÌìƽҲÄܽÏ׼ȷµØ²â³öÑùÆ·ÖÐNa2C03µÄÖÊÁ¿·ÖÊý£¬ÇëÄã·ÖÎö¶¡Í¬Ñ§·½°¸ÖУ¬³ýÁ˲ⶨÑùÆ·µÄÖÊÁ¿£¬»¹Ðè²â¶¨µÄÁíÁ½¸öÊý¾ÝÊÇ·´Ó¦Ç°×°ÓÐÒ©Æ·µÄÕû¸ö×°ÖõÄÖÊÁ¿¡¢·´Ó¦ºóÕû¸ö×°ÖõÄÖÊÁ¿£®
£¨5£©»¹¿ÉÒÔÓÃÆäËûʵÑé·½·¨²â¶¨ÊÔÑùÖд¿¼îµÄÖÊÁ¿·ÖÊý£¬Çë¼òÊöÒ»ÖÖÓëÉÏÊö¸÷·½·¨²»Í¬µÄʵÑé·½·¨³ÆÁ¿Ò»¶¨ÖÊÁ¿µÄÑùÆ·Åä³ÉÈÜÒº£¬¼ÓÈë×ãÁ¿µÄBaCl2ÈÜÒººóµÃµ½³Áµí£¬½«³Áµí½øÐйýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡¢ÀäÈ´¡¢³ÆÁ¿£®ÓɳÁµí£¨BaCO3£©ÖÊÁ¿¼ÆËã³öNa2CO3µÄÖÊÁ¿£¬×îºó¼ÆËã³öNa2CO3µÄÖÊÁ¿·ÖÊý£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

16£®ÇëÌîÈëÏàÓ¦»¯Ñ§ÊÔ¼ÁÒÔ³ýÈ¥À¨ºÅÖеÄÔÓÖÊ£¬²¢Ð´³öÏàÓ¦µÄ»¯Ñ§·´Ó¦·½³Ìʽ£®£¨¼ÓÈëµÄÊÔ¼ÁΪÊÊÁ¿£¬·´Ó¦Ìõ¼þ×ÔÐÐÑ¡Ôñ£©
£¨1£©Na2SO4ÈÜÒº£¨MgSO4£©ÊÔ¼Á£ºÇâÑõ»¯ÄÆÈÜÒº»¯Ñ§·½³Ìʽ£º2NaOH+MgSO4¨TNa2SO4+Mg£¨OH£©2¡ý
£¨2£©Cu£¨CuO£©ÊÔ¼Á£ºÏ¡ÁòËữѧ·½³Ìʽ£ºH2SO4+CuO=CuSO4+H2O
£¨3£©CO£¨CO2£©ÊÔ¼Á£ºÇâÑõ»¯ÄÆÈÜÒº»¯Ñ§·½³Ìʽ£ºCO2+2NaOH=Na2CO3+H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸