¡¾ÌâÄ¿¡¿¢ñ£©
£¨1£©ÊµÑéÖÐÄܹ۲쵽µÄÏÖÏóÊÇ____________________¡££¨Ñ¡Ìî´úºÅ£©
A£®ÊÔ¹ÜÖÐþƬÖð½¥Èܽâ B£®ÊÔ¹ÜÖвúÉúÎÞÉ«ÆøÅÝ
C£®ÉÕ±Íâ±Ú±äÀä D£®ÉÕ±µ×²¿Îö³öÉÙÁ¿°×É«¹ÌÌå
£¨2£©ÓÉʵÑéÍÆÖª£¬MgCl2ÈÜÒººÍH2µÄ×ÜÄÜÁ¿________£¨Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±¡°µÈÓÚ¡±£©Ã¾Æ¬ºÍÑÎËáµÄ×ÜÄÜÁ¿¡£
¢ò£©ÓÃ50mL0.50mol/LÑÎËáÓë50mL0.55mol/LNaOHÈÜÒºÔÚÈçͼËùʾµÄ×°ÖÃÖнøÐÐÖкͷ´Ó¦¡£Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©´ÓʵÑé×°ÖÃÉÏ¿´£¬Í¼ÖÐÉÐȱÉÙµÄÒ»ÖÖ²£Á§ÓÃÆ·ÊÇ___________£¬³ý´ËÖ®Í⻹ÓÐÒ»´¦´íÎóµÄÊÇ____________________¡£
£¨2£©´óÉÕ±ÉÏÈç²»¸ÇÓ²Ö½°å£¬ÇóµÃµÄÖкÍÈÈÊýÖµ£¨¾ø¶ÔÖµ£©__________£¨Ìî¡°Æ«´ó¡¢Æ«Ð¡¡¢ÎÞÓ°Ï족£©¡£
¢ó£©£¨1£©ÃºÈ¼Éյķ´Ó¦ÈÈ¿Éͨ¹ýÒÔÏÂÁ½¸ö;¾¶À´ÀûÓãºa.ÀûÓÃúÔÚ³ä×ãµÄ¿ÕÆøÖÐÖ±½ÓȼÉÕ²úÉúµÄ·´Ó¦ÈÈ£»b.ÏÈʹúÓëË®ÕôÆø·´Ó¦µÃµ½ÇâÆøºÍÒ»Ñõ»¯Ì¼£¬È»ºóʹµÃµ½µÄÇâÆøºÍÒ»Ñõ»¯Ì¼ÔÚ³ä×ãµÄ¿ÕÆøÖÐȼÉÕ¡£ÕâÁ½¸ö¹ý³ÌµÄÈÈ»¯Ñ§·½³ÌʽΪ£º
a£®C£¨s£©£«O2£¨g£©===CO2£¨g£© ¦¤H£½E1 ¢Ù
b£®C£¨s£©£«H2O£¨g£©===CO£¨g£©£«H2£¨g£©¦¤H£½E2 ¢Ú
H2£¨g£©£«1/2O2£¨g£©===H2O£¨g£©¦¤H£½E3 ¢Û
CO£¨g£©£«1/2O2£¨g£©===CO2£¨g£©¦¤H£½E4 ¢Ü
Çë±í´ïE1¡¢E2¡¢E3¡¢E4Ö®¼äµÄ¹ØϵΪE2£½_________________¡£
£¨2£©ÈçͼËùʾÔÚ³£Î³£Ñ¹Ï£¬1Ħ¶ûNO2 ºÍ1Ħ¶ûCOÍêÈ«·´Ó¦Éú³ÉCO2ºÍNO¹ý³ÌÖÐÄÜÁ¿±ä»¯Ê¾Òâͼ£¬Çëд³öNO2ºÍCO·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º___________________¡£
£¨3£©»¯Ñ§·´Ó¦µÄìʱäÓë·´Ó¦ÎïºÍÉú³ÉÎïµÄ¼üÄÜÓйء£ÒÑ֪ijЩ»¯Ñ§¼üµÄ¼üÄÜÈçϱíËùʾ£º
¹²¼Û¼ü | H¡ªH | Cl¡ªCl | H¡ªCl |
¼üÄÜ/£¨kJ¡¤mol£1£© | 436 | 247 | 434 |
Ôò·´Ó¦£ºH2£¨g£©+Cl2£¨g£©=2HCl£¨g£©µÄìʱ䦤H £½ ____________________¡£
¡¾´ð°¸¡¿A B DСÓÚ»·Ðβ£Á§½Á°è°ôСÉÕ±¿ÚºÍ´óÉÕ±¿ÚûÓÐƽÆ루ÆäËüºÏÊʴ𰸸ø·Ö£©Æ«Ð¡E2£½E1£E3£E4NO2£¨g£©+CO£¨g£©=NO£¨g£©+CO2£¨g£© ¦¤H£½£234 kJ¡¤mol£1¦¤H £½£185 kJ¡¤mol£1
¡¾½âÎö¡¿
¢ñ)(1)þÓëÑÎËá¾çÁÒ·´Ó¦£¬²úÉúÇâÆø²¢·Å³ö´óÁ¿µÄÈÈ£¬ÓÉÓÚÇâÑõ»¯¸ÆµÄÈܽâ¶ÈËæζÈÉý¸ß¶ø¼õС£¬ËùÒÔ±¥ºÍʯ»ÒË®ÉýκóÎö³öµÄÇâÑõ»¯¸ÆʹÈÜÒº³Ê»ë×Ç×´£¬Ã¾ÌõÈܽ⣬¹ÊABDÑ¡ÏîÖеÄÏÖÏó·ûºÏ£»¹Ê´ð°¸Îª£ºABD£»
(2)µ±·´Ó¦ÎïµÄÄÜÁ¿¸ßÓÚÉú³ÉÎïµÄÄÜÁ¿Ê±£¬·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬¹ÊMgCl2ÈÜÒººÍH2µÄ×ÜÄÜÁ¿Ð¡ÓÚþƬµÄÑÎËáµÄ×ÜÄÜÁ¿£¬¹Ê´ð°¸Îª£ºÐ¡ÓÚ£»
¢ò)(1)ÓÉÁ¿ÈȼƵĹ¹Ôì¿ÉÖª¸Ã×°ÖõÄȱÉÙÒÇÆ÷ÊÇ»·Ðβ£Á§½Á°è°ô£»ÖкÍÈȲⶨʵÑé³É°ÜµÄ¹Ø¼üÊDZ£Î¹¤×÷£¬ÄÚÍâÉձΪһÑù¸ß£¬·ñÔò£¬ÈÈÁ¿É¢Ê§´ó£»¹Ê´ð°¸Îª£º»·Ðβ£Á§½Á°è°ô£»Ð¡ÉÕ±¿ÚºÍ´óÉÕ±¿ÚûÓÐƽÆ룻
(2)´óÉÕ±ÉÏÈç²»¸ÇÓ²Ö½°å£¬»áÓÐÒ»²¿·ÖÈÈÁ¿É¢Ê§£¬ÇóµÃµÄÖкÍÈÈÊýÖµ½«»á¼õС£¬¹Ê´ð°¸Îª£ºÆ«Ð¡£»
¢ó)(1)¢ÙC(s)+O2(g)¨TCO2(g)¡÷H=E1£¬¢ÛH2(g)+O2(g)¨TH2O(g)¡÷H=E3£¬¢ÜCO(g)+O2(g)¨TCO2(g)¡÷H=E4£¬¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ù-¢Û-¢Ü¿ÉµÃ£ºC(s)+H2O(g)¨TCO(g)+H2(g)£¬¹ÊE2=E1-E3-E4£¬¹Ê´ð°¸Îª£ºE1-E3-E4£»
(2)ÓÉͼ¿ÉÖª£¬1Ħ¶ûNO2ºÍ1Ħ¶ûCOÍêÈ«·´Ó¦Éú³ÉCO2ºÍNO·Å³öÈÈÁ¿Îª(368-134)kJ=234kJ£¬·´Ó¦ÈÈ»¯Ñ§·½³ÌʽΪ£ºNO2(g)+CO(g)¨TNO(g)+CO2(g)¡÷H=-234 kJmol-1£¬¹Ê´ð°¸Îª£ºNO2(g)+CO(g)¨TNO(g)+CO2(g)¡÷H=-234 kJmol-1£»
(3)·´Ó¦ÈÈ=·´Ó¦Îï×ܼüÄÜ-Éú³ÉÎï×ܼüÄÜ£¬¹Ê·´Ó¦£ºH2(g)+Cl2(g)¨T2HCl(g)µÄìʱä¡÷H=436kJ/mol+247kJ/mol-2¡Á434kJ/mol=-185kJ/mol£¬¹Ê´ð°¸Îª£º-185 kJ/mol¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A. ³ÁµíÈܽâ´ïµ½Æ½ºâʱ£¬ÈÜÒºÖÐÈÜÖʵÄÀë×ÓŨ¶ÈÏàµÈ£¬ÇÒ±£³Ö²»±ä
B. ÒÑÖªBaSO4ÈܶȻý³£Êý1.0¡Á10£10£¬ÔòBaSO4ÔÚË®ÖÐÈܽâ¶ÈΪ2.33¡Á10£4g/100gË®
C. Ksp(AgCl)£½1.8¡Á10£10£¬Ksp(AgI)£½1.0¡Á10£16,³£ÎÂÏ£¬AgClÈôÒªÔÚNaIÈÜÒºÖпªÊ¼×ª»¯ÎªAgI£¬ÔòNaIµÄŨ¶ÈÒ»¶¨ÊÇ¡Á10£11mol¡¤L£1
D. Ïòº¬ÓÐMgCO3¹ÌÌåµÄÈÜÒºÖеμÓÉÙÐíŨÑÎËá(ºöÂÔÌå»ý±ä»¯),ÈÜÒºÖеÄc(Mg2+)¼õС¶øKsp(MgCO3) ²»±ä
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¼ìÑéÈÜÒºÖÐÊÇ·ñº¬ÓÐCl-£¬Ñ¡ÓõÄÊÔ¼ÁÊÇÏ¡ÏõËáºÍAgNO3ÈÜÒº£¬ÆäÖÐÏ¡ÏõËáµÄ×÷ÓÃÊÇ£¨ £©
A.¼ÓËÙ·´Ó¦µÄ½øÐÐB.ÅųýijЩÔÓÖÊÀë×ӵĸÉÈÅ
C.Éú³É²»ÈÜÓÚË®µÄÎïÖÊD.¼Ó´ó³ÁµíµÄÉú³ÉÁ¿
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁйØÓÚÓлúÎïÐÔÖʵÄ˵·¨ÕýÈ·µÄÊÇ £¨ £©
A. ÒÒÏ©ºÍ¼×Í鶼¿ÉÒÔÓëÂÈÆø·´Ó¦ B. ÒÒÏ©ºÍ¾ÛÒÒÏ©¶¼ÄÜʹäåË®ÍÊÉ«
C. ÒÒÏ©ºÍ±½¶¼ÄÜʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ« D. ÒÒËáÄÜÓëÄƲúÉúÇâÆø£¬¶øÒÒ´¼²»¿ÉÒÔ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ä³»¯Ñ§¿ÎÍâÐËȤС×éΪ̽¾¿±½ÓëÒºäåÖ®¼äµÄ·´Ó¦£¬½øÐÐʵÑ飬װÖÃÈçͼËùʾ£¬Çë¸ù¾ÝÒªÇó»Ø´ðÏà¹ØÎÊÌâ¡£(ÒÑÖª±½µÄ·ÐµãΪ80.1 ¡æ£¬ÃܶÈΪ0.9 g/mL )
£¨1£©Ð´³ö±½ÓëÒºäå·´Ó¦µÄ»¯Ñ§·½³Ìʽ_____________________£¬·´Ó¦ÀàÐÍΪ___________
£¨2£©×¶ÐÎÆ¿ÖÐÓе»ÆÉ«»ë×ÇÉú³É£¬¸Ã×éͬѧ¾¹ýÌÖÂÛºóÈÏΪ£¬ÒÀ¾Ý¸ÃÏÖÏó²»ÄÜÈ·¶¨·¢ÉúÁËÒÔÉÏ·´Ó¦£¬ÀíÓÉÊÇ£º______________________________________£¬Òò´ËÓбØÒª¶ÔʵÑé½øÐиĽø¡£
£¨3£©ÇëÒÀ¾ÝÏÂÃæµÄʵÑéÁ÷³ÌͼѡȡºÏÊʵÄ×°ÖúÍÊÔ¼Á¶ÔÉÏÊöʵÑé½øÐиĽø£º
I _____________£¨Ìî×Öĸ£¬ÏÂͬ£©, II _______________
A£®×°ÓÐNaOHÈÜÒºµÄÏ´ÆøÆ¿ B£®×°ÓÐCC14µÄÏ´ÆøÆ¿
C£®×°ÓÐKIÈÜÒºµÄÏ´ÆøÆ¿ D£®×°ÓÐʪÈóµí·ÛKIÊÔÖ½µÄ¼¯ÆøÆ¿
¢ÙС×éͬѧ¶Ô¸Ä½øʵÑéºóµÄB×°ÖÃÖвúÉúµÄµ»ÆÉ«³Áµí£¬½øÐйýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¡£ÔÚÒÔÉϲ½ÖèÖУºÏ´µÓ³ÁµíµÄ²Ù×÷Ϊ___________________________________________________
¢ÚÈôʵÑéÖÐÈ¡Óõı½Îª17.3 mL£¬ÒºäåÉÔ¹ýÁ¿£¬×îÖÕ²âµÃµÄ³ÁµíÖÊÁ¿Îª18.8 g£¬Ôò±½Ôڸ÷´Ó¦ÖеÄת»¯ÂÊΪ ____________£¨±£ÁôÁ½Î»ÓÐЧÊý×Ö£©£¬¸Ã×éͬѧÈÏΪת»¯ÂʹýµÍ£¬³ýÁË¿ÉÄÜ·¢Éú¸±·´Ó¦ºÍ·´Ó¦¿ÉÄܽøÐв»ÍêÈ«Í⣬Äã·ÖÎö»¹¿ÉÄܵÄÔÒòÊÇ__________________________£¬__________________£¨Çë´ð³ö2Ìõ£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÔÚÂÈÑõ»¯·¨´¦Àíº¬CN£µÄ·ÏË®¹ý³ÌÖУ¬ÒºÂÈÔÚ¼îÐÔÌõ¼þÏ¿ÉÒÔ½«Ç軯ÎïÑõ»¯³ÉÇèËáÑÎ(Æ䶾ÐÔ½öΪÇ軯ÎïµÄǧ·ÖÖ®Ò»)£¬ÇèËáÑνøÒ»²½±»Ñõ»¯ÎªÎÞ¶¾ÎïÖÊ¡£
£¨1£©Ä³³§·ÏË®Öк¬KCN£¬ÆäŨ¶ÈΪ650mg/L¡£ÏÖÓÃÂÈÑõ»¯·¨´¦Àí£¬·¢ÉúÈçÏ·´Ó¦(ÆäÖÐN¾ùΪ-3¼Û)£ºKCN£«2KOH£«Cl2£½KOCN£«2KCl£«H2O£¬±»Ñõ»¯µÄÔªËØÊÇ_______¡£
£¨2£©Í¶Èë¹ýÁ¿ÒºÂÈ£¬¿É½«ÇèËáÑνøÒ»²½Ñõ»¯ÎªµªÆø¡£ÇëÅäƽÏÂÁл¯Ñ§·½³Ìʽ£¬²¢±ê³öµç×ÓתÒÆ·½ÏòºÍÊýÄ¿_______________________________£º
KOCN£«KOH£«Cl2¡úCO2£«N2£«KCl£«H2O
£¨3£©Èô´¦ÀíÉÏÊö·ÏË®100L£¬Ê¹KCNÍêȫת»¯ÎªÎÞ¶¾ÎïÖÊ£¬ÖÁÉÙÐèÒºÂÈ_______g¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÑÖªA¡«FÊÇÖÐѧ»¯Ñ§Öг£¼ûÎïÖÊ£¬ÆäÖÐA¡¢C¡¢EΪÆøÌ壬ÇÒAÄÜʹƷºìÈÜÒºÍÊÉ«£»B¡¢DΪҺÌ壬DµÄŨÈÜÒºÔÚ³£ÎÂÏÂÄÜʹÌú¶Û»¯£»FµÄŨÈÜÒºÓëX¹²ÈÈͨ³£ÓÃÓÚʵÑéÊÒÖƱ¸µ¥ÖÊC£»XÊÇÒ»ÖÖºÚÉ«·ÛÄ©£¬B·Ö×ÓÖÐÓÐ18¸öµç×Ó¡£(·´Ó¦Öв¿·ÖÉú³ÉÎïÒÑÂÔÈ¥£©
(1)д³ö·´Ó¦¢ÚµÄ»¯Ñ§·½³Ìʽ£º___________________¡£
(2)д³ö·´Ó¦¢Ù¡¢¢ÝµÄÀë×Ó·½³Ìʽ£º¢Ù____________£»¢Ý________________£®
(3)¸ù¾ÝͼÖÐÐÅÏ¢£¬B¡¢C¡¢D¡¢XÑõ»¯ÐÔ´ÓÇ¿µ½ÈõµÄ˳ÐòÊÇ_________________£®
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Áò¼°Æ仯ºÏÎïÓÐÐí¶àÓÃ;£¬Ïà¹ØÎïÖʵÄÎïÀí³£ÊýÈçϱíËùʾ£º
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)»ù̬FeÔ×Ó¼Û²ãµç×ӵĵç×ÓÅŲ¼Í¼(¹ìµÀ±í´ïʽ)Ϊ____________________________£¬»ù̬SÔ×Óµç×ÓÕ¼¾Ý×î¸ßÄܼ¶µÄµç×ÓÔÆÂÖÀªÍ¼Îª________ÐΡ£
(2)¸ù¾Ý¼Û²ãµç×Ó¶Ô»¥³âÀíÂÛ£¬H2S¡¢SO2¡¢SO3µÄÆø̬·Ö×ÓÖУ¬ÖÐÐÄÔ×Ó¼Û²ãµç×Ó¶ÔÊý²»Í¬ÓÚÆäËû·Ö×ÓµÄÊÇ__¡£
(3)ͼ(a)ΪS8µÄ½á¹¹£¬ÆäÈÛµãºÍ·ÐµãÒª±È¶þÑõ»¯ÁòµÄÈÛµãºÍ·Ðµã¸ßºÜ¶à£¬Ö÷ÒªÔÒòΪ__________________¡£
(4)Æø̬ÈýÑõ»¯ÁòÒÔµ¥·Ö×ÓÐÎʽ´æÔÚ£¬Æä·Ö×ÓµÄÁ¢Ìå¹¹ÐÍΪ________ÐΣ¬ÆäÖй²¼Û¼üµÄÀàÐÍÓÐ________ÖÖ£»¹ÌÌåÈýÑõ»¯ÁòÖдæÔÚÈçͼ(b)ËùʾµÄÈý¾Û·Ö×Ó£¬¸Ã·Ö×ÓÖÐSÔ×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪ________¡£
(5)FeS2¾§ÌåµÄ¾§°ûÈçͼ(c)Ëùʾ¡£¾§°û±ß³¤Îªanm¡¢FeS2Ïà¶ÔʽÁ¿ÎªM¡¢°¢·ü¼ÓµÂÂÞ³£ÊýµÄֵΪNA£¬Æ侧ÌåÃܶȵļÆËã±í´ïʽΪ______________________________g¡¤cm£3£»¾§°ûÖÐFe2£«Î»ÓÚËùÐγɵÄÕý°ËÃæÌåµÄÌåÐÄ£¬¸ÃÕý°ËÃæÌåµÄ±ß³¤Îª________nm¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Îª²â¶¨Na2CO3ÓëNa2SO3»ìºÏÎïÖи÷×é·ÖµÄº¬Á¿£¬Éè¼ÆÈçÏÂʵÑé·½°¸£º
£¨1£©·½°¸Ò»£º³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÑùÆ·(30g)£¬ÖÃÓÚÛáÛöÖмÓÈÈÖÁºãÖØ£¬ÀäÈ´£¬³ÆÈ¡Ê£Óà¹ÌÌåÖÊÁ¿Îª31.6g£¬¼ÆËã¡£¢ÙʵÑéÖмÓÈÈÖÁºãÖصÄÄ¿µÄÊÇ_________________________¡£¢ÚÑùÆ·ÖÐNa2CO3µÄÖÊÁ¿·ÖÊýΪ_______________________________
£¨2£©·½°¸¶þ£ºÒÔÏÂͼËùʾװÖýøÐÐʵÑ飺£¨Ìú¼Ų̈¡¢Ìú¼ÐµÈÒÇÆ÷δÔÚͼÖл³ö£©
¢ÙÒÑÖªÒÇÆ÷CÖÐ×°ÓÐÆ·ºìÈÜÒº£¬Æä×÷ÓÃÊÇ______________£¬ÓÐÈËÌá³ö¸ÃÈÜÒº¿ÉÄÜÒýÆðNa2CO3º¬Á¿µÄ²âÁ¿½á¹û±Èʵ¼ÊֵƫµÍ£¬ÀíÓÉÊÇ____________________¡£
¢ÚʵÑéÊÒÖб¸ÓÐÒÔϳ£ÓÃÊÔ¼Á£ºa.ŨÁòËá b.Æ·ºìÈÜÒº c.ËáÐÔ¸ßÃÌËá¼ØÈÜÒºd.ÇâÑõ»¯ÄÆÈÜÒº e.ÎÞË®ÁòËáÍ f.¼îʯ»Ò g.ÎåÑõ»¯¶þÁ× h.ÎÞË®ÂÈ»¯¸Æ Ç뽫ÏÂÁÐÈÝÆ÷ÖÐӦʢ·ÅµÄÊÔ¼ÁÐòºÅÌîÈëÏàÓ¦¿Õ¸ñ£ºBÖÐ______£¬DÖÐ________£¬EÖÐ________¡£
¢ÛʵÑé¹ý³ÌÖУ¬µ±ÒÇÆ÷AÄڵĹÌÌå·´Ó¦ÍêÈ«ºó£¬Ðè´ò¿ª»îÈûK£¬ÏòAÖÐͨÈë´óÁ¿µÄµªÆø¡£ÕâÑù×öµÄÄ¿µÄÊÇ______________¡£
£¨3£©·½°¸Èý£º³ÆÈ¡Ò»¶¨Á¿ÑùÆ·£¬ÖÃÓÚСÉÕ±ÖУ¬¼ÓÊÊÁ¿Ë®Èܽ⣬ÏòСÉÕ±ÖмÓÈë×ãÁ¿BaCl2ÈÜÒº¡£¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï³Áµí£¬³ÆÁ¿¹ÌÌåÖÊÁ¿¡£ÊµÑéÖÐÅжϳÁµíÏ´¸É¾»µÄ·½·¨ÊÇ_________¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com