¡¾ÌâÄ¿¡¿¹ýÑõ»¯¸Æ(CaO2)ÄÑÈÜÓÚË®£¬ÔÚ³£ÎÂÏÂÎȶ¨£¬ÔÚ³±Êª¿ÕÆø¼°Ë®ÖлºÂý·Ö½â·Å³öÑõÆø£¬Òò¶ø¹ã·ºÓ¦ÓÃÓÚÓæÒµ¡¢Å©Òµ¡¢»·±£µÈÐí¶à·½Ãæ¡£ÏÂͼÊÇÒÔ´óÀíʯ(Ö÷ÒªÔÓÖÊÊÇÑõ»¯Ìú)µÈΪÔÁÏÖÆÈ¡¹ýÑõ»¯¸Æ(CaO2)µÄÁ÷³Ì¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)²Ù×÷¢ÙÓ¦°üÀ¨ÏÂÁвÙ×÷ÖеÄ____________¡£(ÌîÐòºÅ)
A.Èܽâ B.¹ýÂË C.ÕôÁó D.·ÖÒº E.Õô·¢½á¾§
(2)Óð±Ë®µ÷½ÚpHÖÁ8¡«9µÄÄ¿µÄÊÇ_________________________________¡£
(3)Èô²âµÃÂËÒºCÖÐc(CO32-)£½10£3mol/L£¬ÔòCa2£«____(Ìî¡°ÊÇ¡±»ò¡°²»¡±)³ÁµíÍêÈ«¡£[ÒÑÖªc(Ca2£«)¡Ü10£5 mol/Lʱ¼´¿ÉÊÓΪ³ÁµíÍêÈ«£»Ksp(CaCO3)£½4.96¡Á10£9]
(4)ÈôÔÚÂËÒºCÖУ¬¼ÓÈëHNO3ʹÈÜÒº³ÊËáÐÔÒԵõ½¸±²úÎïNH4NO3£¬ÔòËữºóÈÜÒºÖÐc(NH4+) _______c(NO3-)(Ìî¡°¡Ý¡±¡¢¡°¡Ü¡±¡¢¡°<¡±¡¢¡°>¡±»ò¡°£½¡±)¡£
(5)²Ù×÷¢ÚÊÇ£ºÔÚµÍÎÂÏ£¬Íù¹ýÑõ»¯ÇâŨÈÜÒºÖÐͶÈëÎÞË®ÂÈ»¯¸Æ½øÐз´Ó¦£¬Ò»¶Îʱ¼äºó£¬ÔÙ¼ÓÈëÇâÑõ»¯ÄÆÈÜÒº£¬µ±µ÷½ÚÈÜÒºpHÖÁ9¡«11£¬²Å³öÏÖ´óÁ¿³Áµí¡£Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ_________________________£»ÓüòÒªµÄÎÄ×Ö½âÊÍÓÃÇâÑõ»¯ÄƵ÷½ÚpHÖÁ9¡«11µÄÔÒò____________¡£
(6)ÒÑÖª´óÀíʯº¬CaCO3µÄÖÊÁ¿·ÖÊýΪa£¬m g´óÀíʯ¿ÉÒÔÖƵÃn g CaO2£¬Çë¼ÆË㣺CaCO3 ת»¯ÎªCaO2¹ý³ÌÖУ¬CaÔ×ÓµÄÀûÓÃÂÊ__________________¡£
¡¾´ð°¸¡¿£¨1£© ABE£¨2·Ö£©£¨2£©³ýÈ¥Fe3£«£¨2·Ö£©
£¨3£©ÊÇ£¨2·Ö£©£¨4£©£¼£¨2·Ö£©
£¨5£©CaCl2£«H2O2CaO2£«2HCl»ò£¨CaCl2£«H2O2£«2NaOH===CaO2¡ý£«2NaCl£«2H2O£©£¨2·Ö£©£»¼ÓÈëNaOHÈÜҺʹÉÏÊöƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬ÓÐÀûÓÚCaO2³ÁµíµÄÉú³É£¨3·Ö£©£¨6£©£¨2·Ö£©
¡¾½âÎö¡¿
ÊÔÌ⣨1£©²Ù×÷¢ÙÊÇÏò̼Ëá¸ÆÖмÓÈëÑÎËáµÃµ½ÂÈ»¯¸ÆÈÜÒº¼Ì¶øµÃµ½ÂÈ»¯¸Æ¾§ÌåµÄ¹ý³Ì£¬ËùÉæ¼°µ½µÄ²Ù×÷²½ÖèÓÐÓÃÑÎËáÈÜÒºÈܽâ̼Ëá¸Æ¹ÌÌ壬¹ýÂËÍêδÍêÈ«·´Ó¦µÄ̼Ëá¸Æ£¬ÒÔ¼°Í¨¹ýÕô·¢½á¾§µÃµ½ÂÈ»¯¸Æ¾§Ìå¡£
£¨2£©¼ÓÈëÏõËáÈÜÒºÖ®ºó£¬ÔÈÜÒºÖеÄÑõ»¯ÌúÒѽüÒÔÈý¼ÛÌúµÄÐÎʽ´æÔÚ£¬µ÷½ÚPHµÄÄ¿µÄ¾ÍÊÇͨ¹ýÐγÉÇâÑõ»¯ÌúÀ´³ýÈ¥ÌúÀë×Ó¡£
£¨3£©¸ù¾ÝKSPÒÔ¼°ÈÜÒºÖеÄ̼Ëá¸ùÀë×ÓµÄŨ¶È²»ÄÑÇó³ö¸ÆÀë×ÓµÄŨ¶ÈΪ4.96¡Á10-6mol/L£¬ËùÒÔ¿ÉÒÔÈÏΪ¸ÆÀë×ÓÒѾ³ÁµíÍêÈ«¡£
£¨4£©¸ù¾ÝÈÜÒºÖÐÕý¸ºµçºÉµÄ´úÊýºÍʼÖÕΪ0£¬¶øÇÒÈÜÒº³ÊËáÐÔ£¬ËùÒԿɵóöÈÜÒºÖÐ笠ùÀë×ÓŨ¶È±ÈÏõËá¸ùÀë×ÓŨ¶ÈҪС¡£
£¨5£©ÂÈ»¯¸ÆÓëË«ÑõË®µÄ·´Ó¦·½³ÌʽΪ£ºCaCl2£«H2O2CaO2£«2HCl£¬µ÷½ÚPH²Å»á³öÏÖ´óÁ¿³ÁµíµÄÔÒòÊÇ£¬¼ÓÈëÇâÑõ»¯ÄÆʹµÃÉÏÊö·´Ó¦µÄƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯²Å»á³öÏÖ´óÁ¿µÄ¹ýÑõ»¯¸Æ³Áµí¡£
£¨6£©ÒÑÖª¹ýÑõ»¯¸ÆµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª72£¬¶ø̼Ëá¸ÆµÄÏà¶ÔÔ×ÓÖÊÁ¿Îª100£¬ËùÒÔ¸ÆÔ×ÓµÄÀûÓÃÂÊΪ£¨n/72£©/£¨ma/100£©ËùÒÔ¿ÉÒÔÇóµÃ¸ÆÔ×ÓµÄÔ×ÓÀûÓÃÂÊΪ
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¿¡¿¢ñ.Ñо¿µªÑõ»¯ÎïÓëÐü¸¡ÔÚ´óÆøÖк£ÑÎÁ£×ÓµÄÏ໥×÷ÓÃʱ£¬Í¬Î¶ÈÏÂÉæ¼°ÈçÏ·´Ó¦£º
¢Ù2NO(g)£«Cl2(g)2ClNO(g) ƽºâ³£ÊýΪK1£»
¢Ú2NO2(g)£«NaCl(s)NaNO3(s)£«ClNO(g) ƽºâ³£ÊýΪK2¡£
(1)4NO2(g)£«2NaCl(s)2NaNO3(s)£«2NO(g)£«Cl2(g)µÄƽºâ³£ÊýK£½____(ÓÃK1¡¢K2±íʾ)¡£
(2)Èô·´Ó¦¢ÙÔÚζÈTÏ´ﵽƽºâʱ£¬Æ½ºâ³£ÊýΪK1£¬Éý¸ßζȺóK1Ôö´ó£¬ÔòÕý·´Ó¦µÄìʱä¡÷H____0£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£ÈôƽºâºóÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬³äÈëÉÙÁ¿Cl2£¬Cl2µÄת»¯ÂÊ___________£¨Ìî¡°±ä´ó¡±¡¢¡°±äС¡±»ò¡°²»±ä¡±,ÏÂͬ£©£¬µÄÖµ___________¡£
¢ò.½üÄ꣬¿Æѧ¼ÒÑо¿ÁËÒÒ´¼´ß»¯ºÏ³ÉÒÒËáÒÒõ¥µÄз½·¨£º2C2H5OH(g)CH3COOC2H5(g)£«2H2(g)ÔÚ³£Ñ¹Ï·´Ó¦£¬ÀäÄýÊÕ¼¯£¬²âµÃ³£ÎÂÏÂҺ̬ÊÕ¼¯ÎïÖÐÖ÷Òª²úÎïµÄÖÊÁ¿·ÖÊýÈçͼ3Ëùʾ¡£¹ØÓڸ÷½·¨£¬ÏÂÁÐÍƲâºÏÀíµÄÊÇ________¡£
A£®·´Ó¦Î¶Ȳ»Ò˳¬¹ý300¡æ
B£®Ôö´óÌåϵѹǿ£¬ÓÐÀûÓÚÌá¸ßÒÒ´¼Æ½ºâת»¯ÂÊ
C£®ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬ÒÒÈ©ÊÇ·´Ó¦Àú³ÌÖеÄÖмä²úÎï
D£®Ìá¸ß´ß»¯¼ÁµÄ»îÐÔºÍÑ¡ÔñÐÔ£¬¼õÉÙÒÒÃÑ¡¢ÒÒÏ©µÈ¸±²úÎïÊǹ¤ÒյĹؼü
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿X¡¢Y¡¢Z¡¢L¡¢MÎåÖÖÔªËصÄÔ×ÓÐòÊýÒÀ´ÎÔö´ó¡£X¡¢Y¡¢Z¡¢LÊÇ×é³Éµ°°×ÖʵĻù´¡ÔªËØ£¬MÊǵؿÇÖк¬Á¿×î¸ßµÄ½ðÊôÔªËØ¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)LµÄÃû³ÆΪ__________________________£»ÎåÖÖÔªËصÄÔ×Ӱ뾶´Ó´óµ½Ð¡µÄ˳ÐòΪ______________________________(ÓÃÔªËØ·ûºÅ±íʾ)¡£
(2)Z¡¢XÁ½ÔªËØ°´Ô×ÓÊýÄ¿±È1¡Ã3ºÍ2¡Ã4¹¹³É·Ö×ÓAºÍB£¬¹¤ÒµºÏ³ÉAµÄ»¯Ñ§·½³ÌʽΪ______________________________£¬BµÄ½á¹¹Ê½Îª_____________________¡£ÔÚ±ê×¼×´¿öÏ£¬½«A³äÂúÒ»¸ö¸ÉÔïÉÕÆ¿£¬½«ÉÕÆ¿µ¹ÖÃÓÚË®ÖУ¬Æ¿ÄÚÒºÃæÉÏÉý£¬(¼ÙÉèÈÜÖʲ»À©É¢)×îºóÉÕÆ¿ÄÚÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿Å¨¶ÈΪ________(¾«È·µ½0.001)¡£
(3)Îø(Se)ÊÇÈËÌå±ØÐèµÄ΢Á¿ÔªËØ£¬ÓëLͬһÖ÷×壬SeÔ×Ó±ÈLÔ×Ó¶àÁ½¸öµç×Ӳ㣬ÔòÆä×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½Îª__________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿£¨1£©¹ãÖÝÑÇÔ˻ᡰ³±Á÷¡±»ð¾æÄÚÐÜÐÜ´ó»ðÀ´Ô´ÓÚ±ûÍéµÄȼÉÕ£¬±ûÍéÊÇÒ»ÖÖÓÅÁ¼µÄȼÁÏ¡£ÊԻشðÏÂÁÐÎÊÌ⣺
¢ÙÈçͼÊÇÒ»¶¨Á¿±ûÍéÍêȫȼÉÕÉú³ÉCO2ºÍ1 mol H2O(l)¹ý³ÌÖеÄÄÜÁ¿±ä»¯Í¼£¬ÇëÔÚͼÖеÄÀ¨ºÅÄÚÌîÈë¡°£«¡±»ò¡°£¡±_____¡£
¢Úд³ö±íʾ±ûÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ£º___________________________________¡£
¢Û¶þ¼×ÃÑ(CH3OCH3)ÊÇÒ»ÖÖÐÂÐÍȼÁÏ£¬Ó¦ÓÃÇ°¾°¹ãÀ«¡£1 mol¶þ¼×ÃÑÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ·Å³ö1 455 kJÈÈÁ¿¡£Èô1 mol±ûÍéºÍ¶þ¼×ÃѵĻìºÏÆøÌåÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ¹²·Å³ö1835 kJÈÈÁ¿£¬Ôò»ìºÏÆøÌåÖУ¬±ûÍéºÍ¶þ¼×ÃѵÄÎïÖʵÄÁ¿Ö®±ÈΪ_______¡£
£¨2£©¿Æѧ¼Ò¸Ç˹ÔøÌá³ö£º¡°²»¹Ü»¯Ñ§¹ý³ÌÊÇÒ»²½Íê³É»ò·Ö¼¸²½Íê³É£¬Õâ¸ö×ܹý³ÌµÄÈÈЧӦÊÇÏàͬµÄ¡£¡±ÀûÓøÇ˹¶¨ÂɿɲâijЩÌرð·´Ó¦µÄÈÈЧӦ¡£
¢ÙP4(s£¬°×Á×)£«5O2(g)===P4O10(s)¡¡¦¤H1£½£2 983.2 kJ¡¤mol£1
¢ÚP(s£¬ºìÁ×)£«5/4O2(g)===1/4P4O10(s)¡¡¦¤H2£½£738.5 kJ¡¤mol£1
Ôò°×Á×ת»¯ÎªºìÁ×µÄÈÈ»¯Ñ§·½³ÌʽΪ_________________________________________¡£ÏàͬµÄ×´¿öÏ£¬ÄÜÁ¿½ÏµÍµÄÊÇ________£»°×Á×µÄÎȶ¨ÐԱȺìÁ×________(Ìî¡°¸ß¡±»ò¡°µÍ¡±)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿300¡æʱ,½«2mol AºÍ2mol BÁ½ÖÖÆøÌå»ìºÏ¼ÓÈë2LÃܱÕÈÝÆ÷ÖÐ,·¢Éú·´Ó¦3A(g)+B(g)2C(g)+2D(g)¡÷H,2minÄ©·´Ó¦´ïµ½Æ½ºâ,Éú³É0.8mol D¡££¨ÓÃÈý¶Îʽ½â´ð¡££©
(1)Çó¸Ã·´Ó¦µÄƽºâ³£Êý____¡£
(2)ÇóƽºâʱAµÄת»¯ÂÊ____¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÀë×Ó·½³ÌʽÖУ¬ÊéдÕýÈ·µÄÊÇ( )¡£
A.ÌúÓëÏ¡ÑÎËá·´Ó¦£º2Fe£«6H£«£½2Fe3£«£«3H2¡ü
B.Ï¡ÁòËáÓëÇâÑõ»¯±µÈÜÒº·´Ó¦£ºBa2£«£«H£«£«OH-£«£½H2O£«BaSO4¡ý
C.̼Ëá¸ÆÓëÏ¡ÑÎËá·´Ó¦£ºCaCO3£«2H£«£½Ca2£«£«CO2¡ü£«H2O
D.ÍƬ¸úÏõËáÒøÈÜÒº·´Ó¦£ºCu£«Ag£«£½Cu2£«£«Ag
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿³£ÎÂÏ£¬ÏÂÁи÷×éÀë×ÓÔÚÖ¸¶¨ÈÜÒºÖÐÄÜ´óÁ¿¹²´æµÄÊÇ£¨ £©
A. ÎÞɫ͸Ã÷µÄÈÜÒºÖУºFe3+¡¢Mg2+¡¢SCN-¡¢Cl-
B. c(Fe2+)=1mol¡¤L-1µÄÈÜÒºÖУºK+¡¢NH4+¡¢MnO4-¡¢SO42-
C. c(H+)/c(OH-)=1¡Á10-12µÄÈÜÒºÖУºK+¡¢Na+¡¢CO32-¡¢NO3-
D. ÄÜʹ¼×»ù³È±äºìµÄÈÜÒºÖУºNa+¡¢NH4+¡¢SO42-¡¢HCO3-
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ZnÔÚÏÖ´ú¹¤ÒµÖжÔÓÚµç³ØÖÆÔìÉÏÓв»¿ÉÄ¥ÃðµÄµØ룬Ã÷³¯Ä©ÄêËÎÓ¦ÐÇËùÖøµÄ¡¶Ì칤¿ªÎï¡·Ò»ÊéÖоÍÓÐÊÀ½çÉÏ×îÔçµÄ¹ØÓÚÁ¶Ð¿¼¼ÊõµÄ¼ÇÔØ¡£»Ø´ðÏÂÁÐÎÊÌ⣺
(1)Zn»ù̬Ô×ӵĵç×ÓÅŲ¼Ê½Îª___________£¬4sÄܼ¶ÉϵijɶԵç×ÓÊýΪ___________¡£
(2)ÆÏÌÑÌÇËáп{[CH2OH( CHOH)4COO]2Zn}ÊÇÄ¿Ç°Êг¡ÉÏÁ÷ÐеIJ¹Ð¿¼Á¡£ÆÏÌÑÌÇËáпÖÐ̼Ô×ÓÔÓ»¯ÐÎʽÓÐ___________£¬C¡¢HÁ½ÔªËصĵÚÒ»µçÀëÄܵĴóС¹ØϵΪ___________________¡£
(3)ZnCl2ÓëNH3ÐγɵÄÅäºÏÎï[Zn(NH3)4]Cl2ÖУ¬´æÔÚ___________(Ìî×Öĸ)¡£
AÀë×Ó¼ü B.¦Ò¼ü C.¦Ð¼ü
(4)пÓëij·Ç½ðÊôÔªËØXÐγɵĻ¯ºÏÎᄃ°ûÈçͼËùʾ£¬ÆäÖÐZnºÍXͨ¹ý¹²¼Û¼ü½áºÏ£¬¸Ã»¯ºÏÎïÖÐZnÓëXµÄÔ×Ó¸öÊýÖ®±ÈΪ___________¡£
(5)ÔÚͼʾ¾§°ûÖÐÈôÖ»¿¼²éXµÄÅÅÁз½Ê½£¬ÔòXµÄ¶Ñ»ý·½Ê½ÊôÓÚ½ðÊô¾§Ìå¶Ñ»ý·½Ê½ÖеÄ___________¶Ñ»ý£»Éè°¢·ü¼ÓµÂÂÞ³£ÊýµÄÊýֵΪNA£¬¸Ã¾§°ûÖÐZnµÄ°ë¾¶Îªr1nm£¬XµÄ°ë¾¶Îªr2nm£¬XµÄÏà¶ÔÔ×ÓÖÊÁ¿ÎªM£¬Ôò¸Ã¾§ÌåµÄÃܶÈΪ___________g¡¤cm£3(Óú¬r1¡¢r2¡¢M¡¢NAµÄ´úÊýʽ±íʾ)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Óñê×¼NaOHÈÜÒºµÎ¶¨³äÂúHClµÄÉÕÆ¿(±ê¿öÏÂ)×öÍêÅçȪʵÑéºóµÃµ½µÄÏ¡ÑÎËᣬÒԲⶨËüµÄ׼ȷŨ¶È£¬ÇëÄã»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÀíÂÛ¼ÆËã¸ÃÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ£º________________________________¡£
(2)ÈôÓü׻ù³È×÷ָʾ¼Á£¬´ïµ½Âú¶¨ÖÕµãʱµÄÏÖÏóÊÇ___________________________¡£
(3)ÏÖÓÐÈýÖÖŨ¶ÈµÄ±ê×¼NaOHÈÜÒº£¬ÄãÈÏΪ×îºÏÊʵÄÊÇÏÂÁеÚ__________ÖÖ¡£
¢Ù5.00mol¡¤L-1 ¢Ú0.500mol¡¤L-1 ¢Û0.0500mol¡¤L-1
(4)Èô²ÉÓÃÉÏÊö×îºÏÊÊŨ¶ÈµÄ±ê×¼NaOHÈÜÒºÂú¶¨£¬µÎ¶¨Ê±ÊµÑéÊý¾ÝÁбíÈçÏ£º
ʵÑé´ÎÊý±àºÅ | ´ý²âÑÎËáÌå»ý(mL) | µÎÈëNaOHÈÜÒºÌå»ý(mL) |
1 | 10.00 | 8.48 |
2 | 10.00 | 8.52 |
3 | 10.00 | 8.00 |
ÇóÕâÖÖ´ý²âÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶Èc(HCl)=___________________¡£
(5)Ôڵ樲Ù×÷¹ý³ÌÖУ¬ÒÔϸ÷Ïî²Ù×÷ʹ²â¶¨ÖµÆ«¸ßµÄÓУº_______________
¢ÙµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»ºó£¬Î´ÓÃÒÑ֪Ũ¶ÈµÄ±ê×¼ÈÜÒºÈóÏ´
¢ÚµÎ¶¨¹Ü(×°±ê×¼ÈÜÒº)Ôڵζ¨Ç°¼â×ì´¦ÓÐÆøÅÝ£¬µÎ¶¨ÖÕÁËÎÞÆøÅÝ
¢ÛµÎ¶¨Ç°Æ½ÊÓ£¬µÎ¶¨ÖÕÁ˸©ÊÓ
¢Ü¿´µ½ÑÕÉ«±ä»¯ºóÁ¢¼´¶ÁÊý
¢ÝÏ´µÓ׶ÐÎƿʱ£¬Îó°ÑϡʳÑÎË®µ±×öÕôÁóË®½øÐÐÏ´µÓ
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