¡¾ÌâÄ¿¡¿ÒÑÖª£ºÖÜÆÚ±íÖÐÇ°ËÄÖÜÆÚµÄÁùÖÖÔªËØA¡¢B¡¢C¡¢D¡¢E¡¢FºËµçºÉÊýÒÀ´ÎÔö´ó£¬ÆäÖÐAÔ×ÓºËÍâÓÐÈý¸öδ³É¶Ôµç×Ó£º»¯ºÏÎïB2EµÄ¾§ÌåΪÀë×Ó¾§Ì壬EÔ×ÓºËÍâµÄM²ãÖÐÓÐÁ½¶Ô³É¶Ôµç×Ó£ºCÔªËØÊǵ׿ÇÖк¬Á¿×î¸ßµÄ½ðÊôÔªËØ£ºDµ¥ÖʵÄÈÛµãÔÚͬÖÜÆÚÔªËØÐγɵĵ¥ÖÊÖÐÊÇ×î¸ßµÄ£ºF2£«Àë×ÓºËÍâ¸÷²ãµç×Ó¾ù³äÂú¡£Çë¸ù¾ÝÒÔÉÏÐÅÏ¢£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©A¡¢B¡¢C¡¢DµÄµÚÒ»µçÀëÄÜÓÉСµ½´óµÄ˳ÐòΪ______£¨ÓÃÔªËØ·ûºÅ±íʾ£©
£¨2£©BµÄÂÈ»¯ÎïµÄÈÛµã±ÈDµÄÂÈ»¯ÎïµÄÈÛµã¸ß£¬ÀíÓÉÊÇ______¡£
£¨3£©EµÄ×î¸ß¼ÛÑõ»¯Îï·Ö×ӵĿռ乹ÐÍÊÇ______¡£ÊÇ______·Ö×Ó£¨Ìî¡°¼«ÐÔ¡±¡°·Ç¼«ÐÔ¡±£©
£¨4£©FÔ×ӵļ۲ãµç×ÓÅŲ¼Ê½ÊÇ_____¡£
£¨5£©E¡¢FÐγÉijÖÖ»¯ºÏÎïÓÐÈçͼËùʾÁ½ÖÖ¾§Ìå½á¹¹£¨ÉîÉ«Çò±íʾFÔ×Ó£©¡£Æ仯ѧʽΪ_____¡££¨a£©ÖÐEÔ×ÓµÄÅäλÊýΪ______£¬ÈôÔÚ£¨b£©µÄ½á¹¹ÖÐÈ¡³öÒ»¸öƽÐÐÁùÃæÌå×÷Ϊ¾§°û£¬Ôòƽ¾ùÒ»¸ö¾§°ûÖк¬ÓÐ_____¸öFÔ×Ó¡£½á¹¹£¨a£©Ó루b£©Öо§°ûµÄÔ×Ó¿Õ¼äÀûÓÃÂÊÏà±È£¬£¨a£©________£¨b£©£¨Ìî¡°>¡±¡°<¡±»ò¡°=¡±£©
¡¾´ð°¸¡¿ NaCl ΪÀë×Ó¾§Ìå¶øSiCl4Ϊ·Ö×Ó¾§Ìå ƽÃæÕýÈý½ÇÐÎ ·Ç¼«ÐÔ ZnS 4 2 >
¡¾½âÎö¡¿CÊǵؿÇÖк¬Á¿×î¸ßµÄ½ðÊôÔªËØ,ÔòCÊÇAlÔªËØ£»AÔ×ÓºËÍâÓÐ3¸öδ³É¶Ôµç×Ó£¬ÇÒAµÄÔ×ÓÐòÊýСÓÚC(Al)£¬ÔòAÔ×ÓºËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p3£¬AÊǵª(N)ÔªËØ£»EÔ×ÓºËÍâµÄM²ãÖÐÓÐÁ½¶Ô³É¶Ôµç×Ó£¬EÔ×ÓºËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p4£¬EÊÇSÔªËØ£»»¯ºÏÎïB2EµÄ¾§ÌåΪÀë×Ó¾§Ì壬BÏÔ+1¼ÛΪµÚIA×åÔªËØ£¬BµÄÔ×ÓÐòÊý´óÓÚµªÔªËØ¡¢Ð¡ÓÚAlÔªËØ£¬ËùÒÔBÊÇNaÔªËØ£»½áºÏÔ×ÓÐòÊý¿ÉÖªD´¦ÓÚµÚÈýÖÜÆÚ£¬Dµ¥Öʵľ§ÌåÈÛµãÔÚͬÖÜÆÚÐγɵĵ¥ÖÊÖÐÊÇ×î¸ßµÄ£¬ËùÒÔDÊÇSiÔªËØ£»F2+Àë×ÓºËÍâ¸÷ÑDzãµç×Ó¾ùÒѳäÂú£¬FÔ×ÓÐòÊý´óÓÚSÔªËØ£¬Ôò´¦ÓÚµÚËÄÖÜÆÚ£¬F2+Àë×ÓºËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p63d10£¬¹ÊFÔ×ÓºËÍâµç×ÓÊýΪ30£¬ÔòFÊÇZnÔªËØ¡£
£¨1£©A¡¢B¡¢C¡¢D·Ö±ðÊÇN¡¢Na¡¢Al¡¢SiÔªËØ£¬Í¬Ò»ÖÜÆÚÖУ¬ÔªËصĵÚÒ»µçÀëÄÜËæ×ÅÔ×ÓÐòÊýµÄÔö´ó¶ø³ÊÔö´óÇ÷ÊÆ£¬Í¬Ò»Ö÷×åÖУ¬ÔªËصĵÚÒ»µçÀëÄÜËæ×ÅÔ×ÓÐòÊýµÄÔö´ó¶ø¼õС£¬ËùÒÔµÚÒ»µçÀëÄÜ£ºNa£¼Al£¼Si£¼P£¬P£¼N£¬ËùÒÔµÚÒ»µçÀëÄÜ´óС˳ÐòΪNa£¼Al£¼Si£¼N¡£
¹Ê´ð°¸Îª£ºNa£¼Al£¼Si£¼N£»
£¨2£©BµÄÂÈ»¯ÎïÊÇNaCl£¬DµÄÂÈ»¯ÎïÊÇSiCl4£¬NaClÊÇÀë×Ó¾§Ì壬SiCl4ÊÇ·Ö×Ó¾§Ì壬Àë×Ó¾§ÌåµÄÈÛµã´óÓÚ·Ö×Ó¾§Ìå¡£
¹Ê´ð°¸Îª£ºNaClÊÇÀë×Ó¾§Ìå,SiCl4ÊÇ·Ö×Ó¾§Ì壻
£¨3£©EÊÇSÔªËØ£¬SO3·Ö×ÓSÔ×Óº¬ÓÐ3¸ö¦Ò¼üÇÒ²»º¬¹Âµç×Ó¶Ô£¬ËùÒÔΪƽÃæÕýÈý½ÇÐνṹ£¬SO3·Ö×ÓÖÐÕý¸ºµçºÉÖØÐÄÖغϣ¬Îª·Ç¼«ÐÔ·Ö×Ó¡£
¹Ê´ð°¸Îª£ºÆ½ÃæÕýÈý½ÇÐΣ»·Ç¼«ÐÔ£»
£¨4£©FΪZnÔªËØ£¬ZnΪ30ºÅÔªËØ£¬Ô×ÓºËÍâµç×ÓÅŲ¼Ê½Îª[Ar]3d104s2£¬¼Û²ãµç×ÓÅŲ¼Ê½Îª£º3d104s2¡£¹Ê´ð°¸Îª£º3d104s2£»
£¨5£©EÊÇSÔªËØ,FÊÇZnÔªËØ£¬SÔ×Ó¸öÊý=8¡Á+6¡Á=4£¬ZnÔ×Ó¸öÊý=4£¬ËùÒԸþ§°ûÖÐS¡¢ZnÔ×Ó¸öÊýÖ®±È=4:4=1:1£¬ËùÒÔÆ仯ѧʽΪZnS¡£ÔÚaͼÖУ¬ÒÔ¶¥µãSÔ×ÓΪÀý£¬ÓëS¾àÀë×î½üµÄZnÔ×ÓÓÐ4¸ö£¬·Ö²¼ÔÚSÔ×ÓÖÜΧ²»ÏàÁÚµÄÁ¢·½ÌåÄÚ£¬¹ÊSÅäλÊýΪ4¡£ÔÚbµÄ½á¹¹ÖÐÈ¡³öÒ»¸öƽÐÐÁùÃæÌå×÷Ϊ¾§°û£¬ÔòÓÐÈçͼËùʾµÄ½á¹¹£¨¿ÕÐÄÇò±íʾZnÔ×Ó£©£¬ÆäÖк¬ÓÐZnÔ×ÓÊýΪ8¡Á+1=2£»ÔÚbͼÖÐÁòÀë×ӵĶѻý·½Ê½ÎªÁù·½×îÃܶѻý£¬Í¼aÖÐÁòÀë×ÓÖ®¼äÐγÉÁËÃæÐÄÁ¢·½¶Ñ»ý£¬Ô×Ó¿Õ¼äÀûÓÃÂÊÓëbͼÖÐÁòÀë×ӵĶѻý·½Ê½Ïàͬ£¬¶øÔÚaͼÖо§°ûÖк¬ÓÐ4¸öпÀë×Ó£¬ÔÚbͼÖÐÔòÓÐ2¸öпÀë×Ó£¬¹Ê½á¹¹aÖеÄÔ×ÓÀûÓÃÂÊ´óÓڽṹbÖеÄÔ×Ó¿Õ¼äÀûÓÃÂÊ¡£
¹Ê´ð°¸Îª£ºZnS£»4£»2£»£¾¡£
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¿¡¿¡°84¡±Ïû¶¾ÒºÊÇÒ»ÖÖÒÔNaClOΪÖ÷µÄÏû¶¾¼Á£¬¹ã·ºÓ¦ÓÃÓÚÒ½Ôº¡¢Ê³Æ·¼Ó¹¤¡¢¼ÒÍ¥µÈµÄÎÀÉúÏû¶¾¡£
£¨1£©¡°84¡±Ïû¶¾ÒºÖÐͨÈëCO2ÄÜÔöÇ¿Ïû¶¾Ð§¹û£¬Ð´³öÏò¡°84¡±Ïû¶¾ÒºÖÐͨÈë¹ýÁ¿CO2µÄÀë×Ó·½³Ìʽ£º___________________¡£
£¨2£©²â¶¨¡°84¡±Ïû¶¾ÒºÖÐNaClOµÄÎïÖʵÄÁ¿Å¨¶ÈµÄ·½·¨ÈçÏ£º
¢ÙÅäÖÆ100.00mL 0.5000 mol¡¤L£1µÄNa2S2O3ÈÜÒº¡£ÅäÖƹý³ÌÖÐÐè׼ȷ³ÆÈ¡Na2S2O3¹ÌÌå___________________g£¬ÐèÒªÓõ½µÄ²£Á§ÒÇÆ÷ÓÐÉÕ±¡¢½ºÍ·µÎ¹Ü¡¢Á¿Í²¡¢___________________¡£
¢Ú׼ȷÁ¿È¡10.00 mLÏû¶¾ÒºÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë¹ýÁ¿µÄKIÈÜÒº£¬ÓÃ×ãÁ¿µÄÒÒËáËữ£¬³ä·Ö·´Ó¦ºóÏòÈÜÒºÖеμÓNa2S2O3ÈÜÒº£¬ÍêÈ«·´Ó¦Ê±ÏûºÄNa2S2O3ÈÜÒº25.00 mL¡£·´Ó¦¹ý³ÌÖеÄÏà¹ØÀë×Ó·½³ÌʽΪ£º
2CH3COOH+2I¡ª+ClO¡ª=I2+Cl¡ª+2CH3COO¡ª+H2O£¬I2+2S2O=2I¡ª+S4O
ͨ¹ý¼ÆËãÇó³ö¸Ã¡°84¡±Ïû¶¾ÒºÖÐNaClOµÄÎïÖʵÄÁ¿Å¨¶È¡££¨Ð´³ö¼ÆËã¹ý³Ì£©__________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÊÒÎÂÏ£¬ÓÃ0.1mo1/LHClÈÜÒºµÎ¶¨10.mL0.1mol¡¤L£1Na2CO3ÈÜÒº£¬µÎ¶¨ÇúÏßÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©
A. Ë®µçÀë³Ì¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ:a>b>c>d
B. aµãʱ£ºc(Na£«)>c(CO32£)>c(HCO3£)>c(OH£)
C. bµãʱ£º3c(Na£«)=2c(CO32£)£«2c(HCO3£)£«2c(H2CO3)
D. dµãʱ£ºc(H£«)>c(HCO3£)=c(CO32£)
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÎïÖʾùΪ³£¼ûÎïÖÊ»òËüÃǵÄÈÜÒº£¬ÆäÖÐAΪµ»ÆÉ«¹ÌÌ壬C¡¢X¾ùΪÎÞÉ«ÆøÌ壬ZΪdzÂÌÉ«ÈÜÒº£¬DΪһÖÖ³£¼ûµÄÇ¿¼î¡£¸ù¾ÝËüÃÇÖ®¼äµÄת»¯¹Øϵ£¨ÏÂͼ£©, Óû¯Ñ§ÓÃÓï»Ø´ðÎÊÌâ¡££¨²¿·Ö²úÎïÒÑÊ¡ÂÔ£©
£¨1£©Ð´³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£ºA______£¬B_______D_______¡£
£¨2£©AÓëSO3(g)µÄ·´Ó¦ÀàËÆÓÚAÓëXµÄ·´Ó¦£¬Çëд³öAÓëS03(g)·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______¡£
£¨3£©ÏòZÈÜÒºÖÐͨÈëÒ»¶¨Á¿µÄCl2£¬Ð´³ö¼ìÑéZÖÐÑôÀë×ÓÊÇ·ñ·´Ó¦³äÈ«ËùÐèÊÔ¼Á£º______¡£
£¨4£©ÎïÖÊEת»¯ÎªÎïÖÊFµÄÏÖÏóΪ_____, »¯Ñ§·½³ÌʽΪ______¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿»¯Ñ§·´Ó¦ÔÀíÔÚ»¯¹¤Éú²úºÍʵÑéÖÐÓÐ׏㷺¶øÖØÒªµÄÓ¦Óá£
¢ñ.ÀûÓú¬ÃÌ·ÏË®(Ö÷Òªº¬Mn2£«¡¢SO¡¢H£«¡¢Fe2£«¡¢Al3£«¡¢Cu2£«)¿ÉÖƱ¸¸ßÐÔÄÜ´ÅÐÔ²ÄÁÏ̼ËáÃÌ(MnCO3)¡£ÆäÖÐÒ»ÖÖ¹¤ÒÕÁ÷³ÌÈçÏ£º
ÒÑ֪ijЩÎïÖÊÍêÈ«³ÁµíµÄpHÈçÏÂ±í£º
³ÁµíÎï | Fe(OH)3 | Al(OH)3 | Cu(OH)2 | Mn(OH)2 | CuS | MnS | MnCO3 |
³ÁµíÍêȫʱµÄpH | 3.2 | 5.4 | 6.4 | 9.8 | ¡Ý0 | ¡Ý7 | ¡Ý7 |
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¹ý³Ì¢ÚÖУ¬ËùµÃÂËÔüWµÄÖ÷Òª³É·ÖÊÇ______________________¡£
£¨2£©¹ý³Ì¢ÛÖУ¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______________________¡£
£¨3£©¹ý³Ì¢ÜÖУ¬ÈôÉú³ÉµÄÆøÌåJ¿Éʹ³ÎÇåʯ»ÒË®±ä»ë×Ç£¬ÔòÉú³ÉMnCO3µÄ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_______________________¡£
£¨4£©ÓÉMnCO3¿ÉÖƵÃÖØÒªµÄ´ß»¯¼ÁMnO2£º2MnCO3£«O2===2MnO2£«2CO2¡£ÏÖÔÚ¿ÕÆøÖмÓÈÈ460.0 g MnCO3£¬µÃµ½332.0 g²úÆ·£¬Èô²úÆ·ÖÐÔÓÖÊÖ»ÓÐMnO£¬Ôò¸Ã²úÆ·ÖÐMnO2µÄÖÊÁ¿·ÖÊýÊÇ________(ÓðٷÖÊý±íʾ£¬Ð¡Êýµãºó±£Áô1λСÊý)¡£
¢ò.³£ÎÂÏ£¬Å¨¶È¾ùΪ0.1 mol¡¤L£1µÄÏÂÁÐÁùÖÖÈÜÒºµÄpHÈçÏÂ±í£º
ÈÜÖÊ | CH3COONa | NaHCO3 | Na2CO3 | NaClO | NaCN/span> | C6H5ONa |
pH | 8.8 | 9.7 | 11.6 | 10.3 | 11.1 | 11.3 |
£¨1£©ÉÏÊöÑÎÈÜÒºÖеÄÒõÀë×Ó£¬½áºÏH£«ÄÜÁ¦×îÇ¿µÄÊÇ_______________________¡£
£¨2£©¸ù¾Ý±íÖÐÊý¾ÝÅжϣ¬Å¨¶È¾ùΪ0.01 mol¡¤L£1µÄÏÂÁÐÎïÖʵÄÈÜÒºÖУ¬ËáÐÔ×îÇ¿µÄÊÇ________(ÌîÐòºÅ)¡£
A. HCN¡¡ B. HClO¡¡ C. C6H5OH¡¡ D. CH3COOH E. H2CO3
¢ó.ÒÑÖª:CO(g)+H2O(g)CO2(g)+H2(g)¡¡¦¤H = Q kJ¡¤mol-1Æäƽºâ³£ÊýËæζȱ仯ÈçϱíËùʾ:
ζÈ/¡æ | 400 | 500 | 850 |
ƽºâ³£Êý | 9.94 | 9 | 1 |
Çë»Ø´ðÏÂÁÐÎÊÌâ:
£¨1£©ÉÏÊö·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪ___,¸Ã·´Ó¦µÄQ____(Ìî¡°>¡±»ò¡°<¡±)0¡£
£¨2£©850 ¡æʱ,ÏòÌå»ýΪ10 LµÄ·´Ó¦Æ÷ÖÐͨÈëÒ»¶¨Á¿µÄCOºÍH2O(g),·¢ÉúÉÏÊö·´Ó¦,COºÍH2O(g)µÄŨ¶È±ä»¯ÈçͼËùʾ,Ôò0~4 minʱƽ¾ù·´Ó¦ËÙÂÊv(CO)=____¡£
£¨3£©ÈôÔÚ500 ¡æʱ½øÐÐÉÏÊö·´Ó¦,ÇÒCO¡¢H2O(g)µÄÆðʼŨ¶È¾ùΪ0.020 mol¡¤L-1,¸ÃÌõ¼þÏÂ,COµÄ×î´óת»¯ÂÊΪ____¡£
£¨4£©ÈôÔÚ850 ¡æʱ½øÐÐÉÏÊö·´Ó¦,ÉèÆðʼʱCOºÍH2O(g)¹²Îª1 mol,ÆäÖÐË®ÕôÆøµÄÌå»ý·ÖÊýΪx,ƽºâʱCOµÄת»¯ÂÊΪy,ÊÔÍƵ¼yËæx±ä»¯µÄ¹Øϵʽ:____¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÓлúºÏ³ÉÖÆÒ©¹¤ÒµÖеÄÖØÒªÊֶΡ£GÊÇij¿¹Ñ×Ö¢Ò©ÎïµÄÖмäÌ壬ÆäºÏ³É·ÏßÈçÏ£º
ÒÑÖª£º
£¨¾ßÓл¹ÔÐÔ£¬¼«Ò×±»Ñõ»¯£©¡£
£¨1£©BµÄ½á¹¹¼òʽΪ ______¡£
£¨2£©·´Ó¦¢ÜµÄÌõ¼þΪ ___________ £»¢ÙµÄ·´Ó¦ÀàÐÍΪ ____________ £»·´Ó¦¢ÚµÄ×÷ÓÃÊÇ_________ ¡£
£¨3£©ÏÂÁжÔÓлúÎïGµÄÐÔÖÊÍƲâÕýÈ·µÄÊÇ __________£¨ÌîÑ¡Ïî×Öĸ£©¡£
A£®¾ßÓÐÁ½ÐÔ£¬¼ÈÄÜÓëËá·´Ó¦Ò²ÄÜÓë¼î·´Ó¦
B£®ÄÜ·¢ÉúÏûÈ¥·´Ó¦¡¢È¡´ú·´Ó¦ºÍÑõ»¯·´Ó¦
C£®Äܾۺϳɸ߷Ö×Ó»¯ºÏÎï
D£®1molGÓë×ãÁ¿NaHCO3ÈÜÒº·´Ó¦·Å³ö2molCO2
£¨4£©DÓë×ãÁ¿µÄNaHCO3ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______¡£
£¨5£©·ûºÏÏÂÁÐÌõ¼þµÄCµÄͬ·ÖÒì¹¹ÌåÓÐ _____ÖÖ¡£
A£®ÊôÓÚ·¼Ïã×廯ºÏÎÇÒº¬ÓÐÁ½¸ö¼×»ù
B£®ÄÜ·¢ÉúÒø¾µ·´Ó¦
C£®ÓëFeCI3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦
ÆäÖк˴Ź²ÕñÇâÆ×ÓÐ4×é·å£¬ÇÒ·åÃæ»ýÖ®±ÈΪ6£º2£º2£º1µÄÊÇ___£¨Ð´³öÆäÖÐÒ»Öֽṹ¼òʽ£©
£¨6£©ÒÑÖª£º±½»·ÉÏÓÐÍéÌþ»ùʱ£¬ÐÂÒýÈëµÄÈ¡´ú»ùÁ¬ÔÚ±½»·µÄÁÚ¶Ôλ£»±½»·ÉÏÓзå»ùʱ£¬ÐÂÒýÈëµÄÈ¡´ú»ùÁ¬ÔÚ±½Ò»Æ½µÄ¼äλ¡£¸ù¾ÝÌâÖеÄÐÅÏ¢£¬Ð´ÏÖÒÔ¼×±½ÎªÔÁϺϳÉÓлúÎ£©µÄÁ÷³Ìͼ£¨ÎÞ»úÊÔ¼ÁÈÎÑ¡£©________¡£ºÏ³É·ÏßÁ÷³ÌͼʾÀýÈçÏ£º Ä¿±ê²úÎï¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¿Æѧ¼ÒÑÐÖƳöµÄ°ËÏõ»ùÁ¢·½Í飨½á¹¹ÈçͼËùʾ£¬Ì¼Ô×Óδ»³ö£©ÊÇÒ»ÖÖÐÂÐ͸ßÄÜÕ¨Ò©£¬±¬Õ¨·Ö½âµÃµ½ÎÞ¶¾¡¢Îȶ¨µÄÆøÌ壬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©
A.·Ö×ÓÖÐC£¬N¼äÐγɷǼ«ÐÔ¼ü
B.1mol¸Ã·Ö×ÓÖк¬8mol ¶þÑõ»¯µª
C.¸ÃÎïÖʼÈÓÐÑõ»¯ÐÔÓÖÓл¹ÔÐÔ
D.¸ÃÎïÖʱ¬Õ¨²úÎïÊÇNO2¡¢CO2¡¢H2O
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÉÆÓÚ¹éÄÉÔªËؼ°Æ仯ºÏÎïÖ®¼äµÄת»¯¹Øϵ£¬¶ÔѧϰԪËØ»¯ºÏÎï¾ßÓÐÖØÒªÒâÒå¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Öû»·´Ó¦ÊÇÖÐѧ³£¼û»ù±¾·´Ó¦ÀàÐÍÖ®Ò»¡£
¢Ùд³ö½ðÊôµ¥ÖÊÖû»³ö·Ç½ðÊôµ¥ÖʵÄÀë×Ó·´Ó¦·½³Ìʽ_____________£»
¢Úд³ö·Ç½ðÊôµ¥ÖÊÖû»³ö·Ç½ðÊôµ¥ÖʵĻ¯Ñ§·½³Ìʽ_____________¡£
£¨2£©¡°Èý½Çת»¯¡±Êǵ¥Öʼ°Æ仯ºÏÎïÖ®¼äÏ໥ת»¯Öг£¼ûµÄת»¯¹Øϵ֮һ¡£ÏÖÓÐÈçͼת»¯¹Øϵ¡£
¢ÙÈôZÊÇÒ»ÖÖÄܹ»ÓëѪºìµ°°×½áºÏÔì³ÉÈËÌåȱÑõµÄÆøÌå¡£ÔòaµÄÃû³ÆΪ___________£»
¢ÚÈôXÊÇÒ»ÖÖ³£¼û½ðÊôµ¥ÖÊ£¬aΪһÖÖº¬ÑõËá¡£Yת»¯ÎªZµÄÀë×Ó·½³Ìʽ___________¡£
¢ÛÈôXÊÇÒ»ÖÖÇ¿¼î£¬aÊÇÒ»ÖÖËáÐÔÑõ»¯Îï¡££Úת»¯Îª£ÙµÄÀë×Ó·´Ó¦·½³Ìʽ____________£»
£¨3£©ÔÚÏÂͼµÄת»¯¹ØϵÖУ¬£Á¡¢£Â¡¢£Ã¡¢£Ä¡¢£ÅΪº¬ÓÐÒ»ÖÖÏàͬԪËصÄÎåÖÖÎïÖÊ¡£
¢ÙÈô£ÁΪµ»ÆÉ«¹ÌÌåµ¥ÖÊ£¬Ôò£Â¡ú£ÄµÄ»¯Ñ§·½³ÌʽΪ_____________£»
¢ÚÈô£Á³£ÎÂÏÂΪÆøÌåµ¥ÖÊ£¬ÊÕ¼¯ÆøÌå£Â²ÉÓõķ½·¨Îª________£»Èô½«32 gÍͶÈëÉÔ¹ýÁ¿µÄ£ÅµÄŨÈÜÒºÖУ¬²úÉúÆøÌåµÄÌå»ýΪ11.2 L(STP),²Î¼Ó·´Ó¦µÄÏõËáµÄÎïÖʵÄÁ¿Îª_____£»
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁи÷×éÖÐ΢Á£ÄÜ´óÁ¿¹²´æ£¬ÇÒµ±¼ÓÈëÊÔ¼Áºó·´Ó¦µÄÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ(¡¡ ¡¡)
Ñ¡Ïî | ΢Á£×é | ¼ÓÈëÊÔ¼Á | ·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ |
A | Fe3£«¡¢I£¡¢Cl£ | NaOHÈÜÒº | Fe3£«£«3OH£===Fe(OH)3¡ý |
B | K£«¡¢NH3¡¤H2O¡¢CO | ͨÈëÉÙÁ¿CO2 | 2OH££«CO2===CO£«H2O |
C | H£«¡¢Fe2£«¡¢SO | Ba(NO3)2ÈÜÒº | SO£«Ba2£«===BaSO4¡ý |
D | Na£«¡¢Al3£«¡¢Cl£ | ÉÙÁ¿³ÎÇåʯ»ÒË® | Al3£«£«3OH£===Al(OH)3¡ý |
A. A B. B C. C D. D
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