1£®Èçͼ±íʾһЩ¾§ÌåÖеÄijЩ½á¹¹£¬ËüÃÇ·Ö±ðÊÇNaCl¡¢¸É±ù¡¢½ð¸Õʯ¡¢Ê¯Ä«½á¹¹ÖеÄijһÖÖµÄijһ²¿·Ö£®

ÏÂÁÐ6ÖÖÎïÖÊ£ºA£®P4£¨°×Á×£©£¬B£®SiO2£¬C£®CaCl2£¬D£®Ca£¨OH£©2£¬E£®NaF£¬F£®CO2£¬¹Ì̬϶¼Îª¾§Ì壬»Ø´ðÏÂÁÐÎÊÌ⣺£¨Ìî×Öĸ±àºÅ£©
£¨1£©ÈÛ»¯Ê±ÓëÆÆ»µ£¨1£©¾§ÌåÀàÐÍÏàͬ»¯Ñ§¼üµÄÓÐCDE£»
£¨2£©ÆäÖÐÓ루2£©¾§ÌåÀàÐÍÏàͬµÄÓÐB£¬ËüÃÇÊôÓÚÔ­×Ó¾§Ì壮
£¨3£©Ó루3£©¾§ÌåÀàÐÍÏàͬµÄÓÐA£¬ÈÛ»¯Ê±ÐèÒªÆÆ»µ·Ö×Ó¼ä×÷ÓÃÁ¦£®
£¨4£©·ÖÎö¶Ô±È£¨2£©£¨4£©¾§Ì壬Ӳ¶È½Ï´óµÄÊÇ£¨2£©£®
£¨5£©ÉÏÊö£¨1£©-£¨3£©ÈýÖÖ¾§Ìåͨ³£ÈÛµãÓɸߵ½µÍµÄÅÅÁÐ˳ÐòΪ£¨ÌîÊý×Ö±àºÅ£©£¨2£©£¨1£©£¨3£©£®

·ÖÎö £¨1£©Í¼ÖУ¨1£©ÊôÓÚÂÈ»¯ÄƵĽṹͼ£¬ÂÈ»¯ÄÆÈÛ»¯Ê±ÆÆ»µÀë×Ó¼ü£»
£¨2£©Í¼ÖУ¨2£©ÊôÓÚ½ð¸ÕʯµÄ½á¹¹Í¼£¬ÊôÓÚÔ­×Ó¾§Ì壻
£¨3£©Í¼ÖУ¨3£©ÊôÓÚ¶þÑõ»¯Ì¼µÄ½á¹¹Í¼£¬¶þÑõ»¯Ì¼ÊôÓÚ·Ö×Ó¾§Ì壬ÈÛ»¯Ê±ÆÆ»µ·Ö×Ó¼ä×÷ÓÃÁ¦£»
£¨4£©Í¼ÖУ¨4£©ÊôÓÚʯīµÄ½á¹¹Í¼£¬Ê¯Ä«ÊôÓÚ»ìºÏ¾§Ì壬½ð¸ÕʯÊôÓÚÔ­×Ó¾§Ì壻
£¨5£©ÈÛµãµÄÒ»°ã¹æÂÉ£ºÔ­×Ó¾§Ì壾Àë×Ó¾§Ì壾·Ö×Ó¾§Ì壮

½â´ð ½â£º£¨1£©Í¼ÖУ¨1£©ÊôÓÚÂÈ»¯ÄƵĽṹͼ£¬ÂÈ»¯ÄÆÈÛ»¯Ê±ÆÆ»µÀë×Ó¼ü£¬º¬ÓÐÀë×Ó¼üµÄ»¯ºÏÎïΪ£ºC£®CaCl2£¬D£®Ca£¨OH£©2£¬E£®NaF£»
¹Ê´ð°¸Îª£ºCDE£»
£¨2£©Í¼ÖУ¨2£©ÊôÓÚ½ð¸ÕʯµÄ½á¹¹Í¼£¬ÊôÓÚÔ­×Ó¾§Ì壬¶þÑõ»¯¹èÒ²ÊôÓÚÔ­×Ó¾§Ì壻
¹Ê´ð°¸Îª£ºB£»Ô­×Ó£»
£¨3£©Í¼ÖУ¨3£©ÊôÓÚ¶þÑõ»¯Ì¼µÄ½á¹¹Í¼£¬¶þÑõ»¯Ì¼ÊôÓÚ·Ö×Ó¾§Ì壬ÈÛ»¯Ê±ÆÆ»µ·Ö×Ó¼ä×÷ÓÃÁ¦£¬°×Á×Ò²ÊÇ·Ö×Ó¾§Ì壬ÔòÊôÓÚ·Ö×Ó¾§ÌåµÄÊÇAF£»
¹Ê´ð°¸Îª£ºA£»·Ö×Ӽ䣻
£¨4£©Í¼ÖУ¨4£©ÊôÓÚʯīµÄ½á¹¹Í¼£¬Ê¯Ä«ÊôÓÚ»ìºÏ¾§Ì壬¾§ÌåÖдæÔÚ²ã×´½á¹¹£¬²ãÓë²ãÖ®¼äÄÜ·¢Éú»¬¶¯£¬ËùÒÔʯīµÄÓ²¶È½ÏС£¬½ð¸ÕʯÊôÓÚÔ­×Ó¾§ÌåÓ²¶ÈºÜ´ó£»
¹Ê´ð°¸Îª£º£¨2£©£»
£¨5£©ÈÛµãµÄÒ»°ã¹æÂÉ£ºÔ­×Ó¾§Ì壾Àë×Ó¾§Ì壾·Ö×Ó¾§Ì壬£¨2£©ÊôÓÚ½ð¸ÕʯµÄ½á¹¹Í¼£¬ÊôÓÚÔ­×Ó¾§Ì壬£¨1£©ÊôÓÚÂÈ»¯ÄƵĽṹͼ£¬ÂÈ»¯ÄÆÊôÓÚÀë×Ó¾§Ì壬£¨3£©ÊôÓÚ¶þÑõ»¯Ì¼µÄ½á¹¹Í¼£¬¶þÑõ»¯Ì¼ÊôÓÚ·Ö×Ó¾§Ì壬ÔòÈÛµãÓɸߵ½µÍµÄÅÅÁÐ˳ÐòΪ£¨2£©£¨1£©£¨3£©£»
¹Ê´ð°¸Îª£º£¨2£©£¨1£©£¨3£©£®

µãÆÀ ±¾Ì⿼²éÁ˾§ÌåÀàÐ͵ÄÅжϡ¢»¯Ñ§¼ü¡¢ÈÛµã±È½Ï£¬ÄѶȲ»´ó£¬¸ù¾Ý²»Í¬ÎïÖʾ§ÌåµÄ½á¹¹ÌصãÀ´±æ±ðͼÐÎËù´ú±íµÄÎïÖÊÀ´½â´ð¼´¿É£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

11£®Ä³°×É«¹ÌÌå¿ÉÄÜÓÉ¢ÙNH4Cl¡¢¢ÚAlCl3¡¢¢ÛKCl¡¢¢ÜAgNO3¡¢¢ÝNaOHÖеÄÒ»ÖÖ»ò¼¸ÖÖ×é³É£¬´Ë¹ÌÌåͶÈëË®Öеõ½³ÎÇåÈÜÒº£¬¸ÃÈÜÒº¿Éʹ·Ó̪³ÊºìÉ«£¬ÈôÏòÈÜÒºÖмÓÏ¡ÏõËáµ½¹ýÁ¿£¬Óа×É«³Áµí²úÉú£®Ôò¶ÔÔ­¹ÌÌåµÄÅжϲ»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¿Ï¶¨´æÔÚ¢ÙB£®ÖÁÉÙ´æÔڢں͢ÝC£®ÎÞ·¨È·¶¨ÊÇ·ñÓТÛD£®ÖÁÉÙ´æÔÚ¢Ù¡¢¢Ü¡¢¢Ý

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

12£®CuSO4?5H2OÊÇÍ­µÄÖØÒª»¯ºÏÎÓÐ׏㷺µÄÓ¦Óã®ÈçͼÊÇCuSO4?5H2OµÄʵÑéÊÒÖƱ¸Á÷³ÌȦ£®

¸ù¾ÝÌâÒâÍê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©Ïòº¬Í­·ÛµÄÏ¡ÁòËáÖеμÓÉÙÁ¿Å¨ÏõËᣨ¿É¼ÓÈÈ£©£¬ÔÚÍ­·ÛÈܽâʱ¿ÉÒԹ۲쵽µÄʵÑéÏÖÏó£ºÈÜÒº³ÊÀ¶É«£»ÓÐÎÞÉ«ÆøÌåÉú³É£®
£¨2£©¸ù¾Ý·´Ó¦Ô­Àí£¬ÏõËáÓëÁòËáµÄÀíÂÛÅä±È£¨ÎïÖʵÄÁ¿Ö®±È£©Îª2£º3£®
£¨3£©ÒÑÖª£ºCuSO4+2NaOH¡úCu£¨OH£©2+Na2SO4
³ÆÈ¡0.1000gÌá´¿ºóµÄCuSO4?5H2OÊÔÑùÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë0.1000mol/LÇâÑõ»¯ÄÆÈÜÒº28.00mL£¬·´Ó¦ÍêÈ«ºó£¬¹ýÁ¿µÄÇâÑõ»¯ÄÆÓÃ0.1000mol/LÑÎËáµÎ¶¨ÖÁÖյ㣬ºÄÓÃÑÎËá20.16mL£¬Ôò0.1000g¸ÃÊÔÑùÖк¬CuSO4?5H2O0.098g£®
£¨4£©Ôڵζ¨ÖУ¬ÑÛ¾¦Ó¦×¢ÊÓ׶ÐÎÆ¿ÖÐÈÜÒºÑÕÉ«±ä»¯£»µÎ¶¨ÖÕµãʱ£¬×¼È·¶ÁÊýÓ¦¸ÃÊǵζ¨¹ÜÉÏÀ¶Ïß´Öϸ½»½çµãËù¶ÔÓ¦µÄ¿Ì¶È£®
£¨5£©ÈôÉÏÊöµÎ¶¨²Ù×÷ÖУ¬µÎ¶¨¹Ü¼ÓÑÎËá֮ǰδ½øÐÐÈóÏ´£¬Ôò²âµÃÊÔÑùÖÐËùº¬CuSO4?5H2OµÄÖÊÁ¿Æ«Ð¡£®£¨Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©
£¨7£©Èç¹ûl.040gÌá´¿ºóµÄÊÔÑùÖк¬CuSO4?5H2OµÄ׼ȷֵΪ1.015g£¬¶øʵÑé²â¶¨½á¹ûÊÇ1.000g£¬²â¶¨µÄÏà¶ÔÎó²îΪ-1.48%£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

9£®X¡¢Y¡¢Z¾ùΪ¶ÌÖÜÆÚÔªËØ£¬X¡¢YͬһÖÜÆÚ£¬X¡¢ZµÄ×îµÍ¼ÛÀë×Ó·Ö±ðΪX2-ºÍZ-£¬Y+ºÍZ-¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Ô­×ÓµÄ×îÍâ²ãµç×ÓÊý£ºX£¾Y£¾ZB£®µ¥Öʷе㣺Z£¾Y
C£®Àë×Ӱ뾶£ºX2-£¾Y+£¾Z-D£®Ô­×ÓÐòÊý£ºX£¾Y£¾Z

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

16£®£¨1£©ÏÂÁÐËÄÖÖÁ£×ÓÖУ¬°ë¾¶°´ÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòÊÇS2-£¾S£¾Cl£¾N£®
¢Ù»ù̬XµÄÔ­×ӽṹʾÒâͼ£º
¢Ú»ù̬YµÄ¼Ûµç×ÓÅŲ¼Ê½£º3s23p5
¢Û»ù̬Z2-µÄµç×ÓÅŲ¼Í¼£º

¢ÜW»ù̬ԭ×ÓÓÐ2¸öÄܲ㣬µç×Óʽ£º
£¨2£©ÒÑÖªAn+¡¢B£¨n+1£©+¡¢Cn-¡¢D£¨n+1£©-¶¼¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£¬ÔòA¡¢B¡¢C¡¢DµÄÔ­×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇA£¾B£¾D£¾C£¬Àë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇD£¾C£¾A£¾B£¬Ô­×ÓÐòÊýÓÉ´óµ½Ð¡µÄ˳ÐòÊÇB£¾A£¾C£¾D£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

6£®ÈçͼËùʾ£¬×¶ÐÎÆ¿ÄÚÊ¢ÓÐÆøÌåX£¬µÎ¹ÜÄÚÊ¢ÓÐÒºÌåY£®Èô¼·Ñ¹½ºÍ·µÎ¹Ü£¬Ê¹ÒºÌåYµÎÈë׶ÐÎÆ¿ÖУ¬Õñµ´£¬¹ýÒ»»á¶ù£¬¿É¼ûСÆøÇòa¹ÄÕÍÆðÀ´£®ÆøÌåXºÍÒºÌåY²»¿ÉÄÜÊÇ£¨¡¡¡¡£©
XY
ANH3H2O
BSO2NaOHÈÜÒº
CCO26mol•L-1 H2SO4ÈÜÒº
DHCl6mol•L-1 Na2SO4ÈÜÒº
A£®AB£®BC£®CD£®D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

13£®¶ÔÓÚZn£¨s£©+H2SO4£¨aq£©¨TZnSO4£¨aq£©+H2£¨g£©¡÷H£¼0µÄ»¯Ñ§·´Ó¦ÏÂÁÐÐðÊö²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®
·´Ó¦¹ý³ÌÖÐÄÜÁ¿¹Øϵ¿ÉÓÃÈçͼ±íʾ
B£®¡÷HµÄÖµÓë·´Ó¦·½³ÌʽµÄ¼ÆÁ¿ÏµÊýÓйØ
C£®Èô½«¸Ã·´Ó¦Éè¼Æ³ÉÔ­µç³Ø£¬Ð¿Îª¸º¼«
D£®Èô½«ÆäÉè¼ÆΪԭµç³Ø£¬µ±ÓÐ32.5 gпÈܽâʱ£¬Õý¼«·Å³öÆøÌåÒ»¶¨Îª11.2 L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

10£®Ï±íÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬»Ø´ðÓйØÎÊÌ⣺
    Ö÷×å
ÖÜÆÚ
IAIIAIIIAIVAVAVIAVIIA0
¶þ£¨1£©£¨2£©
Èý£¨3£©£¨4£©£¨5£©£¨6£©£¨7£©£¨8£©£¨9£©
ËÄ£¨10£©£¨11£©£¨12£©
£¨1£©Ð´³öÏÂÁÐÔªËØ·ûºÅ£º£¨1£©N£¬£¨6£©Si
£¨2£©ÔÚÕâЩԪËØÖУ¬×î²»»îÆõÄÔªËصĽṹʾÒâͼÊÇ£®
£¨3£©ÔÚÕâЩԪËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÖУ¬ËáÐÔ×îÇ¿µÄÓë³ÊÁ½ÐԵķ¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ3HClO4+Al£¨OH£©3=Al£¨ClO4£©3+3H2O£»¼îÐÔ×îÇ¿µÄÓë³ÊÁ½ÐԵķ¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪAl£¨OH£©3+OH-=AlO2-+2H2O£®
£¨4£©ÔÚ£¨2£©£¨3£©£¨8£©£¨10£©ÕâЩԪËؼòµ¥Àë×ÓÖУ¬Àë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇCl-£¾K+£¾F-£¾Na+£¨ÌîÀë×Ó·ûºÅ£©
£¨5£©ÔªËØ£¨8£©¡¢£¨11£©ºÍÑõÔªËØÐγÉÒ»Öֹ㷺ʹÓÃɱ¾úÏû¶¾¼Á£¬¸ÃÎïÖÊÖдæÔڵĻ¯Ñ§¼üÓÐÀë×Ó¼ü¡¢¹²¼Û¼ü£®
£¨6£©Óõç×Óʽ±íʾÓÉÔªËØ£¨8£©ºÍ£¨10£©Ðγɻ¯ºÏÎïµÄ¹ý³Ì£º£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

11£®ÓлúÎïCH2=CH-CH2-OOC-CH3²»¿ÉÄܾßÓеÄÐÔÖÊÊÇ£¨¡¡¡¡£©
A£®ÄÜ·¢ÉúÈ¡´ú·´Ó¦B£®ÄÜ·¢Éú¼Ó³É·´Ó¦C£®ÄÜ·¢Éú¾ÛºÏ·´Ó¦D£®ÄÜÈÜÓÚË®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸