ÌâÄ¿ÄÚÈÝ
ÏÖÓÐÔ×ÓÐòÊýÒÀ´ÎÔö´óµÄX¡¢Y¡¢Z¡¢WËÄÖÖ¶ÌÖÜÆÚÔªËØ£®XÓëWͬÖ÷×壮YÓëZͬÖÜÆÚÇÒλÖÃÏàÁÚ£¬Æäµ¥ÖÊÔÚ³£ÎÂϾùΪÎÞÉ«ÆøÌ壮WµÄÖÊ×ÓÊýµÈÓÚYÓëZÁ½Ô×ÓµÄ×îÍâ²ãµç×ÓÊýÖ®ºÍ£®
ÒÑÖªZÓëW¿ÉÐγÉÀë×Ó»¯ºÏÎïW2Z2£¬XÓëY¿ÉÐγÉÑôÀë×ÓYX4+£¬X¡¢Y¡¢Z¿ÉÐγɻ¯ºÏÎïXYZ3£®Çë»Ø´ð£º
£¨1£©WÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃΪ £®
£¨2£©W2Z2ÓëË®·´Ó¦µÄÀë×Ó·½³ÌʽΪ £®
£¨3£©ÊµÑéÊÒÖмìÑéYX4+µÄ²Ù×÷·½·¨¼°ÓйØÊµÑéÏÖÏóÊÇ£ºÈ¡ÉÙÁ¿º¬YX4+µÄÈÜÒº£¬ £¬ÔòÖ¤Ã÷ÈÜÒºÖк¬ÓÐYX4+£®
£¨4£©XYZ3ÔÚÒ»¶¨Ìõ¼þÏÂÒ×·¢Éú·Ö½â·´Ó¦£¬Èô·´Ó¦ÖÐ×ªÒÆµç×ÓÊýΪ9.03¡Á1022£¬ÔòÓÐ mol XYZ3·¢Éú·Ö½â£®
ÒÑÖªZÓëW¿ÉÐγÉÀë×Ó»¯ºÏÎïW2Z2£¬XÓëY¿ÉÐγÉÑôÀë×ÓYX4+£¬X¡¢Y¡¢Z¿ÉÐγɻ¯ºÏÎïXYZ3£®Çë»Ø´ð£º
£¨1£©WÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃΪ
£¨2£©W2Z2ÓëË®·´Ó¦µÄÀë×Ó·½³ÌʽΪ
£¨3£©ÊµÑéÊÒÖмìÑéYX4+µÄ²Ù×÷·½·¨¼°ÓйØÊµÑéÏÖÏóÊÇ£ºÈ¡ÉÙÁ¿º¬YX4+µÄÈÜÒº£¬
£¨4£©XYZ3ÔÚÒ»¶¨Ìõ¼þÏÂÒ×·¢Éú·Ö½â·´Ó¦£¬Èô·´Ó¦ÖÐ×ªÒÆµç×ÓÊýΪ9.03¡Á1022£¬ÔòÓÐ
¿¼µã£ºÎ»ÖýṹÐÔÖʵÄÏ໥¹ØÏµÓ¦ÓÃ
רÌâ£ºÔªËØÖÜÆÚÂÉÓëÔªËØÖÜÆÚ±íרÌâ
·ÖÎö£ºÔ×ÓÐòÊýÒÀ´ÎÔö´óµÄX¡¢Y¡¢Z¡¢WËÄÖÖ¶ÌÖÜÆÚÔªËØ£®YÓëZͬÖÜÆÚÇÒλÖÃÏàÁÚ£¬Æäµ¥ÖÊÔÚ³£ÎÂϾùΪÎÞÉ«ÆøÌ壬ÔòYΪµªÔªËØ¡¢ZΪÑõÔªËØ£»WµÄÖÊ×ÓÊýµÈÓÚYÓëZÁ½Ô×ÓµÄ×îÍâ²ãµç×ÓÊýÖ®ºÍ£¬ÔòWµÄÖÊ×ÓÊý=5+6=11£¬¹ÊWΪNa£»XÓëWͬÖ÷×壬ÇÒXÓëY¿ÉÐγÉÑôÀë×ÓYX4+£¬ÔòXΪÇâÔªËØ£¬ÑôÀë×ÓYX4+ΪNH4+£¬ZÓëWÐγÉÀë×Ó»¯ºÏÎïW2Z2ΪNa2O2£¬X¡¢Y¡¢Z¿ÉÐγɻ¯ºÏÎïXYZ3ΪHNO3£¬¾Ý´Ë½â´ð£®
½â´ð£º
½â£ºÔ×ÓÐòÊýÒÀ´ÎÔö´óµÄX¡¢Y¡¢Z¡¢WËÄÖÖ¶ÌÖÜÆÚÔªËØ£®YÓëZͬÖÜÆÚÇÒλÖÃÏàÁÚ£¬Æäµ¥ÖÊÔÚ³£ÎÂϾùΪÎÞÉ«ÆøÌ壬ÔòYΪµªÔªËØ¡¢ZΪÑõÔªËØ£»WµÄÖÊ×ÓÊýµÈÓÚYÓëZÁ½Ô×ÓµÄ×îÍâ²ãµç×ÓÊýÖ®ºÍ£¬ÔòWµÄÖÊ×ÓÊý=5+6=11£¬¹ÊWΪNa£»XÓëWͬÖ÷×壬ÇÒXÓëY¿ÉÐγÉÑôÀë×ÓYX4+£¬ÔòXΪÇâÔªËØ£¬ÑôÀë×ÓYX4+ΪNH4+£¬ZÓëWÐγÉÀë×Ó»¯ºÏÎïW2Z2ΪNa2O2£¬X¡¢Y¡¢Z¿ÉÐγɻ¯ºÏÎïXYZ3ΪHNO3£¬
£¨1£©WΪNaÔªËØ£¬ÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃΪµÚÈýÖÜÆÚIA×壬
¹Ê´ð°¸Îª£ºµÚÈýÖÜÆÚIA×壻
£¨2£©Na2O2ÓëË®·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2Na2O2+2H2O¨T4Na++4OH-+O2¡ü£¬
¹Ê´ð°¸Îª£º2Na2O2+2H2O¨T4Na++4OH-+O2¡ü£»
£¨3£©ÊµÑéÊÒÖмìÑéNH4+·½·¨ÊÇ£ºÈ¡ÉÙÁ¿º¬NH4+µÄÈÜÒº£¬¼ÓÈëÇâÑõ»¯ÄÆÈÜÒº²¢¼ÓÈÈ£¬²úÉúÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壬ÔòÖ¤Ã÷ÈÜÒºÖк¬ÓÐNH4+£¬
¹Ê´ð°¸Îª£º¼ÓÈëÇâÑõ»¯ÄÆÈÜÒº²¢¼ÓÈÈ£¬²úÉúÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壻
£¨4£©·¢Éú·´Ó¦£º4HNO3=4NO2¡ü+O2+2H2O£¬Èô·´Ó¦ÖÐ×ªÒÆµç×ÓÊýΪ9.03¡Á1022£¬ÔòÉú³É¶þÑõ»¯µªµÄÎïÖʵÄÁ¿=
=0.15mol£¬ÔòÓÐ0.15mol HNO3·¢Éú·Ö½â£¬
¹Ê´ð°¸Îª£º0.15£®
£¨1£©WΪNaÔªËØ£¬ÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃΪµÚÈýÖÜÆÚIA×壬
¹Ê´ð°¸Îª£ºµÚÈýÖÜÆÚIA×壻
£¨2£©Na2O2ÓëË®·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2Na2O2+2H2O¨T4Na++4OH-+O2¡ü£¬
¹Ê´ð°¸Îª£º2Na2O2+2H2O¨T4Na++4OH-+O2¡ü£»
£¨3£©ÊµÑéÊÒÖмìÑéNH4+·½·¨ÊÇ£ºÈ¡ÉÙÁ¿º¬NH4+µÄÈÜÒº£¬¼ÓÈëÇâÑõ»¯ÄÆÈÜÒº²¢¼ÓÈÈ£¬²úÉúÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壬ÔòÖ¤Ã÷ÈÜÒºÖк¬ÓÐNH4+£¬
¹Ê´ð°¸Îª£º¼ÓÈëÇâÑõ»¯ÄÆÈÜÒº²¢¼ÓÈÈ£¬²úÉúÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壻
£¨4£©·¢Éú·´Ó¦£º4HNO3=4NO2¡ü+O2+2H2O£¬Èô·´Ó¦ÖÐ×ªÒÆµç×ÓÊýΪ9.03¡Á1022£¬ÔòÉú³É¶þÑõ»¯µªµÄÎïÖʵÄÁ¿=
| 9.03¡Á1022 |
| 6.03¡Á1023mol-1 |
¹Ê´ð°¸Îª£º0.15£®
µãÆÀ£º±¾Ì⿼²é½á¹¹ÐÔÖÊλÖùØÏµÓ¦Óã¬ÍƶÏÔªËØÊǽâÌâ¹Ø¼ü£¬²àÖØ¶Ô»ù´¡ÖªÊ¶µÄ¹®¹Ì£¬×¢ÒâÕÆÎÕ笠ùÀë×ӵļìÑ飬Àí½âÑõ»¯»¹Ô·´Ó¦Öеç×Ó×ªÒÆÊØºãµÄÓ¦Óã®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
| A¡¢XºÍY²»Á¬½Óʱ£¬Í°ôÉÏ»áÓнðÊôÒøÎö³ö |
| B¡¢XºÍYÓõ¼ÏßÁ¬½Óʱ£¬Òø°ôÉÏ·¢ÉúµÄ·´Ó¦ÎªAg++e-=Ag |
| C¡¢ÈôX½ÓÖ±Á÷µçÔ´µÄÕý¼«Y½Ó¸º¼«£¬Ag+ÏòÒøµç¼«Òƶ¯ |
| D¡¢ÎÞÂÛXºÍYÊÇ·ñÁ¬½Ó£¬Í°ô¾ù»áÈܽ⣬ÈÜÒº¶¼´ÓÎÞÉ«Öð½¥±ä³ÉÀ¶É« |
ÔÚÏÂÁÐÈÜÒºÖУ¬¸÷×éÀë×ÓÒ»¶¨Äܹ»´óÁ¿¹²´æµÄÊÇ£¨¡¡¡¡£©
| A¡¢º¬ÓдóÁ¿[Al£¨OH£©4]-µÄÈÜÒºÖУºNa+¡¢OH-¡¢Cl-¡¢CO32- | ||
B¡¢³£ÎÂÏÂ
| ||
| C¡¢ÄÜʹpHÊÔÖ½ÏÔºìÉ«µÄÈÜÒºÖУºNa+¡¢ClO-¡¢Fe2+¡¢SO42- | ||
| D¡¢ÄÜʹµí·Ûµâ»¯¼ØÊÔÖ½ÏÔÀ¶É«µÄÈÜÒº£ºK+¡¢SO42-¡¢S2-¡¢SO32- |
| A¡¢ÔÚpH=13ʱ£¬ÈÜÒºÖÐÖ÷Òªº¬Á×ÎïÖÖŨ¶È´óС¹ØÏµÎªc£¨HPO42-£©£¾c£¨PO43-£© |
| B¡¢Îª»ñµÃ¾¡¿ÉÄÜ´¿µÄNaH2PO4£¬pHÓ¦¿ØÖÆÔÚ4¡«5.5×óÓÒ |
| C¡¢ÔÚpH=7.2ʱ£¬HPO42-¡¢H2PO4-µÄ·Ö²¼·ÖÊý¸÷Ϊ0.5£¬ÔòH3PO4µÄKa2=10-7.2 |
| D¡¢Na2HPO4ÈÜÒºÏÔ¼îÐÔ£¬ÈôÏòÈÜÒºÖмÓÈë×ãÁ¿µÄCaCl2ÈÜÒº£¬ÈÜÒºÔòÏÔËáÐÔ |