ÌâÄ¿ÄÚÈÝ
îæÔªËØ£¨Ce£©ÊÇïçϵ½ðÊôÖÐ×ÔÈ»·á¶È×î¸ßµÄÒ»ÖÖ£¬³£¼û¼Û̬ÓÐ+3¡¢+4£¬îæµÄºÏ½ðÄ͸ßΣ¬¿ÉÒÔÓÃÀ´ÖÆÔìÅçÆøÍƽøÆ÷Áã¼þ¡£
£¨1£©Îíö²Öк¬ÓдóÁ¿µÄÎÛȾÎïNO£¬¿ÉÒÔ±»Ce4+ÈÜÒºÎüÊÕ£¬Éú³ÉNO2£¡¢NO3££¨¶þÕßÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã1£©£¬¸Ã·´Ó¦Ñõ»¯¼ÁÓ뻹ԼÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ___________¡£
£¨2£©¿É²ÉÓõç½â·¨½«ÉÏÊöÎüÊÕÒºÖеÄNO2£×ª»¯ÎªÎÞ¶¾ÎïÖÊ£¬Í¬Ê±ÔÙÉúCe4+£¬ÆäÔÀíÈçÓÒͼËùʾ¡£
¢ÙCe4+´Óµç½â²ÛµÄ__________£¨Ìî×ÖĸÐòºÅ)¿ÚÁ÷³ö¡£
¢Úд³öÒõ¼«µÄµç¼«·´Ó¦Ê½____________________________¡£Ã¿ÏûºÄ1mol NO2££¬Òõ¼«ÇøH+ÎïÖʵÄÁ¿¼õÉÙ______mol¡£
£¨3£©îæÔªËØÔÚ×ÔÈ»ÖÐÖ÷ÒªÒÔ·ú̼¿óÐÎʽ´æÔÚ£¬Ö÷Òª»¯Ñ§³É·ÖΪCeFCO3£¬¹¤ÒµÉÏÀûÓ÷ú̼îæ¿óÌáÈ¡CeCl3µÄÒ»ÖÖ¹¤ÒÕÁ÷³ÌÈçÏ£º
¢Ù±ºÉÕ¹ý³ÌÖз¢ÉúµÄÖ÷Òª·´Ó¦·½³ÌʽΪ______________________________________¡£
¢ÚËá½þ¹ý³ÌÖÐÓÐͬѧÈÏΪÓÃÏ¡ÁòËáºÍH2O2Ìæ»»ÑÎËá¸üºÃ£¬ËûµÄÀíÓÉÊÇ_________________________¡£
¢ÛCe(BF4)3¡¢KBF4µÄKsp·Ö±ðΪa¡¢b£¬ÔòCe(BF4)3(s) + 3KCl(aq)3KBF4(s) + CeCl3 (aq)ƽºâ³£ÊýΪ______________________¡£
¢Ü¼ÓÈÈCeCl3¡¤6H2OºÍNH4ClµÄ¹ÌÌå»ìºÏÎï¿ÉµÃµ½ÎÞË®CeCl3£¬ÆäÖÐNH4ClµÄ×÷ÓÃÊÇ______________________¡£
ÏÂÁÐʵÑéÄÜ´ïµ½Ô¤ÆÚÄ¿µÄµÄÊÇ
±àºÅ | ʵÑéÄÚÈÝ | ʵÑéÄ¿µÄ |
A | Ïòij´ý²âÒºÖÐͨÈëCl2 ,ÔÙµÎÈë2µÎKSCNÈÜÒººó£¬ÈÜÒº±äΪѪºìÉ« | Ö¤Ã÷¸Ã´ý²âÒºÖÐÒ»¶¨º¬ÓÐFe2+ |
B | Ïòij´ý²âÒºÖмÓÈëÑÎËᣬ²úÉúÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×ǵÄÆøÌå | Ö¤Ã÷¸Ã´ý²âÒºÖÐÒ»¶¨º¬ÓÐCO32- |
C | °ÑÁòËáËữµÄH2O2ÈÜÒºµÎÈëFeCl2ÈÜÒºÖУ¬ÈÜÒº±ä³É»ÆÉ« | Ö¤Ã÷H2O2Ñõ»¯ÐÔ´óÓÚFe3+ |
D | ÏòAl(0H)3³ÁµíÖзֱð¼ÓÈëÑÎËáºÍ°±Ë®£¬³Áµí¶¼»áÈܽâ | Ö¤Ã÷Al(0H)3ÊÇÁ½ÐÔÇâÑõ»¯Îï |
A. A B. B C. C D. D
Ò½ÓÃÂÈ»¯¸Æ¿ÉÓÃÓÚÉú²ú²¹¸Æ¡¢¿¹¹ýÃôºÍÏûÑ׵ȣ¬ÒÔ¹¤ÒµÌ¼Ëá¸Æ£¨º¬ÉÙÁ¿Na+¡¢Al3+¡¢Fe3+µÈÔÓÖÊ)Éú²úÒ½Ò©¼¶¶þË®ºÏÂÈ»¯¸Æ(CaCl2¡¢2H2OµÄÖÊÁ¿·ÖÊýΪ97.0%¡«103.0%)µÄÖ÷ÒªÁ÷³ÌÈçÏ£º
ÒÑÖª£º
ÇâÑõ»¯Îï | Fe£¨OH£©3 | Al£¨OH£©3 | Al£¨OH£©3 | |
¿ªÊ¼³ÁµíʱµÄpH | 2.3 | 4.0 | ¿ªÊ¼ÈܽâʱµÄpH | 7.8 |
ÍêÈ«³ÁµíʱµÄpH | 3.7 | 5.2 | ÍêÈ«ÈܽâʱµÄpH | 10.8 |
£¨1£©CaCO3ÓëÑÎËá·´Ó¦µÄÀë×Ó·½³Ìʽ___________¡£
£¨2£©¡°³ýÔÓ¡±²Ù×÷ÊǼÓÈëÇâÑõ»¯¸Æ£¬µ÷½ÚÈÜÒºµÄpH·¶Î§Îª________£¬Ä¿µÄÊdzýÈ¥ÈÜÒºÖеÄÉÙÁ¿Al3+¡¢Fe2+¡£
£¨3£©¹ýÂËʱÐèÓõıȲ£Á§Æ÷ÓÐ__________¡£
£¨4£©¡°Ëữ¡±²Ù×÷ÊǼÓÈëÑÎËᣬµ÷½ÚÈÜÒºµÄpHԼΪ4.0£¬ÆäÄ¿µÄÓУº¢Ù·ÀÖ¹ÇâÑõ»¯¸ÆÎüÊÕ¿ÕÆøÖеĶþÑõ»¯Ì¼£»¢Ú·ÀÖ¹Ca2+ÔÚÕô·¢Ê±Ë®½â£»¢Û_______¡£
£¨5£©Õô·¢½á¾§Òª±£³ÖÔÚ160¡æµÄÔÒòÊÇ__________¡£
£¨6£©²â¶¨ÑùÆ·ÖÐCl-º¬Á¿µÄ·½·¨ÊÇ£º³ÆÈ¡0.750 0 gCaCl2¡¤2H2OÑùÆ·£¬Èܽ⣬ÔÚ250 mLÈÝÁ¿Æ¿Öж¨ÈÝ£»Á¿È¡25.00 mL´ý²âÈÜÒºÓÚ׶ÐÎÆ¿ÖУ»ÓÃ0.050 00 mol/L AgNO3ÈÜÒºµÎ¶¨ÖÁÖյ㣨ÓÃK2Cr2O2£©£¬ÏûºÄAgNO3ÈÜÒºÌå»ýµÄƽ¾ùֵΪ20.39 mL¡£
¢ÙÉÏÊö²â¶¨¹ý³ÌÖÐÐèÓÃÈÜÒºÈóÏ´µÄÒÇÆ÷ÓÐ________¡£
¢Ú¼ÆËãÉÏÊöÑùÆ·ÖÐCaCl2¡¤2H2OµÄÖÊÁ¿·ÖÊýΪ_______¡££¨±£ÁôËÄλÓÐЧÊý×Ö£©
¢ÛÈôÓÃÉÏÊö·½·¨²â¶¨µÄÑùÆ·ÖÐCaCl2¡¤2H2OµÄÖÊÁ¿·ÖÊýÆ«¸ß(²â¶¨¹ý³ÌÖвúÉúµÄÎó²î¿ÉºöÂÔ)£¬Æä¿ÉÄÜÔÒòÓÐ________£»__________¡£