ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿A¡¢B¡¢C¡¢D¡¢E¡¢FÊÇÖÐѧ»¯Ñ§Öг£¼ûµÄÁùÖÖ¶ÌÖÜÆÚÔªËØ£¬ÓйØλÖü°ÐÅÏ¢ÈçÏ£ºAµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÓëÆäÇ⻯Îï·´Ó¦Éú³ÉÀë×Ó»¯ºÏÎCµ¥ÖÊÒ»°ã±£´æÔÚúÓÍÖУ»FµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¼ÈÄÜÓëÇ¿Ëá·´Ó¦ÓÖÄÜÓëÇ¿¼î·´Ó¦£¬Gµ¥ÖÊÊÇÈÕ³£Éú»îÖÐÓÃÁ¿×î´óµÄ½ðÊô£¬Ò×±»¸¯Ê´»òË𻵡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AÔªËصÄÇ⻯ÎïË®ÈÜÒºÄÜʹ·Ó̪±äºìµÄÔÒòÓõçÀë·½³Ìʽ½âÊÍΪ_____________¡£
£¨2£©Í¬ÎÂͬѹÏ£¬½«a L AÇ⻯ÎïµÄÆøÌåºÍb L DµÄÇ⻯ÎïÆøÌåÏȺóͨÈëһʢˮµÄÉÕ±ÖУ¬ÈôËùµÃÈÜÒºµÄpH=7£¬Ôòa_________b(Ìî¡°>"»ò¡°<¡±»ò¡°=¡±)¡£
£¨3£©³£ÎÂÏ£¬ÏàͬŨ¶ÈF¡¢G¼òµ¥Àë×ÓµÄÈÜÒºÖеμÓNaOHÈÜÒº£¬F¡¢GÁ½ÔªËØÏȺó³Áµí£¬F(OH)nÍêÈ«³ÁµíµÄpHÊÇ4.7£¬G(OH)nÍêÈ«³ÁµíµÄpHÊÇ2.8£¬ÔòÔÚÏàͬÌõ¼þÏ£¬Èܽâ¶È½Ï´óµÄÊÇ£º___£¨Ìѧʽ£©¡£
£¨4£©AÓëB¿É×é³ÉÖÊÁ¿±ÈΪ7:16µÄÈýÔ×Ó·Ö×Ó£¬¸Ã·Ö×ÓÊÍ·ÅÔÚ¿ÕÆøÖÐÆ仯ѧ×÷ÓÿÉÄÜÒý·¢µÄºó¹ûÓУº_______________¡£
¢ÙËáÓê ¢ÚÎÂÊÒЧӦ ¢Û¹â»¯Ñ§ÑÌÎí ¢Ü³ôÑõ²ãÆÆ»µ
£¨5£©AºÍC×é³ÉµÄÒ»ÖÖÀë×Ó»¯ºÏÎÄÜÓëË®·´Ó¦Éú³ÉÁ½Öּ¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ______¡£
£¨6£©Óõç×Óʽ±íʾB,CÐγɻ¯ºÏÎïC2B2µÄ¹ý³Ì_______________¡£
£¨7£©ÒÑÖªÒ»¶¨Á¿µÄEµ¥ÖÊÄÜÔÚB2(g)ÖÐȼÉÕ£¬Æä¿ÉÄܵIJúÎï¼°ÄÜÁ¿¹ØϵÈçÏÂͼËùʾ£ºÇëд³öÒ»¶¨Ìõ¼þÏÂEB2(g) ÓëE(s)·´Ó¦Éú³ÉEB(g)µÄÈÈ»¯Ñ§·½³Ìʽ__________________¡£
¡¾´ð°¸¡¿ NH3¡¤H2ONH4+ + OH£ £¾ Al(OH)3 ¢Ù¢Û Na3N + 4H2O£½3NaOH + NH3¡¤H2O CO2(g) + C(s) = 2CO(g) ;¡÷H= +172.5kJ/mol
¡¾½âÎö¡¿AµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÓëÆäÇ⻯Îï·´Ó¦Éú³ÉÀë×Ó»¯ºÏÎAÊǵªÔªËØ£»Cµ¥ÖÊÒ»°ã±£´æÔÚúÓÍÖУ¬CÊÇÄÆÔªËØ£»FµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¼ÈÄÜÓëËá·´Ó¦ÓÖÄÜÓë¼î·´Ó¦£¬FÊÇÂÁÔªËØ£¬Gµ¥ÖÊÊÇÈÕ³£Éú»îÖÐÓÃÁ¿×î´óµÄ½ðÊô£¬Ò×±»¸¯Ê´»òË𻵣¬GÊÇÌúÔªËØ£¬¸ù¾Ý±í¸ñBÊÇÑõÔªËØ£»EÊÇ̼ԪËØ£»DÊÇÂÈÔªËØ¡££¨1£©AÔªËصÄÇ⻯ÎïÊÇNH3£¬ÈÜÓÚË®µÃµ½µÄÈÜÒº°±Ë®³ÊÈõ¼îÐÔ£¬ÄܵçÀë³öÇâÑõ¸ù£¬Ê¹·Ó̪±äºì£ºNH3¡¤H2ONH4+ + OH££»£¨2£©Í¬ÎÂͬѹÏ£¬½«a L NH3ÆøÌåºÍb L HClÆøÌåͨÈëË®ÖУ¬Éú³ÉÂÈ»¯ï§£¬Ç¿ËáÈõ¼îÑΣ¬ÈÜÒº³ÊËáÐÔ£¬ÈôËùµÃÈÜÒºµÄpH=7£¬°±ÆøÒª¹ýÁ¿£»Ôòa>b£»£¨3£©³£ÎÂÏ£¬ÏàͬŨ¶È£¨ÉèŨ¶È¾ùΪc£©Al¡¢Fe¼òµ¥Àë×ÓµÄÈÜÒºÖеμÓNaOHÈÜÒº£¬Al(OH)3ÍêÈ«³ÁµíµÄpHÊÇ4.7£¬ÆäKsp=c¡Ác3(OH-)=(10-9.3)3¡Ác=10-27.9¡Ác£»Fe(OH)3ÍêÈ«³ÁµíµÄpHÊÇ2.8£¬ÆäKsp= c¡Ác3(OH-)= (10-11.2)3¡Ác=10-33.6¡Ác£¬Ôòksp½Ï´óµÄÊÇAl(OH)3£»£¨4£©AÓëB×é³ÉÖÊÁ¿±ÈΪ7:16µÄÈýÔ×Ó·Ö×ÓÊÇNO2£¬Ò×ÈÜÓÚË®£¬ÊÍ·ÅÔÚ¿ÕÆøÖпÉÄÜÒý·¢ËáÓêºÍ¹â»¯Ñ§ÑÌÎí£»¶øÎÂÊÒЧӦÊÇÒòΪ¶þÑõ»¯Ì¼µÄÅÅ·Å£¬³ôÑõ²ãÆÆ»µÊÇ·úÀû°ºµÈÔì³ÉµÄ¡£¹ÊÑ¡¢Ù¢Û£»£¨5£©NºÍNa×é³ÉµÄÀë×Ó»¯ºÏÎïÊÇNa3N£¬ÄÜÓëË®·´Ó¦Éú³ÉÁ½Öּ¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇNa3N + 4H2O£½3NaOH + NH3¡¤H2O£»£¨6£©Óõç×Óʽ±íʾO¡¢NaÐγɻ¯ºÏÎïNa2O2µÄ¹ý³ÌΪ£º£»£¨6£©¸ù¾ÝÄÜÁ¿Í¼£¬¢ÙC(s)+O2(g)=CO2(g) ¡÷H1=-393.5kJ/mol£»¢ÚCO(g)+ O2(g)=CO2(g) ¡÷H2=-283.0kJ/mol£»¸ù¾Ý¸Ç˹¶¨ÂÉÓÉ¢Ù-2¡Á¢ÚµÃ£ºCO2(g) + C(s) = 2CO(g) ¡÷H= +172.5kJ/mol¡£
¡¾ÌâÄ¿¡¿¹¤ÒµÉÏÒÔï®»ÔʯΪÔÁÏÉú²ú̼Ëá﮵IJ¿·Ö¹¤ÒµÁ÷³ÌÈçÏ£º
ÒÑÖª£º
¢Ùï®»ÔʯµÄÖ÷Òª³É·ÖΪLi2O¡¤Al2O3¡¤4SiO2£¬ÆäÖк¬ÉÙÁ¿Ca¡¢MgÔªËØ¡£
¢ÚLi2O¡¤Al2O3¡¤4SiO2 + H2SO4£¨Å¨£© Li2SO4 + Al2O3¡¤4SiO2¡¤H2O
¢ÛijЩÎïÖʵÄÈܽâ¶È£¨s£©ÈçϱíËùʾ¡£
T/¡æ | 20 | 40 | 60 | 80 |
s£¨Li2CO3£©/g | 1.33 | 1.17 | 1.01 | 0.85 |
s£¨Li2SO4£©/g | 34.2 | 32.8 | 31.9 | 30.7 |
£¨1£©´ÓÂËÔü¢ñÖзÖÀë³öAl2O3µÄÁ÷³ÌÈçÏÂͼËùʾ¡£Çëд³öÉú³É³ÁµíµÄÀë×Ó·½³Ìʽ______¡£
£¨2£©ÒÑÖªÂËÔü2µÄÖ÷Òª³É·ÖÓÐMg£¨OH£©2ºÍCaCO3¡£ÏòÂËÒº1ÖмÓÈëʯ»ÒÈéµÄ×÷ÓÃÊÇ£¨ÔËÓû¯Ñ§Æ½ºâÔÀí¼òÊö£©________________________________________________¡£
£¨3£©×îºóÒ»¸ö²½ÖèÖУ¬Óá°ÈÈˮϴµÓ¡±µÄÔÒòÊÇ______________________________¡£
£¨4£©¹¤ÒµÉÏ£¬½«Li2CO3´ÖÆ·ÖƱ¸³É¸ß´¿Li2CO3µÄ²¿·Ö¹¤ÒÕÈçÏ£º
a.½«Li2CO3ÈÜÓÚÑÎËá×÷µç½â²ÛµÄÑô¼«Òº£¬LiOHÈÜÒº×öÒõ¼«Òº£¬Á½ÕßÓÃÀë×ÓÑ¡Ôñ͸¹ýĤ¸ô¿ª£¬ÓöèÐԵ缫µç½â¡£
b.µç½âºóÏòLiOHÈÜÒºÖмÓÈëÉÙÁ¿NH4HCO3ÈÜÒº²¢¹²ÈÈ£¬¹ýÂË¡¢ºæ¸ÉµÃ¸ß´¿Li2CO3¡£
¢ÙaÖУ¬Ñô¼«µÄµç¼«·´Ó¦Ê½ÊÇ_________________________
¢Úµç½âºó£¬LiOHÈÜҺŨ¶ÈÔö´óµÄÔÒò_________________£¬bÖÐÉú³ÉLi2CO3·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ___________________________________________¡£
£¨5£©Á×ËáÑÇÌú﮵ç³Ø×Ü·´Ó¦Îª£ºFePO4+LiLiFePO4£¬µç³ØÖеĹÌÌåµç½âÖÊ¿É´«µ¼Li+£¬ÊÔд³ö¸Ãµç³Ø·ÅµçʱµÄÕý¼«·´Ó¦£º__________________¡£