ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿Á×´æÔÚÓÚÈËÌåËùÓÐϸ°ûÖУ¬ÊÇά³Ö¹Ç÷ÀºÍÑÀ³ÝµÄ±ØÒªÎïÖÊ£¬¼¸ºõ²ÎÓëËùÓÐÉúÀíÉϵĻ¯Ñ§·´Ó¦¡£Á×»¹ÄÜÈÃÐÄÔàÓйæÂɵØÌø¶¯¡¢Î¬³ÖÉöÔàÕý³£»úÄܺʹ«´ïÉñ¾´Ì¼¤µÄÖØÒªÎïÖÊ¡£Ðí¶àÁ׵Ļ¯ºÏÎïÊǹ¤ÒµÉϵÄÖØÒªÔÁÏ¡£ÇëÍê³ÉÏÂÁÐÌî¿Õ£º
P(s£¬ºìÁ×)+O2(g)=P4O10(s)+738.5 kJ
P4(s£¬°×Á×)+ 5O2(g)=P4O10(s)+2983.2 kJ
(1)Ôò°×Á×ת»¯ÎªºìÁ×µÄÈÈ»¯Ñ§·½³Ìʽ_____________________¡£
(2)ºìÁ×ÔÚKOHÈÜÒºµÄÐü×ÇÒºÖкÍKOCl×÷Ó㬿ÉÒÔÉú³ÉK6P6O12µÄ¼ØÑκÍKClµÈ²úÎï¡£Çëд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¬²¢±ê³öµç×ÓתÒÆ·½ÏòºÍÊýÄ¿¡£____
(3)ÖƱ¸MgNH4PO4ʱͨ³£ÔÚþÑΣ¨ÈçÂÈ»¯Ã¾£©ÈÜÒºÖмÓNa2HPO4 ¡¢°±Ë®ºÍï§ÑÎÈÜÒº£¨ÈçÂÈ»¯ï§£©£¬·´Ó¦Ê½ÎªMgCl2+Na2HPO4+NH3=2NaCl+MgNH4PO4¡ý¡£ÔÚ´Ë·´Ó¦ÖУ¬Èç¹û²»¼Óï§ÑΣ¬ÔÚ¼îÐÔÈÜÒºÖУ¬Mg2+½«»á±»°±Ë®³Áµí£¬ÆäÀë×Ó·½³ÌʽΪ____________________£»¼ÓÈëï§ÑÎÖ®ºó£¬¿É·ÀÖ¹²úÉúMg(OH)2³Áµí¡£ÇëÓõçÀëƽºâÔÀí½âÊÍÔÒò£º_________
(4)H3PO3ÓëµâË®·´Ó¦£¬µâË®µÄ×Ø»ÆÉ«ÍÊÈ¥£¬Éú³ÉÁ×ËᣬÏòÉÏÊö·´Ó¦ºóµÄÈÜÒºÖмÓÈë¹ýÁ¿µÄAgNO3ÈÜÒº£¬Éú³É»ÆÉ«³Áµí¡£ÔòH3PO3ÓëµâË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________£»»ÆÉ«³ÁµíÊÇ_________£¬¿ÉÓÃ×÷ÓÚ_____________________£¨Ð´³öÈÎÒâÒ»ÖÖÓÃ;¼´¿É£©¡£
¡¾´ð°¸¡¿P4(s£¬°×Á×)=4P(s£¬ºìÁ×) +29.2kJ Mg2++2NH3.H2O=Mg(OH)2¡ý+2NH4+ ¼ÓÁËï§ÑÎÖ®ºó£¬ÓÉÓÚÈÜÒºÖÐ笠ùÀë×ÓµÄŨ¶ÈÔö´ó£¬Ê¹Æ½ºâNH3.H2ONH4++OH-Ïò×óÒƶ¯£¬´ó´ó½µµÍÁËÈÜÒºÖÐOH-Ũ¶È(Ôö´óNH4+Ũ¶È£¬ÒÖÖÆNH3.H2OµçÀë) H3PO3+I2+H2O=H3PO4+2HI AgI ¿ÉÓÃÒÔÖÆÕÕÏàµ×Ƭ»ò½º¾íµÄ¸Ð¹â²ã¡¢È˹¤½µÓê
¡¾½âÎö¡¿
(1)ÀûÓøÇ˹¶¨ÂɼÆËãµÃµ½ÈÈ»¯Ñ§·½³Ìʽ£»
(2)¸ù¾ÝÌâÒâд³ö»¯Ñ§·½³Ìʽ£¬ÀûÓû¯ºÏ¼ÛµÄ±ä»¯±ê³öµç×ÓµÄתÒÆÊýÄ¿£»
(3)þÀë×ӺͰ±Ë®·´Ó¦Éú³ÉÇâÑõ»¯Ã¾ºÍ笠ùÀë×Ó£¬Ð´³ö»¯Ñ§·½³Ìʽ£¬¼ÓÈëï§ÑΣ¬Ôö¼Ó笠ùÀë×ÓµÄŨ¶È£¬ÀûÓð±Ë®µÄµçÀëƽºâÒƶ¯·ÖÎö£»
(4)H3PO3ÓëµâË®·´Ó¦£¬µâË®µÄ×Ø»ÆÉ«ÍÊÈ¥£¬Éú³ÉÁ×Ëᣬµâ±»»¹ÔΪµâ»¯Ç⣬д³ö»¯Ñ§·½³Ìʽ£»µâ»¯Çâ¼ÓÈëÏõËáÒøÉú³ÉµÄµâ»¯ÒøÊÇ»ÆÉ«³Áµí£¬ÀûÓõ⻯ÒøµÄÓÃ;»Ø´ð¡£
(1)P(s£¬ºìÁ×)+O2(g)=P4O10(s)H1=+738.5 kJ/mol ¢Ù
P4(s£¬°×Á×)+ 5O2(g)=P4O10(s)H2=+2983.2 kJ/mol ¢Ú
¢Ú-¢Ù¡Á4µÃµ½P4(s£¬°×Á×)=4P(s£¬ºìÁ×) H=-29£®2kJ/mol £»
(2)ºìÁ×ÔÚKOHÈÜÒºµÄÐü×ÇÒºÖкÍKOCl×÷Ó㬿ÉÒÔÉú³ÉK6P6O12£¬KClºÍË®£¬»¯Ñ§·½³ÌʽΪ6P+9KOCl+6KOH= K6P6O12+9KCl+3H2O£¬Óõ¥ÏßÇűíʾ³öµç×ÓµÄתÒƺͷ½Ïò£¬Á×ÊÇ»¹Ô¼Á£¬»¯ºÏ¼Û´Ó0Éý¸ßµ½+3¼Û£¬ÓÐ6¸öÁ×Ô×Ó×öÁË»¹Ô¼Á£¬KOClÊÇÑõ»¯¼Á£¬ÂÈÔ×Ó´Ó+1¼Û½µµÍµ½-1¼Û£¬¹ÊתÒƵĵç×ÓÊýΪ18£¬
(3)þÀë×Ó»áºÍ°±Ë®·´Ó¦Éú³ÉÇâÑõ»¯Ã¾ºÍ笠ùÀë×Ó£¬Àë×Ó·½³ÌʽΪMg2++2NH3.H2O=Mg(OH)2¡ý+2NH4+£»¼ÓÁËï§ÑÎÖ®ºó£¬ÓÉÓÚÈÜÒºÖÐ笠ùÀë×ÓµÄŨ¶ÈÔö´ó£¬Ê¹Æ½ºâNH3.H2ONH4++OH-Ïò×óÒƶ¯£¬´ó´ó½µµÍÁËÈÜÒºÖÐOH-Ũ¶È£¨Ôö´óNH4+Ũ¶È£¬ÒÖÖÆNH3.H2OµçÀ룩£»
(4)H3PO3ÓëµâË®·´Ó¦£¬µâË®µÄ×Ø»ÆÉ«ÍÊÈ¥£¬Éú³ÉÁ×ËáºÍµâ»¯Ç⣬»¯Ñ§·½³ÌʽΪH3PO3+I2+H2O=H3PO4+2HI£¬µâ»¯ÇâÓëÏõËáÒø·´Ó¦Éú³Éµâ»¯ÒøºÍµâ»¯Ç⣬µâ»¯ÒøÊÇ»ÆÉ«³Áµí£¬Ëü¿ÉÓÃÒÔÖÆÕÕÏàµ×Ƭ»ò½º¾íµÄ¸Ð¹â²ã¡¢È˹¤½µÓê¡£
¡¾ÌâÄ¿¡¿Ä³»¯Ñ§Ð¡×é²ÉÓÃÀàËÆÖÆÒÒËáÒÒõ¥µÄ×°Öã¨Èçͼ£©£¬ÒÔ»·¼º´¼ÖƱ¸»·¼ºÏ©¡£
ÒÑÖª£º+H2O
(1)Èçͼ1ÊÇʵÑéÊÒÖÆÈ¡ÒÒËáÒÒõ¥µÄ×°Öá£ÔòÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨____£©
A£®Óұߵ¼¹Ü²»²åÈë±¥ºÍ̼ËáÄÆÈÜÒº
B£®ÊÔ¹ÜDÖÐÓÍ×´ÎïÔÚϲ㣬±¥ºÍ̼ËáÄÆÈÜÒºÖ÷ҪΪÁËÖкͻӷ¢³öµÄÒÒËá
C£®ÊÔ¹ÜCÖмÓÈëÊÔ¼ÁµÄ˳ÐòÊÇ£º2mLŨÁòËá¡ú3mLÒÒ´¼¡ú2mL±ù´×Ëá
D£®·´Ó¦½áÊøºóÊÔ¹ÜCÒºÌå¿ÉÄÜ»á±äºÚÉ«
(2)ÔÚÖÆÈ¡ÒÒËáÒÒõ¥µÄ¹ý³ÌÖУ¬ÓÃŨÁòËá×÷Ϊ´ß»¯¼ÁÓÐȱµã£¬²»ÄÜÖظ´Ê¹Ó㬶øÇÒ¸±·´Ó¦½Ï¶à¡£Ä¿Ç°¶Ô¸Ã·´Ó¦µÄ´ß»¯¼Á½øÐÐÁËеÄ̽Ë÷£¬³õ²½±íÃ÷ÖÊ×ÓËáÀë×ÓÒºÌå¿ÉÓÃ×÷´Ë·´Ó¦µÄ´ß»¯¼Á£¬ÇÒÄÜÖظ´Ê¹Óá£ÊµÑéÊý¾ÝÈçϱíËùʾ£¨ÒÒËáºÍÒÒ´¼ÒÔµÈÎïÖʵÄÁ¿»ìºÏ£©¡£
ͬһ·´Ó¦Ê±¼ä | ͬһ·´Ó¦ÎÂ¶È | ||||
·´Ó¦Î¶È/¡æ | ת»¯ÂÊ(%) | Ñ¡ÔñÐÔ(%)* | ·´Ó¦Ê±¼ä/h | ת»¯ÂÊ(%) | Ñ¡ÔñÐÔ(%)* |
40 | 77.8 | 100 | 2 | 80.2 | 100 |
60 | 92.3 | 100 | 3 | 87.8 | 100 |
80 | 92.6 | 100 | 4 | 92.3 | 100 |
120 | 94.5 | 98.7 | 6 | 93.0 | 100 |
Ñ¡ÔñÐÔ100%±íʾ·´Ó¦Éú³ÉµÄ²úÎïÊÇÒÒËáÒÒõ¥ºÍË® |
¸ù¾Ý±íÖÐÊý¾Ý£¬ÏÂÁÐ________£¨Ìî×Öĸ£©Îª¸Ã·´Ó¦µÄ×î¼ÑÌõ¼þ¡£
A.120¡æ,4h B.80¡æ,6h C. 60¡æ,4h D.80¡æ,4h
(3)ÖƱ¸»·¼ºÏ©´ÖÆ·£¨×°ÖÃÈçͼ2£©
½«12.5mL»·¼º´¼¼ÓÈëÊÔ¹ÜAÖУ¬ÔÙ¼ÓÈë1mLŨÁòËᣬҡÔȺó·ÅÈëËé´ÉƬ£¬»ºÂý¼ÓÈÈÖÁ·´Ó¦ÍêÈ«£¬ÔÚÊÔ¹ÜCÄڵõ½»·¼ºÏ©´ÖÆ·¡£
¢ÙAÖÐËé´ÉƬµÄ×÷ÓÃÊÇ____________£¬µ¼¹ÜB³ýÁ˵¼ÆøÍ⻹¾ßÓеÄ×÷ÓÃÊÇ____________¡£
¢ÚÊÔ¹ÜCÖÃÓÚ±ùˮԡÖеÄÄ¿µÄÊÇ_____________¡£
(4)ÖƱ¸»·¼ºÏ©¾«Æ·
¢Ù»·¼ºÏ©´ÖÆ·Öк¬Óл·¼º´¼ºÍÉÙÁ¿ËáÐÔÔÓÖʵȡ£¼ÓÈë±¥ºÍʳÑÎË®£¬Õñµ´¡¢¾²Öᢷֲ㣬»·¼ºÏ©ÔÚ_________²ã(ÌîÉÏ»òÏÂ)£¬·ÅҺʱ£¬Èô·¢ÏÖÒºÌåÁ÷²»³öÀ´£¬Æä¿ÉÄÜÔÒò³ý·ÖҺ©¶·»îÈû¶ÂÈûÍ⣬»¹ÓÐ________£»·ÖÒººóÓÃ_________ (ÌîÈë±àºÅ)Ï´µÓ£»
a£®NaHSO4ÈÜÒº b£®Na2CO3ÈÜÒº c£®Ï¡H2SO4 d.äåË®
ÔÚ´ËÖƱ¸¹ý³ÌÖУ¬¼ÓÈë±¥ºÍʳÑÎË®µÄ×÷ÓÃÊÇ__________________¡£
¢Ú¶Ô·ÖÀë³öÀ´µÄ»·¼ºÏ©ÔÙ½øÐÐÕôÁóµÃµ½»·¼ºÏ©¾«Æ·£¬ÎªÁËÇø·Ö»·¼ºÏ©¾«Æ·ºÍ´ÖÆ·£¬Ä³¸öС×éÉè¼ÆÁËÒÔϼ¸ÖÖ·½·¨£¬ºÏÀíµÄÊÇ_________¡£
a£®ÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒº b£®ÓýðÊôK c£®ÓÃNa2CO3ÈÜÒº d.ÓÃNaOHÈÜÒº