ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ëæ×ÅÉç»á¾­¼ÃµÄ·¢Õ¹£¬ÈËÃÇÉú»îˮƽµÄÌá¸ßºÍ¶Ô»·¾³ÒªÇóµÄ¼ÓÇ¿£¬À´Ô´¹ã·ºµÄ¸ß°±µª·ÏË®(Ö÷Òªº¬ÓÐNH4+)´¦ÀíÔ½À´Ô½Êܵ½ÖØÊÓ¡£¶ÔÓڸ߰±µª·ÏË®µÄ´¦ÀíÓжàÖÖ·½·¨¡£

£¨1£©´µÍÑ·¨:

ʹÓôµÍÑ·¨Ê±ÐèÒªÔÚ¢ÙÖмÓÈë¼î£¬Ð´³ö·¢ÉúµÄÀë×Ó·´Ó¦·½³Ìʽ______________________¡£

£¨2£©MAP³Áµí·¨£º

¢ÙʹÓû¯Ñ§³Áµí¼Á´¦Àí¸ß°±µª·Ïˮʱ£¬Ïò¸ß°±µª·ÏË®ÖÐͶÈ뺬ÓÐMg2+µÄÎïÖʺÍH3PO4£¬µ÷½ÚÈÜÒºpH£¬ÓëNH4+·´Ó¦Éú³ÉMgNH4PO4(MAP)³Áµí¡£ÎªÓÐЧ¿ØÖÆÈÜÒºPH£¬²¢¿¼ÂÇÎÛË®´¦ÀíЧ¹û£¬Ôò×îºÃÑ¡ÓÃÏÂÁÐÎïÖÊÖÐ_____¡£

A MgO B MgSO4 C MgCl2

¢Ú¿ØÖÆÈÜÒºPHµÄºÏÀí·¶Î§Îª____________________

¢Û´ÓÈܽâƽºâ½Ç¶È½âÊÍPH¹ý¸ß»ò¹ýµÍ²»Ò×ÐγɳÁµíMAPµÄÔ­Òò£¨ÒÑÖªPO43-ÔÚËáÐÔ½ÏÇ¿Ìõ¼þÏÂÒÔHPO42-ÐÎʽ´æÔÚ£©_______________

£¨3£©ÉúÎïÍѵª´«Í³¹¤ÒÕ£º

¢ÙÔÚÓÐÑõÆøµÄÌõ¼þÏ£¬½èÖúÓÚºÃÑõ΢ÉúÎÖ÷ÒªÊǺÃÑõ¾ú£©µÄ×÷ÓÃÉú³ÉNO3-£¬Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ_________________________¡£

¢ÚÔÚÎÞÑõµÄËáÐÔÌõ¼þÏ£¬ÀûÓÃÑáÑõ΢ÉúÎ·´Ïõ»¯¾ú£©µÄ×÷ÓÃʹNO3-Óë¼×´¼×÷ÓÃÉú³ÉN2£¬´ïµ½¾»»¯Ë®µÄÄ¿µÄ¡£Ð´³öÀë×Ó·½³Ìʽ____________________________¡£

¡¾´ð°¸¡¿ NH4++OH-=NH3¡¤H2O A 9-9.5£¨9-10Ò²ÐУ© ÌåϵÖдæÔÚƽºâMgNH4PO4(s)Mg2++NH4++PO43- £¬µ±PH¹ýµÍʱPO43-+H+=HPO42-,c(PO43-)½µµÍ£¬Æ½ºâÕýÒÆ£¬²»ÀûÓÚÉú³É³Áµí¡£µ±PH¹ý¸ßʱ Mg2++2OH-=Mg(OH)2¡ý,c(Mg2+)½µµÍ£¬Æ½ºâÕýÒÆ£¬²»ÀûÓÚÉú³É³Áµí

¡¾½âÎö¡¿(1) ¸ß°±µª·ÏË®(Ö÷Òªº¬ÓÐNH4+)¼Ó¼îºó·¢Éú·´Ó¦µÄÀë×Ó·´Ó¦·½³ÌʽΪNH4++OH-=NH3¡¤H2O¡£

£¨2£©¢ÙʹÓû¯Ñ§³Áµí¼Á´¦Àí¸ß°±µª·Ïˮʱ£¬Ïò¸ß°±µª·ÏË®ÖÐͶÈ뺬ÓÐMg2+µÄÎïÖʺÍH3PO4£¬µ÷½ÚÈÜÒºpH£¬ÓëNH4+·´Ó¦Éú³ÉMgNH4PO4(MAP)³Áµí¡£ÎªÓÐЧ¿ØÖÆÈÜÒºPH£¬²¢¿¼ÂÇÎÛË®´¦ÀíЧ¹û£¬Ôò×îºÃÑ¡-ÓÃMgO£¬ÈÜÓÚË®ÏÔ¼îÐÔ£¬ÓÖ²»Ôö¼ÓÐÂÔÓÖÊ£¬¶øMgSO4 ºÍMgCl2»áÔö¼ÓSO42- Cl-ÔÓÖÊÀë×Ó£¬¹Ê²»Ñ¡£¬ËùÒÔ´ð°¸ÎªA¡£

¢ÚÓÉͼÏñ·ÖÎö£¬µ±PHΪ9-9.5ʱ£¬°±µªÈ¥³ýÂÊ×î¸ß£¬ÁײÐÓàŨ¶È×îµÍ£¬ËùÒÔ¿ØÖÆÈÜÒºPH Ϊ9-9.5ΪºÏÀí·¶Î§¡£

¢ÛÌåϵÖдæÔÚƽºâMgNH4PO4(s)Mg2++NH4++PO43- £¬µ±PH¹ýµÍʱPO43-+H+=HPO42-,c(PO43-)½µµÍ£¬Æ½ºâÕýÒÆ£¬²»ÀûÓÚÉú³É³Áµí¡£µ±PH¹ý¸ßʱ Mg2++2OH-=Mg(OH)2¡ý,c(Mg2+)½µµÍ£¬Æ½ºâÕýÒÆ£¬²»ÀûÓÚÉú³É³Áµí ¡£

£¨3£©¢ÙÑõÆø½èÖúÓÚºÃÑõ΢ÉúÎÖ÷ÒªÊǺÃÑõ¾ú£©µÄ×÷Óý«NH4+Éú³ÉNO3-£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ

¢ÚÔÚÎÞÑõµÄËáÐÔÌõ¼þÏ£¬ÀûÓÃÑáÑõ΢ÉúÎ·´Ïõ»¯¾ú£©µÄ×÷ÓÃʹNO3-Óë¼×´¼×÷ÓÃÉú³ÉN2£¬´ïµ½¾»»¯Ë®µÄÄ¿µÄ¡£Æä·´Ó¦µÄÀë×Ó·½³Ìʽ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÓÃ50 mL 1.0 mol/LÑÎËá¸ú50 mL 1.1 mol/L ÇâÑõ»¯ÄÆÈÜÒºÔÚÈçͼËùʾµÄ×°ÖÃÖнøÐÐÖкͷ´Ó¦¡£Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©´ÓʵÑé×°ÖÃÉÏ¿´£¬Í¼ÖÐÉÐȱÉÙµÄÒ»ÖÖ²£Á§ÓÃÆ·ÊÇ________£»

£¨2£©´óÉÕ±­ÉÏÈç²»¸ÇÓ²Ö½°å£¬ÇóµÃµÄÖкÍÈÈÊýÖµ__________(Ìî¡°Æ«´ó¡¢Æ«Ð¡¡¢ÎÞÓ°Ï족)£»ÑÎËáÔÚ·´Ó¦ÖÐÈôÒòΪÓзÅÈÈÏÖÏ󣬶øÔì³ÉÉÙÁ¿ÑÎËáÔÚ·´Ó¦Öлӷ¢£¬Ôò²âµÃµÄÖкÍÈÈ____________(Ìî¡°Æ«´ó¡¢Æ«Ð¡¡¢ÎÞÓ°Ï족)£»ÔÚÖкÍÈȲⶨʵÑéÖдæÔÚÓÃˮϴµÓζȼÆÉϵÄÑÎËáµÄ²½Ö裬ÈôÎ޴˲Ù×÷²½Ö裬Ôò²âµÃµÄÖкÍÈÈ»á____________(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±)¡£

£¨3£©ÈôÓõÈŨ¶ÈµÄ´×ËáÓëNaOHÈÜÒº·´Ó¦£¬Ôò²âµÃµÄÖкÍÈÈ»á____________(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±)£¬ÆäÔ­ÒòÊÇ_______________________________________________¡£

£¨4£©¸ÃʵÑéС×é×öÁËÈý´ÎʵÑ飬ÿ´ÎÈ¡ÈÜÒº¸÷50 mL£¬²¢¼Ç¼ÏÂԭʼÊý¾Ý(¼ûϱí)¡£

ʵÑéÐòºÅ

ÆðʼζÈt1/¡æ

ÖÕֹζÈ(t2)/¡æ

βî(t2£­t1)/¡æ

ÑÎËá

NaOHÈÜÒº

ƽ¾ùÖµ

1

25.1

24.9

25.0

31.6

6.6

2

25.1

25.1

25.1

31.8

6.7

3

25.1

25.1

25.1

31.9

6.8

ÒÑÖªÑÎËá¡¢NaOHÈÜÒºÃܶȽüËÆΪ1.00g/cm3£¬Öкͺó»ìºÏÒºµÄ±ÈÈÈÈÝc£½4.18¡Á10£­3kJ/(g¡¤¡æ)£¬Ôò¸Ã·´Ó¦µÄÖкÍÈÈΪ¦¤H£½_______¡£¸ù¾Ý¼ÆËã½á¹û£¬Ð´³ö¸ÃÖкͷ´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ______________________¡£

£¨5£©ÊµÑéÖиÄÓÃ60 mL 1.0 mol¡¤L-1µÄÑÎËá¸ú50 mL 1.1mol¡¤L-1µÄNaOHÈÜÒº½øÐз´Ó¦£¬ÓëÉÏÊöʵÑéÏà±È£¬Ëù·Å³öµÄÈÈÁ¿_________£¨Ìî¡°ÏàµÈ¡±»ò¡°²»ÏàµÈ¡±£©£¬ËùÇóÖкÍÈÈ__________ (Ìî¡°ÏàµÈ¡±»ò¡°²»ÏàµÈ¡±)¡£

¡¾ÌâÄ¿¡¿ÂÁÇ⻯ÄÆ(NaAlH4)ÊÇÓлúºÏ³ÉµÄÖØÒª»¹Ô­¼Á£¬ÆäºÏ³É·ÏßÈçͼËùʾ¡£

£¨1£©ÒÑÖªAlCl3µÄÈÛµãΪ190¡æ£¬·ÐµãΪ178¡æ£¬ÔÚºþʪµÄ¿ÕÆøÖÐÒ×Ë®½â¡£Ä³ÊµÑéС×éÀûÓÃÏÂͼÖÐ×°ÖÃÖƱ¸ÎÞË®AlCl3¡£

¢Ùд³öÔ²µ×ÉÕÆ¿Öз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ:_________________¡£

¢Ú°´ÆøÁ÷·½ÏòÁ¬½Ó¸÷ÒÇÆ÷½Ó¿Ú£¬Ë³ÐòΪa¡ú__________b¡úc¡ú_______¡£ (Ìî½Ó¿Ú×Öĸ)

¢Û×°ÖÃFÖÐӦʢװµÄÊÔ¼ÁÊÇ__________£¬×°ÖÃDµÄ×÷ÓÃÊÇ______________¡£

(2)AlCl3ÓëNaH·´Ó¦Ê±£¬ÐèÏȽ«AlCl3ÈÜÓÚÓлúÈܼÁ£¬ÔÙ½«µÃµ½µÄÈÜÒºµÎ¼Óµ½NaH·ÛÄ©ÖÐ,´Ë·´Ó¦ÖÐNaHµÄת»¯Âʽϵͣ¬ÆäÔ­Òò¿ÉÄÜÊÇ______________¡£

(3)ͨ¹ý²â¶¨ÂÁÇ⻯ÄÆÓëË®·´Ó¦Éú³ÉÇâÆøµÄÌå»ýÀ´²â¶¨ÂÁÇ⻯ÄÆÑùÆ·µÄ´¿¶È¡£

¢ÙÂÁÇ⻯µÄÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________¡£

¢ÚÉè¼ÆÈçÏÂËÄÖÖ×°ÖòⶨÂÁÇ⻯ÄÆÑùÆ·µÄ´¿¶È(ÔÓÖÊÖ»ÓÐÇ⻯ÄÆ)¡£´Ó¼òÔ¼ÐÔ¡¢×¼È·ÐÔ¿¼ÂÇ£¬×îÇ¡µ±µÄ×°ÖÃÊÇ___(Ìî±àºÅ)¡£

¢ÛijͬѧѡÓÃÉÏÊö×îÇ¡µ±µÄÅáÖÃ,³ÆÈ¡mgÂÁÇ⻯ÄÆÑùÆ·£¬²âµÃÉú³ÉÆøÌåµÄÌå»ýΪVL(ÒÑÕÛËãΪ±ê×¼×´¿ö),Öظ´ÊµÑé²Ù×÷Èý´Î£¬²âµÃÓйØÊý¾ÝÈçϱíËùʾ£¬ÔòÑùÆ·ÖÐÂÁÇ⻯ÄƵĴ¿¶ÈΪ____¡£

ʵÑé´ÎÊý

ÑùÆ·ÖÊÁ¿m(g)

ÆøÌåÐÝ»ýV(L)

I

1.20

1.907

II

1.20

1.904

III

1.20

1,901

¡¾ÌâÄ¿¡¿ÊµÊ©ÒÔ½ÚÔ¼ÄÜÔ´ºÍ¼õÉÙ·ÏÆøÅÅ·ÅΪ»ù±¾ÄÚÈݵĽÚÄܼõÅÅÕþ²ß£¬ÊÇÓ¦¶ÔÈ«ÇòÆøºòÎÊÌâ¡¢½¨Éè×ÊÔ´½ÚÔ¼ÐÍ¡¢»·¾³ÓѺÃÐÍÉç»áµÄ±ØȻѡÔñ¡£»¯¹¤ÐÐÒµµÄ·¢Õ¹±ØÐë·ûºÏ¹ú¼Ò½ÚÄܼõÅŵÄ×ÜÌåÒªÇó¡£ÊÔÔËÓÃËùѧ֪ʶ£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÒÑ֪ijζÈÏÂij·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪ£ºK=c(H2O)/[ c(CO)¡¤c(H2)]Ëù¶ÔÓ¦µÄ»¯Ñ§·´Ó¦·½³ÌʽΪ£º________________________¡£

£¨2£©ÒÑÖªÔÚÒ»¶¨Î¶ÈÏ£¬¢ÙC(s)+CO2(g)2CO(g) ¡÷H1=a kJ/mol ƽºâ³£ÊýK1£»

¢ÚCO(g)+H2O(g)H2(g)+CO2(g) ¡÷H2=b kJ/mol ƽºâ³£ÊýK2£»

¢ÛC(s)+H2O(g)CO(g)+H2(g) ¡÷H3 ƽºâ³£ÊýK3¡£

ÔòK1¡¢K2¡¢K3Ö®¼äµÄ¹ØϵÊÇ£º_____________£¬¡÷H3=__________£¨Óú¬a¡¢bµÄ´úÊýʽ±íʾ£©¡£

£¨3£©Ãº»¯¹¤Í¨³£Í¨¹ýÑо¿²»Í¬Î¶ÈÏÂƽºâ³£ÊýÒÔ½â¾ö¸÷ÖÖʵ¼ÊÎÊÌâ¡£ÒÑÖªµÈÌå»ýµÄÒ»Ñõ»¯Ì¼ºÍË®ÕôÆø½øÈë·´Ó¦Æ÷ʱ£¬·¢ÉúÈçÏ·´Ó¦£ºCO(g)+H2O(g)H2(g)+CO2(g)£¬¸Ã·´Ó¦Æ½ºâ³£ÊýËæζȵı仯ÈçϱíËùʾ£º

ζÈ/¡æ

400

500

800

ƽºâ³£ÊýK

9.94

9

1

¸Ã·´Ó¦µÄÕý·´Ó¦·½ÏòÊÇ_________·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©£¬ÈôÔÚ500¡æʱ½øÐУ¬ÉèÆðʼʱCOºÍH2OµÄÆðʼŨ¶È¾ùΪ0.020 mol¡¤L-1£¬ÔÚ¸ÃÌõ¼þÏ£¬COµÄƽºâת»¯ÂÊΪ£º______________¡£

£¨4£©ÔÚ´ß»¯¼Á´æÔÚÌõ¼þÏ·´Ó¦£ºH2O(g)+CO(g)CO2(g)+H2(g)£¬COת»¯ÂÊËæÕôÆøÌí¼ÓÁ¿µÄѹǿ±È¼°Î¶ȱ仯¹ØϵÈçÏÂ×óͼËùʾ£»

¶ÔÓÚÆøÏà·´Ó¦£¬ÓÃij×é·Ö(B)µÄƽºâ·Öѹǿ(PB)´úÌæÎïÖʵÄÁ¿Å¨¶È(CB)Ò²¿ÉÒÔ±íʾƽºâ³£Êý£¨¼Ç×÷KP£©£¬Ôò¸Ã·´Ó¦µÄKPµÄ±í´ïʽ£ºKP=____________£¬Ìá¸ßp[H2O(g)]/p(CO)±È£¬ÔòKP__________£¨Ìî¡°±ä´ó¡±¡¢¡°±äС¡±»ò¡°²»±ä¡±£©¡£Êµ¼ÊÉÏ£¬ÔÚʹÓÃÌúþ´ß»¯¼ÁµÄ¹¤ÒµÁ÷³ÌÖУ¬Ò»°ã²ÉÓÃ400¡æ×óÓÒ¡¢p[H2O(g)]/p(CO)=3¡«5¡£ÆäÔ­Òò¿ÉÄÜÊÇ_________________________________¡£

£¨5£©¹¤ÒµÉÏ¿ÉÀûÓÃÔ­µç³ØÔ­Àí³ýÈ¥¹¤ÒµÎ²ÆøÖеÄCO²¢ÀûÓÃÆäµçÄÜ£¬·´Ó¦×°ÖÃÈçÉÏÓÒͼËùʾ£¬Çëд³ö¸º¼«µÄµç¼«·´Ó¦Ê½£º___________________________________¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø