ÏÂͼ±íʾµÄÊÇÄÑÈÜÇâÑõ»¯ÎïÔÚ²»Í¬pHϵÄÈܽâ¶È£¨S/mol¡¤L£­1£©£¬ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ

A£®pH£½3ʱÈÜÒºÖÐÌúÔªËصÄÖ÷Òª´æÔÚÐÎʽÊÇFe3£«
B£®ÈôNi(NO3)2ÈÜÒºÖк¬ÓÐÉÙÁ¿µÄCo2£«ÔÓÖÊ£¬¿Éͨ¹ýµ÷½ÚÈÜÒºpHµÄ·½·¨À´³ýÈ¥
C£®Èô·ÖÀëÈÜÒºÖеÄFe3£«ºÍCu2£«£¬¿Éµ÷½ÚÈÜÒºµÄpHÔÚ4×óÓÒ
D£®ÈôÔÚº¬ÓÐCu2£«ºÍNi2£«µÄÈÜÒºÖмÓÈëÉռNi(OH)2ÓÅÏȳÁµí

C

½âÎöÊÔÌâ·ÖÎö£º¸ù¾ÝͼÏñ¿ÉÖª£¬pH£½3ʱÈÜÒºÖÐÌúÔªËصÄÖ÷Òª´æÔÚÐÎʽÊÇÇâÑõ»¯Ìú£¬A²»ÕýÈ·£»¸ù¾ÝͼÏñ¿ÉÖª£¬ÇâÑõ»¯îÜÍêÈ«³ÁµíʱµÄpH´óÓÚÇâÑõ»¯ÄøÍêÈ«³ÁµíʱµÄpH£¬ËùÒÔµ±Co2£«ÍêÈ«³ÁµíʱNi2£«Ò²ÒѾ­ÍêÈ«³Áµí£¬B²»ÕýÈ·£»pH£½4ʱ£¬ÌúÀë×ÓÍêȫת»¯ÎªÇâÑõ»¯Ìú³Áµí£¬¶øÍ­Àë×Ó»¹ÁôÔÚÈÜÒºÖУ¬CÕýÈ·£»D²»ÕýÈ·£¬ÒòΪͭÀë×ÓÍêÈ«³ÁµíʱµÄpHСÓÚÇâÑõ»¯ÄøÍêÈ«³ÁµíʱµÄpH£¬´ð°¸Ñ¡C¡£
¿¼µã£º¿¼²éÈܶȻý³£ÊýµÄÓйؼÆËã¡¢ÅжϺÍÓ¦ÓÃ
µãÆÀ£º¸ÃÌâÊǸ߿¼Öеij£¼ûͼÏñ£¬ÄѶȽϴó£¬ÊÔÌâ×ÛºÏÐÔÇ¿¡£±¾ÌâÓÃͼ±í±íÊö»¯Ñ§¹ý³Ì»ò³ÊÏÖ±³¾°ÐÅÏ¢ÊÇ»¯Ñ§³£Óõıí´ï·½Ê½¡£Í¼±íÍùÍùÓµÓнϴóµÄÐÅÏ¢´æ´¢Á¿£¬Äܹ»ºÜÈ«ÃæµØ¿¼²éѧÉú·ÖÎö¡¢±È½Ï¡¢¸ÅÀ¨¡¢¹éÄÉÎÊÌâµÄÄÜÁ¦¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013ÄêÆÕͨ¸ßµÈѧУÕÐÉúͳһ¿¼ÊÔ(½­ËÕ¾í)»¯Ñ§ÊÔÌâ ÌâÐÍ£º022

Á×ÊǵؿÇÖк¬Á¿½ÏΪ·á¸»µÄ·Ç½ðÊôÔªËØ£¬Ö÷ÒªÒÔÄÑÈÜÓÚË®µÄÁ×ËáÑÎÈçCa3(PO4)2µÈÐÎʽ´æÔÚ£®ËüµÄµ¥Öʺͻ¯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Óã®

(1)°×Á×(P4)¿ÉÓÉCa3(PO4)2¡¢½¹Ì¿ºÍSiO2ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦»ñµÃ£®Ïà¹ØÈÈ»¯Ñ§·½³ÌʽÈçÏ£º2Ca3(PO4)2(s)£«10C(s)6CaO(s)£«P4(s)£«10CO(g)¡¡¦¤H1£½£«3359.26 kJ¡¤mol£­1

CaO(s)£«SiO2(s)CaSiO3(s)¡¡¦¤H2£½£­89.61 kJ¡¤mol£­1

2Ca3(PO4)2(s)£«6SiO2(s)£«10C(s) 6CaSiO3(s)£«P4(s)£«10CO(g)¡¡¦¤H3Ôò¦¤H3£½________kJ¡¤mol£­1£®

(2)°×Á×Öж¾ºó¿ÉÓÃCuSO4ÈÜÒº½â¶¾£¬½â¶¾Ô­Àí¿ÉÓÃÏÂÁл¯Ñ§·½³Ìʽ±íʾ£º

11P4£«60CuSO4£«96H2O20Cu3P£«24H3PO4£«60H2SO4

60 mol¡¡CuSO4ÄÜÑõ»¯°×Á×µÄÎïÖʵÄÁ¿ÊÇ________£®

(3)Á×µÄÖØÒª»¯ºÏÎïNaH2PO4¡¢Na2HPO4ºÍNa3PO4¿Éͨ¹ýH3PO4ÓëNaOHÈÜÒº·´Ó¦»ñµÃ£¬º¬Á׸÷ÎïÖֵķֲ¼·ÖÊý(ƽºâʱijÎïÖÖµÄŨ¶ÈÕ¼¸÷ÎïÖÖŨ¶ÈÖ®ºÍµÄ·ÖÊý)ÓëpHµÄ¹ØϵÈçÏÂͼËùʾ£®

¢ÙΪ»ñµÃ¾¡¿ÉÄÜ´¿µÄNaH2PO4£¬pHÓ¦¿ØÖÆÔÚ________£»pH£½8ʱ£¬ÈÜÒºÖÐÖ÷Òªº¬Á×ÎïÖÖŨ¶È´óС¹ØϵΪ________£®

¢ÚNa2HPO4ÈÜÒºÏÔ¼îÐÔ£¬ÈôÏòÈÜÒºÖмÓÈë×ãÁ¿µÄCaCl2ÈÜÒº£¬ÈÜÒºÔòÏÔËáÐÔ£¬ÆäÔ­ÒòÊÇ________(ÓÃÀë×Ó·½³Ìʽ±íʾ)£®

(4)Á׵Ļ¯ºÏÎïÈýÂÈÑõÁ×()Óë¼¾ÎìËÄ´¼()ÒÔÎïÖʵÄÁ¿Ö®±È2¡Ã1·´Ó¦Ê±£¬¿É»ñµÃÒ»ÖÖÐÂÐÍ×èȼ¼ÁÖмäÌåX£¬²¢ÊͷųöÒ»ÖÖËáÐÔÆøÌ壮¼¾ÎìËÄ´¼ÓëXµÄºË´Å¹²ÕñÇâÆ×ÈçÏÂͼËùʾ£®

¢ÙËáÐÔÆøÌåÊÇ________(Ìѧʽ)£®

¢ÚXµÄ½á¹¹¼òʽΪ________£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013ÄêÈ«¹úÆÕͨ¸ßµÈѧУÕÐÉúͳһ¿¼ÊÔ»¯Ñ§£¨½­ËÕ¾í´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

Á×ÊǵؿÇÖк¬Á¿½ÏΪ·á¸»µÄ·Ç½ðÊôÔªËØ£¬Ö÷ÒªÒÔÄÑÈÜÓÚË®µÄÁ×ËáÑÎÈçCa3(PO4)2µÈÐÎʽ´æÔÚ¡£ËüµÄµ¥Öʺͻ¯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Óá£
£¨1£©°×Á×(P4)¿ÉÓÉCa3(PO4)2¡¢½¹Ì¿ºÍSiO2ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦»ñµÃ¡£Ïà¹ØÈÈ»¯Ñ§·½³ÌʽÈçÏ£º
2Ca3(PO4)2(s)+10C(s)="==" 6CaO(s)+P4(s)+10CO(g) ¡÷H1 ="+3359.26" kJ¡¤mol£­1
CaO(s)+SiO2(s)="==" CaSiO3(s)                  ¡÷H2 ="-89." 61 kJ¡¤mol£­1
2Ca3(PO4)2(s)+6SiO2(s)+10C(s)="==" 6CaSiO3(s)+P4(s)+10CO(g)   ¡÷H3
Ôò¡÷H3 =           kJ¡¤mol£­1¡£
£¨2£©°×Á×Öж¾ºó¿ÉÓÃCuSO4ÈÜÒº½â¶¾£¬½â¶¾Ô­Àí¿ÉÓÃÏÂÁл¯Ñ§·½³Ìʽ±íʾ£º
11P 4+60CuSO4+96H2O="==" 20Cu3P+24H3PO4+60H2SO4
60molCuSO4ÄÜÑõ»¯°×Á×µÄÎïÖʵÄÁ¿ÊÇ      ¡£
£¨3£©Á×µÄÖØÒª»¯ºÏÎïNaH2PO4¡¢Na2HPO4ºÍNa3PO4¿Éͨ¹ýH3PO4ÓëNaOHÈÜÒº·´Ó¦»ñµÃ£¬º¬Á׸÷ÎïÖֵķֲ¼·ÖÊý(ƽºâʱijÎïÖÖµÄŨ¶ÈÕ¼¸÷ÎïÖÖŨ¶ÈÖ®ºÍµÄ·ÖÊý)ÓëpH µÄ¹ØϵÈçͼËùʾ¡£

¢ÙΪ»ñµÃ¾¡¿ÉÄÜ´¿µÄNaH2PO4£¬pHÓ¦¿ØÖÆÔÚ       £»pH=8ʱ£¬ÈÜÒºÖÐÖ÷Òªº¬Á×ÎïÖÖŨ¶È´óС¹ØϵΪ                          ¡£ 
¢ÚNa2HPO4ÈÜÒºÏÔ¼îÐÔ£¬ÈôÏòÈÜÒºÖмÓÈë×ãÁ¿µÄCaCl2ÈÜÒº£¬ÈÜÒºÔòÏÔËáÐÔ£¬ÆäÔ­ÒòÊÇ
                                  (ÓÃÀë×Ó·½³Ìʽ±íʾ)¡£
£¨4£©Á׵Ļ¯ºÏÎïÈýÂÈÑõÁ×()Óë¼¾ÎìËÄ´¼()ÒÔÎïÖʵÄÁ¿Ö®±È2£º1 ·´Ó¦Ê±£¬¿É»ñµÃÒ»ÖÖÐÂÐÍ×èȼ¼ÁÖмäÌåX£¬²¢ÊͷųöÒ»ÖÖËáÐÔÆøÌå¡£¼¾ÎìËÄ´¼ÓëX µÄºË´Å¹²ÕñÇâÆ×ÈçÏÂͼËùʾ¡£

¢ÙËáÐÔÆøÌåÊÇ        (Ìѧʽ)¡£
¢ÚXµÄ½á¹¹¼òʽΪ                   ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013ÄêÈ«¹úÆÕͨ¸ßµÈѧУÕÐÉúͳһ¿¼ÊÔ»¯Ñ§£¨½­ËÕ¾í½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

Á×ÊǵؿÇÖк¬Á¿½ÏΪ·á¸»µÄ·Ç½ðÊôÔªËØ£¬Ö÷ÒªÒÔÄÑÈÜÓÚË®µÄÁ×ËáÑÎÈçCa3(PO4)2µÈÐÎʽ´æÔÚ¡£ËüµÄµ¥Öʺͻ¯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Óá£

£¨1£©°×Á×(P4)¿ÉÓÉCa3(PO4)2¡¢½¹Ì¿ºÍSiO2ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦»ñµÃ¡£Ïà¹ØÈÈ»¯Ñ§·½³ÌʽÈçÏ£º

2Ca3(PO4)2(s)+10C(s)="==" 6CaO(s)+P4(s)+10CO(g) ¡÷H1 ="+3359.26" kJ¡¤mol£­1

CaO(s)+SiO2(s)="==" CaSiO3(s)                  ¡÷H2 ="-89." 61 kJ¡¤mol£­1

2Ca3(PO4)2(s)+6SiO2(s)+10C(s)="==" 6CaSiO3(s)+P4(s)+10CO(g)   ¡÷H3

Ôò¡÷H3 =           kJ¡¤mol£­1¡£

£¨2£©°×Á×Öж¾ºó¿ÉÓÃCuSO4ÈÜÒº½â¶¾£¬½â¶¾Ô­Àí¿ÉÓÃÏÂÁл¯Ñ§·½³Ìʽ±íʾ£º

11P 4+60CuSO4+96H2O="==" 20Cu3P+24H3PO4+60H2SO4

60molCuSO4ÄÜÑõ»¯°×Á×µÄÎïÖʵÄÁ¿ÊÇ      ¡£

£¨3£©Á×µÄÖØÒª»¯ºÏÎïNaH2PO4¡¢Na2HPO4ºÍNa3PO4¿Éͨ¹ýH3PO4ÓëNaOHÈÜÒº·´Ó¦»ñµÃ£¬º¬Á׸÷ÎïÖֵķֲ¼·ÖÊý(ƽºâʱijÎïÖÖµÄŨ¶ÈÕ¼¸÷ÎïÖÖŨ¶ÈÖ®ºÍµÄ·ÖÊý)ÓëpH µÄ¹ØϵÈçͼËùʾ¡£

¢ÙΪ»ñµÃ¾¡¿ÉÄÜ´¿µÄNaH2PO4£¬pHÓ¦¿ØÖÆÔÚ       £»pH=8ʱ£¬ÈÜÒºÖÐÖ÷Òªº¬Á×ÎïÖÖŨ¶È´óС¹ØϵΪ                          ¡£ 

¢ÚNa2HPO4ÈÜÒºÏÔ¼îÐÔ£¬ÈôÏòÈÜÒºÖмÓÈë×ãÁ¿µÄCaCl2ÈÜÒº£¬ÈÜÒºÔòÏÔËáÐÔ£¬ÆäÔ­ÒòÊÇ

                                  (ÓÃÀë×Ó·½³Ìʽ±íʾ)¡£

£¨4£©Á׵Ļ¯ºÏÎïÈýÂÈÑõÁ×()Óë¼¾ÎìËÄ´¼()ÒÔÎïÖʵÄÁ¿Ö®±È2£º1 ·´Ó¦Ê±£¬¿É»ñµÃÒ»ÖÖÐÂÐÍ×èȼ¼ÁÖмäÌåX£¬²¢ÊͷųöÒ»ÖÖËáÐÔÆøÌå¡£¼¾ÎìËÄ´¼ÓëX µÄºË´Å¹²ÕñÇâÆ×ÈçÏÂͼËùʾ¡£

¢ÙËáÐÔÆøÌåÊÇ        (Ìѧʽ)¡£

¢ÚXµÄ½á¹¹¼òʽΪ                   ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Á×ÊǵؿÇÖк¬Á¿½ÏΪ·á¸»µÄ·Ç½ðÊôÔªËØ£¬Ö÷ÒªÒÔÄÑÈÜÓÚË®µÄÁ×ËáÑÎÈçCa3(PO4)2µÈÐÎʽ´æÔÚ¡£ËüµÄµ¥Öʺͻ¯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Óá£

£¨1£©°×Á×(P4)¿ÉÓÉCa3(PO4)2¡¢½¹Ì¿ºÍSiO2 ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦»ñµÃ¡£Ïà¹ØÈÈ»¯Ñ§·½³ÌʽÈçÏ£º

2Ca3(PO4)2(s)+10C(s)=== 6CaO(s)+P4(s)+10CO(g) ¡÷H1 =+3359.26 kJ¡¤mol£­1

CaO(s)+SiO2(s)=== CaSiO3(s)                  ¡÷H2 =-89. 61 kJ¡¤mol£­1

2Ca3(PO4)2(s)+6SiO2(s)+10C(s)=== 6CaSiO3(s)+P4(s)+10CO(g)    ¡÷H3

Ôò¡÷H3 =            kJ¡¤mol£­1¡£

£¨2£©°×Á×Öж¾ºó¿ÉÓÃCuSO4ÈÜÒº½â¶¾£¬½â¶¾Ô­Àí¿ÉÓÃÏÂÁл¯Ñ§·½³Ìʽ±íʾ£º

11P 4+60CuSO4+96H2O=== 20Cu3P+24H3PO4+60H2SO4

60molCuSO4ÄÜÑõ»¯°×Á×µÄÎïÖʵÄÁ¿ÊÇ       ¡£

£¨3£©Á×µÄÖØÒª»¯ºÏÎïNaH2PO4¡¢Na2HPO4ºÍNa3PO4¿Éͨ¹ýH3PO4ÓëNaOHÈÜÒº·´Ó¦»ñµÃ£¬º¬Á׸÷ÎïÖֵķֲ¼·ÖÊý(ƽºâʱijÎïÖÖµÄŨ¶ÈÕ¼¸÷ÎïÖÖŨ¶ÈÖ®ºÍµÄ·ÖÊý)ÓëpH µÄ¹ØϵÈçÓÒͼËùʾ¡£

   ¢ÙΪ»ñµÃ¾¡¿ÉÄÜ´¿µÄNaH2PO4£¬pHÓ¦¿ØÖÆÔÚ        £»pH=8ʱ£¬ÈÜÒºÖÐÖ÷Òªº¬Á×ÎïÖÖŨ¶È´óС¹ØϵΪ                           ¡£ 

   ¢ÚNa2HPO4 ÈÜÒºÏÔ¼îÐÔ£¬ÈôÏòÈÜÒºÖмÓÈë×ãÁ¿µÄCaCl2 ÈÜÒº£¬ÈÜÒºÔòÏÔËáÐÔ£¬ÆäÔ­ÒòÊÇ

                                   (ÓÃÀë×Ó·½³Ìʽ±íʾ)¡£

£¨4£©Á׵Ļ¯ºÏÎïÈýÂÈÑõÁ×()Óë¼¾ÎìËÄ´¼()ÒÔÎïÖʵÄÁ¿Ö®±È2£º1 ·´Ó¦Ê±£¬¿É»ñµÃÒ»ÖÖÐÂÐÍ×èȼ¼ÁÖмäÌåX£¬²¢ÊͷųöÒ»ÖÖËáÐÔÆøÌå¡£¼¾ÎìËÄ´¼ÓëX µÄºË´Å¹²ÕñÇâÆ×ÈçÏÂͼËùʾ¡£

    ¢ÙËáÐÔÆøÌåÊÇ         (Ìѧʽ)¡£

    ¢ÚXµÄ½á¹¹¼òʽΪ                    ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸