a¡¢b¡¢c¡¢dΪԭ×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÔªËØ£¬b¡¢c¡¢dͬÖÜÆÚ£¬ÓÉa¡¢b¡¢c¡¢dËÄÖÖÔªËØÐγɵij£¼ûËáʽÑÎAÓÐÈçͼËùʾµÄת»¯¹Øϵ(ͼÖÐÿÖÖ×Öĸ±íʾһÖÖµ¥ÖÊ»ò»¯ºÏÎï)£®

(1)CµÄÔ­×ӽṹʾÒâͼΪ________£®

(2)д³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£ºF________£¬I________£®

(3)д³ö·´Ó¦¢ÙµÄ»¯Ñ§·½³Ìʽ£º________£®

(4)½«AºÍE×é³ÉµÄ¹ÌÌå»ìºÏÎïX g ÈÜÓÚË®Åä³ÉÈÜÒº£¬ÏòÆäÖÐÂýÂýµÎÈëIµÄÏ¡ÈÜÒº£¬²âµÃ¼ÓÈëIÈÜÒºµÄÌå»ýÓëÉú³ÉCµÄÌå»ý(±ê×¼×´¿ö)ÈçϱíËùʾ£º

¢ÙIÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ________£®

¢ÚxµÄֵΪ________£®

´ð°¸£º
½âÎö£º

¡¡¡¡(1)(1·Ö)

¡¡¡¡(2)Al(OH)3¡¡HNO3(¸÷1·Ö)

¡¡¡¡(3)4NH3£«5O24NO£«6H2O(2·Ö)

¡¡¡¡(4)1 mol/L(2·Ö)¡¡8.17(3·Ö)


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©ÂÈËá¼ØÈÛ»¯£¬Á£×Ó¼ä¿Ë·þÁË
Àë×Ó¼ü
Àë×Ó¼ü
µÄ×÷ÓÃÁ¦£»¶þÑõ»¯¹èÈÛ»¯£¬Á£×Ó¼ä¿Ë·þÁË
¹²¼Û¼ü
¹²¼Û¼ü
µÄ×÷ÓÃÁ¦£»µâµÄÉý»ª£¬Á£×Ó¼ä¿Ë·þÁË
·Ö×Ó¼ä
·Ö×Ó¼ä
µÄ×÷ÓÃÁ¦£®ÈýÖÖ¾§ÌåµÄÈÛµãÓɸߵ½µÍµÄ˳ÐòÊÇ
SiO2£¾KClO3£¾I2
SiO2£¾KClO3£¾I2
£®
£¨2£©ÏÂÁÐÁùÖÖ¾§Ì壺¢ÙCO2£¬¢ÚNaCl£¬¢ÛNa£¬¢ÜSi£¬¢ÝCS2£¬¢Þ½ð¸Õʯ£¬ËüÃǵÄÈÛµã´ÓµÍµ½¸ßµÄ˳ÐòΪ
¢Ù¢Ý¢Û¢Ú¢Ü¢Þ
¢Ù¢Ý¢Û¢Ú¢Ü¢Þ
£¨ÌîÐòºÅ£©£®
£¨3£©ÔÚH2¡¢£¨NH4£©2SO4¡¢SiC¡¢CO2¡¢HFÖУ¬Óɼ«ÐÔ¼üÐγɵķǼ«ÐÔ·Ö×ÓÓÐ
CO2
CO2
£¬ÓɷǼ«ÐÔ¼üÐγɵķǼ«ÐÔ·Ö×ÓÓÐ
H2
H2
£¬ÄÜÐγɷÖ×Ó¾§ÌåµÄÎïÖÊÊÇ
H2¡¢CO2¡¢HF
H2¡¢CO2¡¢HF
£¬º¬ÓÐÇâ¼üµÄ¾§ÌåµÄ»¯Ñ§Ê½ÊÇ
HF
HF
£¬ÊôÓÚÀë×Ó¾§ÌåµÄÊÇ
£¨NH4£©2SO4
£¨NH4£©2SO4
£¬ÊôÓÚÔ­×Ó¾§ÌåµÄÊÇ
SiC
SiC
£¬ÎåÖÖÎïÖʵÄÈÛµãÓɸߵ½µÍµÄ˳ÐòÊÇ
SiC£¾£¨NH4£©2SO4£¾HF£¾CO2£¾H2
SiC£¾£¨NH4£©2SO4£¾HF£¾CO2£¾H2
£®
£¨4£©A¡¢B¡¢C¡¢DΪËÄÖÖ¾§Ì壬ÐÔÖÊÈçÏ£º
A£®¹Ì̬ʱÄܵ¼µç£¬ÄÜÈÜÓÚÑÎËá
B£®ÄÜÈÜÓÚCS2£¬²»ÈÜÓÚË®
C£®¹Ì̬ʱ²»µ¼µç£¬ÒºÌ¬Ê±Äܵ¼µç£¬¿ÉÈÜÓÚË®
D£®¹Ì̬¡¢ÒºÌ¬Ê±¾ù²»µ¼µç£¬ÈÛµãΪ3 500¡æ
ÊÔÍƶÏËüÃǵľ§ÌåÀàÐÍ£ºA£®
½ðÊô¾§Ìå
½ðÊô¾§Ìå
£»B£®
·Ö×Ó¾§Ìå
·Ö×Ó¾§Ìå
£»C£®
Àë×Ó¾§Ìå
Àë×Ó¾§Ìå
£»D£®
Ô­×Ó¾§Ìå
Ô­×Ó¾§Ìå
£®
£¨5£©Í¼ÖÐA¡«DÊÇÖÐѧ»¯Ñ§½Ì¿ÆÊéÉϳ£¼ûµÄ¼¸ÖÖ¾§Ìå½á¹¹Ä£ÐÍ£¬ÇëÌîдÏàÓ¦ÎïÖʵÄÃû³Æ£º
A£®
CsCl
CsCl
£»B£®
NaCl
NaCl
£»C£®
SiO2
SiO2
£»D£®
½ð¸Õʯ
½ð¸Õʯ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2009?ÄÏƽ¶þÄ££©A¡¢B¡¢C¡¢DΪËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬A¡¢B¡¢DµÄÔ­×ÓÐòÊýºÍÔ­×Ӱ뾶¾ùÒÀ´ÎÔö´ó£¬B¡¢DͬÖ÷×åÇÒÄÜ×é³ÉÒ»ÖÖÄÜÐγɡ°ËáÓꡱµÄ»¯ºÏÎA¡¢B¿ÉÒÔÐγÉA2BºÍA2B2µÄÁ½ÖÖͨ³£Çé¿öϳÊҺ̬µÄ¹²¼Û»¯ºÏÎB¡¢CÐγɵÄÁ½ÖÖÀë×Ó»¯ºÏÎïÈÜÓÚË®£¬ËùµÃµÄÈÜÒº¾ù³ÊÇ¿¼îÐÔ£»CµÄµ¥Öʳ£ÎÂÏ¿ÉÓëA2B¾çÁÒ·´Ó¦£®ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©B¡¢CÁ½ÔªËØÒÔ΢Á£¸öÊý±È1£º1ÐγɵĻ¯ºÏÎïXÖУ¬Òõ¡¢ÑôÀë×Ó¸öÊý±ÈΪ
1£º2
1£º2
£®
£¨2£©ÔÚA2B2×÷ÓÃÏ£¬Í­ÓëÏ¡ÁòËáÖÆÁòËáÍ­µÄ»¯Ñ§·´Ó¦µÄ»¯Ñ§·½³ÌΪ£º
Cu+H2SO4+H2O2=CuSO4+2H2O
Cu+H2SO4+H2O2=CuSO4+2H2O
£®
£¨3£©±íʾÐγÉDB2ÐÍ¡°ËáÓꡱµÄ»¯Ñ§·´Ó¦·½³ÌʽÓжà¸ö£¬ÇëÄãÑ¡ÔñÒ»¸öºÏÊʵķ´Ó¦£¬Ð´³öÕâ¸ö·´Ó¦µÄƽºâ³£Êý±í´ïʽK=
c2(SO3)
c2(SO2)?c(O2)
c2(SO3)
c2(SO2)?c(O2)
£®
£¨4£©ÒÑÖª³£ÎÂÏ 17gA¡¢DÁ½ÔªËØ×é³ÉµÄ»¯ºÏÎïÓë×ãÁ¿µÄDB2ÍêÈ«·´Ó¦Ê±·Å³öÈÈÁ¿Îªa kJ£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ
2H2S£¨g£©+SO2£¨g£©¨T3S£¨s£©+2H2O£¨l£©¡÷H=-4akJ/mol
2H2S£¨g£©+SO2£¨g£©¨T3S£¨s£©+2H2O£¨l£©¡÷H=-4akJ/mol
£®
£¨5£©ÒÑÖª 25¡æʱ£¬Ksp£¨CaCO3£©=1¡Á10-9£®Ksp£¨CaSO4£©=9¡Á10-6£®³¤ÆÚʹÓõĹø¯ÐèÒª¶¨ÆÚ³ýË®¹¸£¬·ñÔò»á½µµÍȼÁϵÄÀûÓÃÂÊ£®Ë®¹¸Öк¬ÓеÄCaSO4£¬¿ÉÏÈÓÃNa2CO3ÈÜÒº´¦Àí£¬Ê¹Ö®×ª»¯ÎªÊèËÉ¡¢Ò×ÈÜÓÚËáµÄCaCO3£¬¶øºóÓÃËá³ýÈ¥£®
¢ÙCaSO4ת»¯ÎªCaCO3µÄÀë×Ó·½³ÌʽΪ
CaSO4£¨s£©+CO32-£¨aq£©=CaCO3£¨s£©+SO42-£¨aq£©
CaSO4£¨s£©+CO32-£¨aq£©=CaCO3£¨s£©+SO42-£¨aq£©
£»
¢ÚÇë·ÖÎöCaSO4ת»¯ÎªCaCO3µÄÔ­Àí
¼ÓÈëNa2CO3ÈÜÒººó£¬CO32-ÓëCa2+½áºÏ£¬×ª»¯ÎªKsp¸üСµÄCaCO3³Áµí£¬Ê¹CaSO4µÄ³ÁµíÈܽâƽºâÏòÈܽⷽÏòÒƶ¯
¼ÓÈëNa2CO3ÈÜÒººó£¬CO32-ÓëCa2+½áºÏ£¬×ª»¯ÎªKsp¸üСµÄCaCO3³Áµí£¬Ê¹CaSO4µÄ³ÁµíÈܽâƽºâÏòÈܽⷽÏòÒƶ¯
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¶ÌÖÜÆÚA£¬B£¬C£¬D£¬EÎåÖÖÔªËØÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬A¡¢BͬÖ÷×壬B¡¢C¡¢DÈýÖÖÔªËØÔ­×ÓÐγɵĽðÊôÑôÀë×Óµç×Ó²ã½á¹¹Ïàͬ£¬EÔªËØÔ­×Ó×îÍâ²ãµç×ÓÊýµÈÓÚA¡¢B¡¢C¡¢DÔªËØÔ­×Ó×îÍâ²ãµç×ÓÊýÖ®ºÍ£®A¡¢BÄÜÐγÉÔ­×Ó¸öÊý±ÈΪ1£º1µÄÀë×Ó»¯ºÏÎïW£¬WÒ×ÓëË®·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©WµÄµç×Óʽ
Na+[£ºH]-
Na+[£ºH]-
£¬WÓëË®·´Ó¦µÄ»¯Ñ§·½³Ìʽ
NaH+H2O=NaOH+H2¡ü
NaH+H2O=NaOH+H2¡ü
£®
£¨2£©ÈôÒÔC¡¢DÁ½ÔªËصĵ¥ÖÊ×÷Ϊµç¼«¢ÙÒÔA¡¢E×é³ÉµÄ»¯ºÏÎïΪµç½âÖʹ¹³ÉÔ­µç³Ø£¬Ôò¸º¼«Îª
þ
þ
£¨ÌîÔªËØÃû³Æ£©£¬Õý¼«·´Ó¦Ê½ÊÇ
2H++2e-=H2¡ü
2H++2e-=H2¡ü
£® ¢ÚÒÔBÔªËØ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïΪµç½âÖʹ¹³ÉÔ­µç³Ø£¬Ôò¸º¼«Îª
ÂÁ
ÂÁ
£¨ÌîÔªËØÃû³Æ£©£¬¸º¼«·´Ó¦Ê½ÊÇ
2Al+8OH--6e-=2AlO2-+4H2O
2Al+8OH--6e-=2AlO2-+4H2O
£®
£¨3£©ÒÑÖªCµ¥ÖÊÓëEµ¥ÖÊ¿É·¢ÉúȼÉÕ·´Ó¦£¬ÒÔH2SO4×÷Ϊµç½âÖÊÈÜÒºÉè¼Æ³ÉȼÁϵç³Ø£¬Ôò¸º¼«·´Ó¦Ê½ÊÇ
Mg-2e-¨TMg2+
Mg-2e-¨TMg2+
£¬Õý¼«·´Ó¦Ê½
Cl2+2e-¨T2Cl-
Cl2+2e-¨T2Cl-
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡¢B¡¢C¡¢DΪ¶ÌÖÜÆÚÔªËØ£¬Çë¸ù¾ÝϱíÐÅÏ¢»Ø´ðÎÊÌ⣮
ÔªËØ A B C D
ÐÔÖÊ»ò½á¹¹ÐÅÏ¢ ¹¤ÒµÉÏͨ¹ý·ÖÀëҺ̬¿ÕÆø»ñµÃÆäµ¥ÖÊ£¬µ¥ÖÊÄÜÖúȼ Æø̬Ç⻯ÎïÏÔ¼îÐÔ +3¼ÛÑôÀë×ӵĺËÍâµç×ÓÅŲ¼ÓëÄÊÔ­×ÓÏàͬ µÚÈýÖÜÆÚÔ­×Ӱ뾶×îС
£¨1£©BÔÚÔªËØÖÜÆÚ±íµÄλÖãº
 
£®
£¨2£©¹¤ÒµÉϵç½â·¨Ò±Á¶µ¥ÖÊCµÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨3£©DµÄµ¥ÖÊÓëAµÄÒ»ÖÖÇ⻯Îï·´Ó¦Éú³É¾ßÓÐƯ°×ÐÔµÄÎïÖÊ£»DµÄµ¥ÖÊÓëAµÄÁíÒ»ÖÖÇ⻯Îï·´Ó¦Éú³ÉAµÄµ¥ÖÊ£®Óйط´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
 
£»
 
£®
£¨4£©DµÄ×î¸ß¼ÛÑõ»¯ÎïΪÎÞÉ«ÒºÌ壬1mol¸ÃÎïÖÊÓëÒ»¶¨Á¿Ë®»ìºÏµÃµ½Ò»ÖÖÏ¡ÈÜÒº£¬²¢·Å³öQkJÈÈÁ¿£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡¢B¡¢C¡¢DΪËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£®AÔ­×Ó×îÍâ²ãÓÐ5¸öµç×Ó£»BµÄÒõÀë×ÓºÍCµÄÑôÀë×Ó¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£¬Á½ÔªËصĵ¥ÖÊ·´Ó¦£¬Éú³ÉÒ»ÖÖµ­»ÆÉ«µÄ¹ÌÌåE£»DÔ­×ÓL²ãÉϵĵç×ÓÊýµÈÓÚK¡¢MÁ½¸öµç×Ó²ãÉϵĵç×ÓÊýÖ®ºÍ£®Çë»Ø´ð£º
£¨1£©AÔªËصÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïµÄ»¯Ñ§Ê½ÊÇ
 
£»
£¨2£©ÎïÖÊEÖÐËù°üº¬µÄ»¯Ñ§¼üÓÐ
 
£»
£¨3£©Ð´³öC¡¢DÁ½ÔªËØÐγɵĻ¯ºÏÎïµÄµç×Óʽ
 
£»
£¨4£©°ÑÊ¢ÓÐ48mL AB¡¢AB2»ìºÏÆøÌåµÄÈÝÆ÷µ¹ÖÃÓÚË®ÖУ¨Í¬ÎÂͬѹÏ£©£¬´ýÒºÃæÎȶ¨ºó£¬ÈÝÆ÷ÄÚÆøÌåÌå»ý±äΪ24mL£¬ÔòÔ­»ìºÏÆøÌåÖÐABµÄÌå»ý·ÖÊýΪ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸