¡¾ÌâÄ¿¡¿¿Æѧ¼ÒÔ¤²â21ÊÀ¼ÍÖÐÒ¶½«½øÈë¡°ÇâÄܾ­¼Ã¡±Ê±´ú£¬ÏÂÁÐÎïÖʶ¼ÊǾßÓйãÀ«Ó¦ÓÃÇ°¾°µÄ´¢Çâ²ÄÁÏ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)Zr(ï¯)ÔÚÔªËØÖÜÆÚ±íÖÐλÓÚµÚÎåÖÜÆÚ£¬ÓëîÑͬ×壬»ù̬ZrµÄ¼Û²ãµç×ÓÅŲ¼Ê½Îª_______¡£

(2)ÑÇ°±»ùï®(Li2NH) Ëùº¬ÔªËصÚÒ»µçÀëÄÜ×îСµÄÊÇ____ £¬µç¸ºÐÔ×î´óµÄÊÇ_____ (ÌîÔªËØ·ûºÅ)¡£

(3)ßÇßò()µÄ·Ðµã±ÈÜÌ()¸ßµÄÖ÷ÒªÔ­ÒòÊÇ________¡£

(4)¢ÙNH3BH3 (°±ÅðÍ飬ÈÛµã104¡æ)ÓëÒÒÍ黥ΪµÈµç×ÓÌå¡£NH3BH3µÄ¾§ÌåÀàÐÍΪ____¡£ÆäÖÐBµÄÔÓ»¯ÀàÐÍΪ____£¬¿Éͨ¹ý_________²â¶¨¸Ã·Ö×ÓµÄÁ¢Ìå¹¹ÐÍ¡£

¢ÚNH3BH3¿Éͨ¹ý»·Åð°±Íé¡¢CH4ÓëH2O½øÐкϳɣ¬¼ü½Ç£º CH4______H2O (Ìî¡°> "»ò¡°<")£¬Ô­ÒòÊÇ________¡£

(5)MgH2¾§ÌåÊôËÄ·½Æ·Ïµ£¬½á¹¹Èçͼ£¬¾§°û²ÎÊýa =b= 450pm£¬ c= 30lpm£¬Ô­×Ó×ø±êΪA(0£¬0£¬0)¡¢B(0.305£¬0.305£¬0)¡¢C(1£¬1£¬1)¡¢D(0.195£¬0.805£¬0.5)¡£

¢ÙMg2+µÄ°ë¾¶Îª72pm£¬ÔòH+µÄ°ë¾¶Îª______pm (Áгö¼ÆËã±í´ïʽ)

¢ÚÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£Êý£¬MgH2¾§ÌåÖÐÇâµÄÃܶÈÊDZê×¼×´¿öÏÂÇâÆøÃܶȵÄ_____±¶(Áгö¼ÆËã±í´ïʽ£¬ÇâÆøÃܶÈΪ0.089g¡¤L-1)¡£

¡¾´ð°¸¡¿4d25s2 Li N ßÇßò·Ö×Ó¼ä´æÔÚÇâ¼ü ·Ö×Ó¾§Ìå sp3 ºìÍâ¹âÆ× £¾ CH4ÖÐÎ޹µç×Ó¶Ô£¬H2OÖÐÓÐÁ½¶Ô¹Âµç×Ó¶Ô£¬¹Âµç×Ó¶ÔÓë¦Ò¼üÖ®¼äµÄÅųâÁ¦´óÓÚ¦Ò ¼üÖ®¼äµÄÅųâÁ¦£¬Ê¹µÃCH4Öеļü½Ç±ÈH2O Öеļü½Ç´ó »ò

¡¾½âÎö¡¿

£¨1£©Zr(ï¯)ÔÚÔªËØÖÜÆÚ±íÖÐλÓÚµÚÎåÖÜÆÚ£¬ÓëîÑͬ×壬ÊôÓÚ¢ôB×壬Ôò»ù̬ZrµÄ¼Û²ãµç×ÓÅŲ¼Ê½Îª£»

£¨2£©LiºÍHÊôÓÚͬÖ÷×壬µÚÒ»µçÀëÄÜÓÉ´ó±äС£¬H>Li£¬µç¸ºÐÔÓÉС±ä´óH<Li £»LiºÍNÊôÓÚͬÖÜÆÚ£¬µÚÒ»µçÀëÄÜÓÉС±ä´óLi<N£¬ µç¸ºÐÔÓÉС±ä´óLi<N£»

£¨3£©ÓÉßÇßò·Ö×ӽṹ¼òʽ¿ÉÖªº¬ÓÐßÇßò·Ö×Ó¼ä´æÔÚNºÍHµÄÇâ¼ü£¬È۷еã½Ï¸ß£»

£¨4£©¢ÙNH3BH3ÈÛµã½ÏµÍ£¬ÔòNH3BH3Ϊ·Ö×Ó¾§Ì壻ÒÀ¾ÝÖÐÐÄÔ­×Ó¼Û²ãµç×Ó¶Ô»¥³âÀíÂÛÅжÏBµÄÔÓ»¯ÀàÐÍ£»¿Éͨ¹ýºìÍâ¹âÆײⶨ¸Ã·Ö×ÓµÄÁ¢Ìå¹¹ÐÍ£»

£¨5£©¢ÙÀûÓÿռ伸ºÎ¼ÆËãABµÄ¾àÀë=Mg2+µÄ°ë¾¶+H+µÄ°ë¾¶£»¢ÚÀûÓþù̯·¨¼ÆË㾧°ûÖÐÇâÃܶȡ£

£¨1£©Zr(ï¯)ÔÚÔªËØÖÜÆÚ±íÖÐλÓÚµÚÎåÖÜÆÚ£¬ÓëîÑͬ×壬ÊôÓÚ¢ôB×壬ÒÀ¾ÝºËÍâµç×ÓÅŲ¼¹æÂÉд³ö»ù̬ZrµÄ¼Û²ãµç×ÓÅŲ¼Ê½Îª4d25s2£¬¹Ê´ð°¸Îª£º4d25s2£»

£¨2£©LiºÍHÊôÓÚͬÖ÷×壬µÚÒ»µçÀëÄÜÓÉ´ó±äС£¬H>Li£¬µç¸ºÐÔÓÉС±ä´óH<Li £»LiºÍNÊôÓÚͬÖÜÆÚ£¬µÚÒ»µçÀëÄÜÓÉС±ä´óLi<N£¬ µç¸ºÐÔÓÉС±ä´óLi<N£¬¹Ê´ð°¸Îª£ºLi£»N£»

£¨3£©ÓÉßÇßò·Ö×ӽṹ¼òʽ¿ÉÖªº¬ÓÐßÇßò·Ö×Ó¼ä´æÔÚNºÍHµÄÇâ¼ü£¬È۷еã½Ï¸ß£¬¹Ê´ð°¸Îª£ºßÇßò·Ö×Ó¼ä´æÔÚÇâ¼ü£»

£¨4£©¢ÙNH3BH3ÈÛµã½ÏµÍ£¬ÔòNH3BH3Ϊ·Ö×Ó¾§Ì壻NH3BH3ÖÐBÔ­×Óº¬ÓÐ3¸ö¦Ò¼ü£¬1¶Ô¹Âµç×Ó¶Ô£¬ÔòÔÓ»¯ÀàÐÍΪsp3ÔÓ»¯£»¿Éͨ¹ýºìÍâ¹âÆײⶨ¸Ã·Ö×ÓµÄÁ¢Ìå¹¹ÐÍ£¬¹Ê´ð°¸Îª£ºsp3£»ºìÍâ¹âÆ×£»

¢ÚÖÐÐÄÔ­×ӹµç×Ó¶ÔÔ½¶à¼ü½ÇԽС£¬CH4ºÍH2O¾ùÊôÓÚsp3ÔÓ»¯£¬CH4ÖÐÎ޹µç×Ó¶Ô£¬H2OÖÐÓÐÁ½¶Ô¹Âµç×Ó¶Ô£¬¹Âµç×Ó¶ÔÓë¦Ò¼üÖ®¼äµÄÅųâÁ¦´óÓÚ¦Ò ¼üÖ®¼äµÄÅųâÁ¦£¬Ê¹µÃCH4Öеļü½Ç±ÈH2O Öеļü½Ç´ó£¬¹Ê´ð°¸Îª£º£¾£»CH4ÖÐÎ޹µç×Ó¶Ô£¬H2OÖÐÓÐÁ½¶Ô¹Âµç×Ó¶Ô£¬¹Âµç×Ó¶ÔÓë¦Ò¼üÖ®¼äµÄÅųâÁ¦´óÓÚ¦Ò ¼üÖ®¼äµÄÅųâÁ¦£¬Ê¹µÃCH4Öеļü½Ç±ÈH2OÖеļü½Ç´ó£»

£¨5£©¢ÙÀûÓÿռ伸ºÎ¼ÆËãABµÄ¾àÀë==Mg2+µÄ°ë¾¶+H+µÄ°ë¾¶£¬ÔòH+µÄ°ë¾¶=£»¢Ú¾§°ûÖÐH+¸öÊýΪ£¬Ôò¾§°ûÖÐÇâµÄÃܶȻò £¬¹Ê´ð°¸Îª£º£» »ò ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓÃNA±íʾ°¢·üÙ¤µÂÂÞ³£ÊýµÄÖµ¡£ÏÂÁÐÅжÏÕýÈ·µÄÊÇ( )

A.46gµÄNO2ºÍN2O4µÄ»ìºÏÎïËùº¬Ô­×Ó×ÜÊýΪ3NA

B.24g Mg±äΪMg2+ʱʧȥµÄµç×ÓÊýĿΪNA

C.1mol/L CaCl2ÈÜÒºÖк¬ÓеÄCl-Àë×ÓÊýĿΪ2NA

D.º¬ÓÐlmol FeCl3µÄÈÜÒº¼ÓÈÈÖÐÍêȫת»¯ÎªÇâÑõ»¯Ìú½ºÌ壬ÆäÖнºÁ£µÄÊýĿΪNA¸ö

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿³£ÎÂÏÂ,Ïò 20 mL 0.2 mol/LH2AÈÜÒºÖеμÓ0. 2 mol/L NaOHÈÜÒº.ÓйØ΢Á£µÄÎïÖʵÄÁ¿±ä»¯ÇúÏßÈçͼËùʾ(ÆäÖÐI´ú±íH2A.£¬II´ú±íHA¡ª£¬III±íA2¡ª)¡£¸ù¾Ýͼʾ.ÅжÏÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A.µ±V(NaOH)=20 mLʱ.ÈÜÒºÖи÷Àë×ÓŨ¶ÈµÄ´óС¹ØϵΪc(Na+)>c(HA¡ª)>c(H+)>c(A2Ò»)>c(OH¡ª)

B.µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒºÓëH2 AÈÜÒº»ìºÏºó£¬ÆäÈÜÒºÖÐË®µÄµçÀë³Ì¶È±È´¿Ë®ÖеĴó

C.NaHAÈÜÒºÖÐ:c(OH¡ª)+2c(A2¡ª) =c(H+) +c(H2A )

D.½«Na2AÈÜÒº¼ÓˮϡÊÍ.ÈÜÒºÖÐËùÓÐÀë×ÓµÄŨ¶È¶¼¼õС.µ«²¿·ÖÀë×ÓµÄÎïÖʵÄÁ¿Ôö¼Ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÀûÓÃÒÒÃÑ¡¢95%ÒÒ´¼½þÅݶÅÖÙ¸ÉÒ¶£¬µÃµ½ÌáÈ¡Òº£¬½ø²½»ñµÃÂÌÔ­Ëá´Ö²úÆ·µÄÖÖ¹¤ÒÕÁ÷³ÌÈçÏ£¬ÏÂÁÐ˵·¨´íÎóµÄÊÇ

A.³£ÎÂÏ£¬ÂÌÔ­ËáÒ×ÈÜÓÚË®

B.½þ¸àµÄÖ÷Òª³É·ÖÊÇÂÌÔ­Ëá

C.¼õѹÕôÁóµÄÄ¿µÄÊǽµµÍÕôÁóζȣ¬ÒÔÃâÂÌÔ­Ëá±äÖÊ

D.ÂÌÔ­Ëá´Ö²úÆ·¿ÉÒÔͨ¹ýÖؽᾧ½øÒ»²½Ìá´¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿CoS2´ß»¯COÑÌÆøÍÑÁò¾ßÓйãÀ«µÄ¹¤Òµ»¯Ç°¾°¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÒÑÖª£º

CoS2(s) +CO(g) CoS(s) +COS(g) H1

2COS(g) +SO2(g) 3S(s) +2CO2(g) H2

S(s) +CoS(S) CoS2 (s) ¡÷H3

Ôò2CO(g)+ SO2(g)2CO2(g)+S(s) H4=____¡£ (ÓÃH1¡¢ H2¡¢H3±íʾ)

(2)ÔÚºãΡ¢ºãѹµÄÈÝÆ÷ÖÐÄ£Äâ»ØÊÕÁò£¬¼ÓÈëSO2ÆðʼÁ¿¾ùΪ1mol£¬²âµÃCO2µÄƽºâÌå»ý·ÖÊýËæCOºÍSO2µÄͶÁϱȱ仯Èçͼ£º

¢Ùµ±Í¶ÁϱÈΪ2ʱ£¬t min ʱ²âµÃSO2ת»¯ÂÊΪ50%£¬ÔòÓÃSµÄÉú³ÉËÙÂʱíʾµÄ·´Ó¦ËÙÂÊv=______g¡¤min-1¡£

¢Úµ±Í¶ÁϱÈΪ3ʱ£¬CO2 µÄƽºâÌå»ý·ÖÊý¶ÔÓ¦µÄµãÊÇ______________¡£

(3)ÏòÌå»ý¾ùΪ1LµÄºãΡ¢ºã¿ÍÃÜÍÅÈÝÆ÷ͨÈë2 mol COºÍ| mol SO2¡£·´Ó¦Ìåϵ×ÜѹǿËæʱ¼äµÄ±ä»¯Èçͼ£º

¢ÙÏà¶ÔÓÚI£¬II¸Ä±äµÄÍâ½çÌõ¼þÊÇ____________________¡£

¢ÚSO2µÄƽºâת»¯ÂÊΪ______£¬Æ½ºâ³£ÊýKp =________(ÓÃƽºâ·Öѹ´úÌæƽºâŨ¶È¼ÆËã)¡£

(4)ÀûÓõç½â·¨´¦ÀíSO2βÆø¿ÉÖƱ¸±£ÏÕ·Û (Na2S2O4).µç½â×°ÖÃÈçͼ£¬Ôòa____ b (Ìî¡°>¡± ¡°=¡±»ò¡°<¡±)£¬Éú³ÉS2O42-µÄµç¼«·´Ó¦Ê½Îª____________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÊµÑéÌâ

»¯Ñ§ÐËȤС×é¶ÔijƷÅÆÑÀ¸àÖеÄĦ²Á¼Á³É·Ö¼°Æ京Á¿½øÐÐÒÔÏÂ̽¾¿£º

²éµÃ×ÊÁÏ£º¸ÃÑÀ¸àĦ²Á¼ÁÓÉ̼Ëá¸Æ£¬ÇâÑõ»¯ÂÁ×é³É£»ÑÀ¸àÖÐÆäËû³É·ÖÓöµ½ÑÎËáʱÎÞÆøÌå²úÉú.

I.Ħ²Á¼ÁÖÐÇâÑõ»¯ÂÁµÄ¶¨ÐÔ¼ìÑ飺ȡÊÊÁ¿ÑÀ¸àÑùÆ·£¬¼ÓË®³ä×ã½Á°è£¬¹ýÂË.

(1)ÍùÂËÔüÖмÓÈë¹ýÁ¿NaOHÈÜÒº£¬¹ýÂË. ÇâÑõ»¯ÂÁÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_______.

(2)Íù(1)ËùµÃÂËÒºÖÐÏÈͨÈë¹ýÁ¿¶þÑõ»¯Ì¼£¬ÔÙ¼ÓÈë¹ýÁ¿Ï¡ÑÎËá.ÕâÒ»¹ý³Ì·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÒÀ´ÎΪ£º__________________________________£¬__________________________________.

II.ÑÀ¸àÑùÆ·ÖÐ̼Ëá¸ÆµÄ¶¨Á¿²â¶¨£ºÀûÓÃÏÂͼËùʾװÖÃ(ͼÖмгÖÒÇÆ÷ÂÔÈ¥)½øÐÐʵÑ飬³ä·Ö·´Ó¦ºó£¬²â¶¨CÖÐÉú³ÉµÄBaCO3³ÁµíÖÊÁ¿£¬ÒÔÈ·¶¨Ì¼Ëá¸ÆµÄÖÊÁ¿·ÖÊý.

ÒÀ¾ÝʵÑé¹ý³Ì»Ø´ðÏÂÁÐÎÊÌ⣺

(3)ʵÑé¹ý³ÌÖÐÐè³ÖÐø»º»ºÍ¨Èë¿ÕÆø.Æä×÷ÓóýÁ˿ɽÁ°èB£¬CÖеķ´Ó¦ÎïÍ⣬»¹ÓУº_________.

(4)CÖз´Ó¦Éú³ÉBaCO3µÄÀë×Ó·½³ÌʽÊÇ___________________________________.

(5)ÏÂÁи÷Ïî´ëÊ©ÖУ¬²»ÄÜÌá¸ß²â¶¨×¼È·¶ÈµÄÊÇ£¨______£©(Ìî±êºÅ).

A.ÔÚ¼ÓÈëÑÎËá֮ǰ£¬Ó¦Åž»×°ÖÃÄÚµÄCO2ÆøÌå

B.µÎ¼ÓÑÎËá²»Ò˹ý¿ì

C.ÔÚA¡«BÖ®¼äÔöÌíÊ¢ÓÐŨÁòËáµÄÏ´Æø×°ÖÃ

D.ÔÚB¡«CÖ®¼äÔöÌíÊ¢Óб¥ºÍ̼ËáÇâÄÆÈÜÒºµÄÏ´Æø×°ÖÃ

(6)ʵÑéÖÐ׼ȷ³ÆÈ¡8.00 gÑùÆ·Èý·Ý£¬½øÐÐÈý´Î²â¶¨£¬²âµÃBaCO3ƽ¾ùÖÊÁ¿Îª3.94 g.ÔòÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ________.

(7)ÓÐÈËÈÏΪ²»±Ø²â¶¨CÖÐÉú³ÉµÄBaCO3ÖÊÁ¿£¬Ö»Òª²â¶¨×°ÖÃCÔÚÎüÊÕCO2Ç°ºóµÄÖÊÁ¿²î£¬Ò»Ñù¿ÉÒÔÈ·¶¨Ì¼Ëá¸ÆµÄÖÊÁ¿·ÖÊý.ʵÑéÖ¤Ã÷°´´Ë·½·¨²â¶¨µÄ½á¹ûÃ÷ÏÔÆ«¸ß£¬Ô­ÒòÊÇ_________________.

(8)×°ÖÃÖÐUÐιÜDÖеļîʯ»ÒµÄ×÷ÓÃÊÇ_____________________________.

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓÒͼÊÇ¿ÉÄæ·´Ó¦A+2B 2C + 3D µÄ»¯Ñ§·´Ó¦ËÙÂÊÓ뻯ѧƽºâËæÍâ½çÌõ¼þ¸Ä±ä¶ø±ä»¯µÄÇé¿ö£¬ÓÉ´ËÍƶϴíÎóµÄÊÇ

A.A¡¢BÒ»¶¨ÊÇÆøÌå

B.C¿ÉÄÜÊÇÆøÌå

C.DÒ»¶¨²»ÊÇÆøÌå

D.Õý·´Ó¦ÊÇ·ÅÈÈ·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÀûÓÃ̼ËáÄƾ§Ì壨Na2CO3¡¤10H2O£¬Ïà¶Ô·Ö×ÓÖÊÁ¿286£©À´ÅäÖÆ0.1mol/LµÄ̼ËáÄÆÈÜÒº980mL£¬¼ÙÈçÆäËû²Ù×÷¾ù׼ȷÎÞÎó£¬ÏÂÁÐÇé¿ö»áÒýÆðÅäÖÆÈÜÒºµÄŨ¶ÈÆ«¸ßµÄÊÇ

A. ³Æȡ̼ËáÄƾ§Ìå28.6g

B. Èܽâʱ½øÐмÓÈÈ£¬²¢½«ÈÈÈÜҺתÒƵ½ÈÝÁ¿Æ¿ÖÐÖÁ¿Ì¶ÈÏß

C. תÒÆʱ£¬¶ÔÓÃÓÚÈܽâ̼ËáÄƾ§ÌåµÄÉÕ±­Ã»ÓнøÐÐÏ´µÓ

D. ¶¨Èݺ󣬽«ÈÝÁ¿Æ¿Õñµ´Ò¡ÔÈ£¬¾²Ö÷¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÓÖ¼ÓÈëÉÙÁ¿Ë®ÖÁ¿Ì¶ÈÏß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Í¨¹ýÒÔÏ·´Ó¦¿É»ñµÃÐÂÐÍÄÜÔ´¶þ¼×ÃÑ()¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

¢Ù

¢Ú

¢Û

¢Ü

A. ·´Ó¦¢Ù¡¢¢ÚΪ·´Ó¦¢ÛÌṩԭÁÏÆø

B. ·´Ó¦¢ÛÒ²ÊÇ×ÊÔ´»¯ÀûÓõķ½·¨Ö®Ò»

C. ·´Ó¦µÄ

D. ·´Ó¦µÄ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸