¡¾ÌâÄ¿¡¿Ä³Ð£»¯Ñ§Ð¡×éµÄͬѧ¿ªÕ¹ÁËһϵÁеĻ¯Ñ§ÊµÑé»î¶¯¡£

ÇëÄã²ÎÓëʵÑé²¢Íê³ÉÏÂÁÐÎÊÌâ:

(1)¼×ͬѧÓÃͼ1ËùʾװÖÃ,²âÁ¿Ð¿ÓëÁòËá·´Ó¦ËÙÂÊ:°Ñ2 gпÁ£·ÅÈë׶ÐÎÆ¿ÄÚ,ͨ¹ý·ÖҺ©¶·¼ÓÈë1 mol¡¤L-1Ï¡ÁòËá40 mL,ÊÕ¼¯10 mLÆøÌå,ͨ¹ý¼Ç¼¡¡¡¡¡¡¡¡¡¡¡¡µÃµ½·´Ó¦ËÙÂÊΪx mol¡¤(L¡¤min)-1¡£ÊµÑ鿪ʼʱ¼ì²é¸Ã×°ÖÃÆøÃÜÐԵķ½·¨ÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

(2)ÒÒͬѧÓÃͼ2×°ÖòⶨNa2CO3ºÍNaClµÄ¹ÌÌå»ìºÏÎïÖÐNa2CO3µÄÖÊÁ¿·ÖÊý:

¢Ù¼×¡¢ÒÒÁ½ÊԹܸ÷ÓÐÁ½¸ùµ¼¹Ü,ÓÃÏ𽺹ÜÁ¬½Ó¶ÔÓ¦½Ó¿Úºó,µ¹ÖÃʢϡÁòËáµÄÊÔ¹Ü,·¢Éú·´Ó¦,·Å³öÆøÌå,Ôò¹ÌÌå»ìºÏÎïºÍÏ¡ÁòËáÓ¦·Ö±ðÖÃÓÚ¡¡¡¡¡¡¡¢¡¡¡¡ÒÇÆ÷ÖÐ(ÌîÒÇÆ÷±àºÅ);

¢ÚG¹Ü¿ÉÒÔÓû¯Ñ§ÊµÑéÊÒÀïµÄÒ»ÖÖ³£ÓÃÒÇÆ÷À´Ìæ´ú,ÕâÖÖÒÇÆ÷ÊÇ¡¡¡¡¡¡;

¢ÛÒÇÆ÷¼×¡¢ÒÒ½Ó¿ÚµÄÁ¬½Ó·½Ê½ÈçÏÂ:AÁ¬½Ó¡¡¡¡,BÁ¬½Ó¡¡¡¡¡¡,CÁ¬½Ó¡¡¡¡¡¡(Ìîд¸÷½Ó¿ÚµÄ±àºÅ);

¢ÜΪÌá¸ß²âÁ¿µÄ׼ȷÐÔ,ÊÕ¼¯ÍêÆøÌåºó,±û×°ÖöÁÊýÇ°Ó¦½øÐеIJÙ×÷ÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

(3)±ûͬѧÓë¼×ͬѧʵÑéÄ¿µÄÏàͬ:ÓÃͼ3×°ÖòⶨÉú³ÉµÄCO2µÄÖÊÁ¿,µ«¸Ã×°ÖôæÔÚÃ÷ÏÔȱÏÝ,´Ó¶øµ¼ÖÂʵÑéÎó²î,ÇëÄã·ÖÎöÆäÖÐʹ²â¶¨½á¹û¿ÉÄÜÆ«´óµÄÖ÷ÒªÔ­Òò¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

¡¾´ð°¸¡¿(1)·´Ó¦Ê±¼ä ¹Ø±ÕA´¦»îÈû,½«×¢ÉäÆ÷»îÈûÀ­³öÒ»¶¨¾àÀë,Ò»¶Îʱ¼äºóËÉ¿ª»îÈû,Èô»îÈûÄܻص½Ô­Î»,Ö¤Ã÷²»Â©Æø,·ñÔò©Æø

(2)¢Ù¼×¡¡ÒÒ¡¡¢Ú¼îʽµÎ¶¨¹Ü¡¡¢ÛD¡¡E¡¡F¡¡¢ÜÉÏÏÂÒƶ¯µÎ¶¨¹Ü,ʹ×óÓÒÁ½±ßÒºÃæÏàƽ¡¡Æ«´ó

(3)CO2ÆøÌåÖлìÓÐË®ÕôÆø»ò¿ÕÆøÖеÄCO2ºÍË®ÕôÆø½øÈë¸ÉÔï¹ÜÖÐ

¡¾½âÎö¡¿(2)ΪÁË˳Àû°ÑÏ¡ÁòËáµ¹ÈëÁíÒ»ÊÔ¹Ü,Ê×ÏÈÓ¦±£³ÖÁ½ÊÔ¹ÜÖÐÆøѹƽºâ,¼×ÓÐÁ½¸öÆøÌå½ø³ö¿Ú,¶øÒÒÖ»ÓÐÒ»¸ö,ËùÒÔ¼×ÊÇÆøÌåµÄ·¢Éú×°ÖÃ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐʵÑéµÄ·´Ó¦Ô­ÀíÓÃÀë×Ó·½³Ìʽ±íʾÕýÈ·µÄÊÇ(¡¡¡¡)

A. ÊÒÎÂÏ£¬²âµÃÂÈ»¯ï§ÈÜÒºpH<7£¬Ö¤Ã÷һˮºÏ°±ÊÇÈõ¼î£ºNH4+£«2H2O===NH3¡¤H2O£«H3O£«

B. ÓÃÇâÑõ»¯ÄÆÈÜÒº³ýȥþ·ÛÖеÄÔÓÖÊÂÁ£º2Al£«2OH£­£«2H2O===2AlO2-£«3H2¡ü

C. ÓÃ̼ËáÇâÄÆÈÜÒº¼ìÑéË®ÑîËáÖеÄôÈ»ù£º£«2HCO3-¡ú£«2H2O£«2CO2¡ü

D. ÓøßÃÌËá¼Ø±ê×¼ÈÜÒºµÎ¶¨²ÝË᣺2MnO4-£«16H£«£«5C2O4-=2Mn2£«£«10CO2¡ü£«8H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿N2O5ÊÇÒ»ÖÖÐÂÐÍÏõ»¯¼Á£¬ÆäÐÔÖʺÍÖƱ¸Êܵ½ÈËÃǵĹØ×¢¡£Ò»¶¨Î¶ÈÏ£¬ÔÚ2 L¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦£º2N2O5(g)4NO2(g)+O2(g) ¦¤H>0¡£·´Ó¦ÎïºÍ²¿·ÖÉú³ÉÎïµÄÎïÖʵÄÁ¿Ë淴Ӧʱ¼ä±ä»¯µÄÇúÏßÈçͼËùʾ¡£ÏÂÁÐÕýÈ·µÄÊÇ£º

A. 0-20 sÄÚƽ¾ù·´Ó¦ËÙÂÊv(N2O5)=0.1 mol¡¤L-1¡¤s-1

B. 10 sʱ£¬Õý¡¢Äæ·´Ó¦ËÙÂÊÏàµÈ£¬´ïµ½Æ½ºâ

C. 20 sʱ£¬Õý·´Ó¦ËÙÂÊ´óÓÚÄæ·´Ó¦ËÙÂÊ

D. ÇúÏßa±íʾNO2µÄÎïÖʵÄÁ¿Ë淴Ӧʱ¼äµÄ±ä»¯

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿²¿·Ö±»Ñõ»¯µÄFe-CuºÏ½ðÑùÆ·£¨Ñõ»¯²úÎïΪFe2O3¡¢CuO£©¹²8.0 g£¬¾­ÈçÏ´¦Àí£º

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®ÂËÒºAÖеÄÑôÀë×ÓΪFe2+¡¢Fe3+¡¢H+

B£®¸ÃÑùÆ·ÖÐCu¡¢OÔªËصÄÖÊÁ¿±ÈΪ10£ºl

C£®V=448

D£®ÈܽâÑùƷʱÏûºÄH2SO4µÄÎïÖʵÄÁ¿Îª0.04 mo1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖªµÄÈÜÒºÖеÄŨ¶ÈΪ£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÏÂÁÐÎïÖÊÄܵ¼µçµÄÊÇ__________£¬ÊôÓÚµç½âÖʵÄÊÇ__________£¬ÊôÓڷǵç½âÖʵÄÊÇ__________£¬£¨ÌîÐòºÅ£©

¢ÙÂÈ»¯Äƾ§Ìå ¢Ú°±Æø ¢ÛË®Òø ¢ÜÕáÌÇ ¢Ýʯī ¢ÞÈÛÈÚ ¢ß¿ÕÆø ¢àBaSO4 ¢áÏ¡ÑÎËá

£¨2£©Ð´³öÔÚË®ÈܽâÖеĵçÀë·½³Ìʽ£º_________________£»Ð´³öNaHSO4ÔÚÈÛÈÚ״̬ϵĵçÀë·½³Ìʽ£º_______________________________¡£

£¨3£©ÈÜÒºÓëNaHCO3ÈÜÒº»ìºÏÓÐÆøÅÝÉú³É£¬·´Ó¦µÄÀë×Ó·½³Ìʽ_______________

£¨4£©¢ÙÈôÓëÈÜÒº»ìºÏºóÈÜÒºÏÔÖÐÐÔ£¬Çëд³ö·´Ó¦µÄÀë×Ó·½³Ìʽ_________________________________________________________________¡£

¢ÚÈôÓëÈÜÒºÖлìºÏÈÜÒº³Ê¼îÐÔ£¬Çëд³ö·´Ó¦µÄÀë×Ó·½³Ìʽ___________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨1£©£«6¼Û¸õµÄ»¯ºÏÎﶾÐԽϴ󣬳£ÓÃNaHSO3½«·ÏÒºÖеÄCr2O»¹Ô­³ÉCr3+£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________________________________¡£

£¨2£©ÏòÊ¢ÓÐH2O2ÈÜÒºµÄÊÔ¹ÜÖмÓÈ뼸µÎËữµÄFeCl2ÈÜÒº£¬ÈÜÒº±ä³É×Ø»ÆÉ«£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________________¡£

£¨3£© ¡°Ëá½þ¡±Ê±V2O5ת»¯ÎªVO£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ________________________________________¡£

£¨4£©ÔÚH2SO4µÄËáÐÔ»·¾³ÖÐClO2Óëµâ»¯¼Ø·´Ó¦µÄÀë×Ó·½³Ìʽ___________________________¡£

£¨5£©ÒÑÖªÔÚËáÐÔÌõ¼þÏÂNaClO2¿É·¢Éú·´Ó¦Éú³ÉNaCl²¢ÊͷųöClO2£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿»¯Ñ§ÊÇÒ»ÃÅÒÔʵÑéΪ»ù´¡µÄ×ÔÈ»¿Æѧ£¬»¯Ñ§ÊµÑéÔÚ»¯Ñ§Ñ§Ï°ÖоßÓм«ÆäÖØÒªµÄ×÷ÓÃ.

(1)ÏÂÁÐÓйØʵÑéµÄÐðÊö£¬ÕýÈ·µÄÊÇ_______.

(A)ÅäÖÆ500mLijÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº£¬¿ÉÓÃÁ½Ö»250mLµÄÈÝÁ¿Æ¿

(B)ÓÃÉøÎö·¨·ÖÀëµí·ÛÖлìÓеÄNaClÔÓÖÊ

(C)ÎÞ·¨Ó÷ÖҺ©¶·½«¸ÊÓͺÍË®µÄ»ìºÏÒºÌå·ÖÀë

(D)ÓÃËáʽµÎ¶¨¹ÜÁ¿È¡20. 00mL¸ßÃÌËá¼ØÈÜÒº

(E)ΪÁ˲ⶨijÈÜÒºµÄpH£¬½«¾­Ë®ÈóʪµÄpHÊÔÖ½½þÈëµ½´ý²âÈÜÒº£¬¹ýÒ»»áÈ¡³ö£¬Óë±ê×¼±ÈÉ«¿¨½øÐжԱÈ

(F)ÓÃŨ°±Ë®Ï´µÓ×ö¹ýÒø¾µ·´Ó¦µÄÊÔ¹Ü

(G)ÅäÖÆÒø°±ÈÜҺʱ£¬½«Ï¡°±Ë®ÂýÂýµÎ¼Óµ½ÏõËáÒøÈÜÒºÖУ¬²úÉú³Áµíºó¼ÌÐøµÎ¼Óµ½³Áµí¸ÕºÃÈܽâΪֹ

(H)ÅäÖÆÒ»¶¨Å¨¶ÈµÄÈÜҺʱ£¬Èô¶¨ÈÝʱ²»Ð¡ÐļÓË®³¬¹ýÈÝÁ¿Æ¿µÄ¿Ì¶ÈÏߣ¬Ó¦Á¢¼´ÓõιÜÎüÈ¥¶àÓàµÄ²¿·Ö¡£

(J)ÔÚÇâÑõ»¯Ìú½ºÌåÖмÓÉÙÁ¿ÁòËá»á²úÉú³Áµí

(K)Óýᾧ·¨¿ÉÒÔ³ýÈ¥ÏõËá¼ØÖлìÓеÄÉÙÁ¿ÂÈ»¯ÄÆ

(2)ÈçͼΪ³£¼ûÒÇÆ÷µÄ²¿·Ö½á¹¹(ÓеÄÒÇÆ÷±»·Å´ó)£º

AͼÖÐÒºÃæËùʾÈÜÒºµÄÌå»ýΪ__________mL£¬ÓÃÉÏÊöËÄÖÖÒÇÆ÷ÖеÄijÖÖ²âÁ¿Ò»ÒºÌåµÄÌå»ý£¬Æ½ÊÓʱ¶ÁÊýΪNmL£¬ÑöÊÓʱ¶ÁÊýΪMmL£¬ÈôM>N£¬ÔòËùʹÓõÄÒÇÆ÷ÊÇ___________(Ìî×Öĸ±êºÅ)¡£

(3)ÏÂͼΪʵÑéÊÒijŨÑÎËáÊÔ¼ÁÆ¿ÉϵıêÇ©£¬ÊÔ¸ù¾ÝÓйØÊý¾Ý»Ø´ðÏÂÁÐÎÊÌ⣺

ÑÎËá

·Ö×Óʽ£ºHCl

Ïà¶Ô·Ö×ÓÁ¿: 36.5

Ãܶȣº1.2g/ml

HClÖÊÁ¿·ÖÊý£º36.5%

¢Ù¸ÃŨÑÎËáÖÐHClµÄÎïÖʵÄÁ¿Å¨¶ÈΪ_________mol/L¡£

¢ÚijѧÉúÓûÓÃÉÏÊöŨÑÎËáºÍÕôÁóË®ÅäÖÆ500 mLÎïÖʵÄÁ¿Å¨¶ÈΪ0.3 mol/LÏ¡ÑÎËá¡£

¢ñ.¸ÃѧÉúÐèÒªÁ¿È¡________mLÉÏÊöŨÑÎËá½øÐÐÅäÖÆ¡£

II.ÔÚÅäÖƹý³ÌÖУ¬ÏÂÁÐʵÑé²Ù×÷»áʹËùÅäÖƵÄÏ¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈÆ«¸ßµÄÊÇ______¡£

A.ÓÃÁ¿Í²Á¿È¡Å¨ÑÎËáʱ¸©Êӹ۲찼ҺÃæ

B.ÈÜҺעÈëÈÝÁ¿Æ¿Ç°Ã»Óлָ´µ½ÊÒξͽøÐж¨ÈÝ

C.¶¨ÈÝʱÑöÊӿ̶ÈÏß

D.ÔÚÅäÁDÇ°ÓÃÒÑ֪Ũ¶ÈµÄÏ¡ÑÎËáÈóÏ´ÈÝÁ¿Æ¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿CË׳ƿàÐÓÈÊËᣬÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬FÊÇ¿¹ÒÖÓôÒ©ÎïÃ×µªÆ½µÄÒ»ÖÖÖØÒªµÄÖмäÌ壬ËüÃÇ¿ÉÓÉÏÂÁзÏߺϳɣº

ÒÑÖª£º(R£¬R£¬´ú±íÇâÔ­×Ó»òÇè»ù)

(1)AµÄÃû³ÆÊÇ_________£»C¡úDµÄ·´Ó¦ÀàÐÍÊÇ________________¡£

(2)CÖеĺ¬Ñõ¹ÙÄÜÍÅÃû³ÆÊÇ_______________¡£

(3)ÏÂÁйØÓÚC¡¢EµÄ˵·¨ÕýÈ·µÄÊÇ__________¡£

a.¶¼ÊôÓÚ·¼Ïã×廯ºÏÎï b.¶¼ÄÜ·¢ÉúÈ¡´ú·´Ó¦

c.Ëùº¬º¬Ñõ¹ÙÄÜÍÅÍêÈ«Ïàͬ d.ÔÚÇâÑõ»¯ÄÆÈÜÒºÖж¼ÄÜ·¢ÉúË®½â·´Ó¦

(4)д³öEÔÚÇâÑõ»¯ÄÆÈÜÒºÖмÓÈÈË®½âµÄ»¯Ñ§·½³Ìʽ_____________________________¡£

(5)д³öÓÉCÉú³É¸ß¾ÛÎïPµÄ»¯Ñ§·½³Ìʽ_____________________¡£

(6)ͬʱÂú×ãÏÂÁÐÌõ¼þµÄCµÄͬ·ÖÒì¹¹Ìå¹²ÓÐ_______ÖÖ¡£

¢ÙÓëŨäåË®×÷ÓÃÓа×É«³ÁµíÉú³É£»

¢Ú¼ÈÄÜ·¢ÉúË®½â·´Ó¦Ò²ÄÜ·¢ÉúÒø¾µ·´Ó¦£¬

ÕâЩͬ·ÖÒì¹¹ÌåÖУ¬ÊôÓÚ¶þÈ¡´ú±½ÇҺ˴Ź²ÕñÇâÆ×ÏÔʾ5¸öÎüÊÕ·åµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽÊÇ_________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÂÈ»¯ÑÇÍ­£¨CuCl)ÊÇ°×É«·ÛÄ©£¬Î¢ÈÜÓÚË®£¬ÄÜÓëÏ¡ÁòËá·´Ó¦£¬¹ã·ºÓ¦ÓÃÓÚ»¯¹¤ºÍӡȾµÈÐÐÒµ¡£Ä³Ñо¿ÐÔѧϰС×éÄâÈÈ·Ö½âCuC122H2OÖƱ¸CuCl,²¢½øÐÐÏà¹Ø̽¾¿£¨ÒÑÖªËáÐÔÌõ¼þÏÂCu+²»Îȶ¨£©¡£

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. XÆøÌå¿ÉÒÔÊÇN2£¬Ä¿µÄÊÇ×ö±£»¤Æø£¬ÒÖÖÆCuCl22H2O¼ÓÈȹý³Ì¿ÉÄܵÄË®½â

B. CuClÓëÏ¡ÁòËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2Cu++4H++SO42-¨T2Cu2++SO2¡ü+2H2O

C. ;¾¶1ÖвúÉúµÄCl2¿ÉÒÔ»ØÊÕÑ­»·ÀûÓã¬Ò²¿ÉÒÔͨÈë±¥ºÍCaCl2ÈÜÒºÖгýÈ¥

D. ;¾¶2ÖÐ200¡æÏ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCu2(OH)2Cl22CuO + 2HCl

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸