¡¾ÌâÄ¿¡¿Ä³Ð£»¯Ñ§ÐËȤС×éͬѧ²ÂÏë×ÔÀ´Ë®ÖпÉÄܺ¬ÓдóÁ¿Ca2£«¡¢Mg2£«ºÍijЩÒõÀë×Ó£¬´Ó¶ø½øÐÐÁËÈý×éʵÑ飺

¢ÙÈ¡ÊÊÁ¿×ÔÀ´Ë®ÓÚÊÔ¹ÜÖУ¬µÎ¼Ó×ãÁ¿µÄNaOHÈÜÒº£¬²úÉú°×É«³Áµí£»

¢Ú¹ýÂ˺óÈ¡ÂËÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó×ãÁ¿µÄNa2CO3ÈÜÒº£¬ÓÖÓа×É«³ÁµíÉú³É£»

¢ÛÁíÈ¡ÊÊÁ¿×ÔÀ´Ë®ÓÚÊÔ¹ÜÖУ¬µÎ¼Ó×ãÁ¿Ï¡ÏõËáºóÔٵμÓAgNO3ÈÜÒº£¬Ò²²úÉú°×É«³Áµí¡£

Çë»Ø´ðÒÔÏÂÎÊÌ⣺

(1) ͨ¹ýʵÑé¿É³õ²½È·¶¨×ÔÀ´Ë®ÖÐ________(Ìî¡°º¬ÓС±»ò¡°²»º¬ÓС±)´óÁ¿Ca2£«¡¢Mg2£«¡£

(2) ×ÔÀ´Ë®ÖÐËùº¬ÒõÀë×Ó¿ÉÒÔÈ·¶¨ÓÐ___________£¬¸ÃÀë×ÓÓöAgNO3ÈÜÒº·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ________

¡¾´ð°¸¡¿º¬ÓÐ Cl-(»òÂÈÀë×Ó)

¡¾½âÎö¡¿

£¨1£©¸ù¾Ý¢Ù¡¢¢ÚµÄÏÖÏó£¬Éú³ÉµÄ°×É«³ÁµíÊÇ̼Ëá¸ÆºÍÇâÑõ»¯Ã¾£¬³õ²½È·¶¨×ÔÀ´Ë®Öк¬ÓÐCa2+¡¢Mg2+ÑôÀë×Ó£»·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ£ºCO32-+Ca2+=CaCO3¡ý¡¢Mg2++2OH-=Mg(OH)2¡ý£¬¹Ê´ð°¸Îªº¬ÓУ»

£¨2£©Cl-+Ag+=AgCl¡ý£¬°×É«³Áµí²»ÈÜÓÚHNO3£¬µÎ¼Ó×ãÁ¿Ï¡ÏõËáºóÔٵμÓAgNO3ÈÜÒº£¬Ò²²úÉú°×É«³Áµí£¬ËµÃ÷×ÔÀ´Ë®Öк¬ÓÐCl-¡£¹Ê´ð°¸ÎªCl-(»òÂÈÀë×Ó)£»Cl-+Ag+=AgCl¡ý¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿1£¬2-¶þäåÒÒÍé¿É×÷¿¹±¬¼ÁµÄÌí¼Ó¼Á¡£ÈçͼΪʵÑéÊÒÖƱ¸1£¬2-¶þäåÒÒÍéµÄ×°ÁDͼ£¬ ͼÖзÖÒºÖƶ·ºÍÉÕÆ¿aÖзֱð×°ÓÐŨH2SO4ºÍÎÞË®ÒÒ´¼£¬d×°ÁDÊÔ¹ÜÖÐ×°ÓÐÒºäå¡£

¼ºÖª£ºCH3CH2OHCH2=CH2¡ü+H2O£»2CH3CH2OHCH3CH2OCH2CH3+H2O

Ïà¹ØÊý¾ÝÁбíÈçÏ£º

ÒÒ´¼

1£¬2-¶þäåÒÒÍé

ÒÒÃÑ

äå

״̬

ÎÞÉ«ÒºÌå

ÎÞÉ«ÒºÌå

ÎÞÉ«ÒºÌå

ºì×ØÉ«ÒºÌå

ÃܶÈ/g¡¤cm-3

0.79

2.18

0.71

3.10

·Ðµã/¡æ

78.5

131.4

34.6

58.8

ÈÛµã/¡æ

-114.3

9.79

- 116.2

-7.2

Ë®ÈÜÐÔ

»ìÈÜ

ÄÑÈÜ

΢ÈÜ

¿ÉÈÜ

£¨1£©ÊµÑéÖÐӦѸËÙ½«Î¶ÈÉý…lµ½170¡æ×óÓÒµÄÔ­ÒòÊÇ______________________________¡£

£¨2£©°²È«Æ¿bÔÚʵÑéÖÐÓжàÖØ×÷Óá£ÆäÒ»¿ÉÒÔ¼ì²éʵÑé½øÐÐÖÐd×°ÁDÖе¼¹ÜÊÇ·ñ·¢Éú¶ÂÈû£¬

Çëд³ö·¢Éú¶ÂÈûʱƿbÖеÄÏÖÏ󣺢Ù_______________________________£»Èç¹ûʵÑéʱd×°ÁDÖе¼¹Ü¶ÂÈû£¬ÄãÈÏΪ¿ÉÄܵÄÔ­ÒòÊÇ¢Ú_______________________________________________£»°²È«Æ¿b»¹¿ÉÒÔÆðµ½µÄ×÷ÓÃÊÇ¢Û__________________¡£

£¨3£©ÈÝÆ÷c¡¢eÖж¼Ê¢ÓÐNaOHÈÜÒº£¬cÖÐNaOHÈÜÒºµÄ×÷ÓÃÊÇ_______________________________¡£

£¨4£©Ä³Ñ§ÉúÔÚ×ö´ËʵÑéʱ£¬Ê¹ÓÃÒ»¶¨Á¿µÄÒºä壬µ±äåÈ«²¿ÍÊɫʱ£¬ËùÏûºÄÒÒ´¼ºÍŨÁòËá»ìºÏÒºµÄÁ¿£¬±ÈÕýÈ·Çé¿öϳ¬¹ýÐí¶à£¬Èç¹û×°ÁDµÄÆøÃÜÐÔûÓÐÎÊÌ⣬ÊÔ·ÖÎö¿ÉÄܵÄÔ­Òò£º______________¡¢______________£¨Ð´³öÁ½Ìõ¼´¿É£©¡£

£¨5£©³ýÈ¥²úÎïÖÐÉÙÁ¿Î´·´Ó¦µÄBr2ºó£¬»¹º¬ÓеÄÖ÷ÒªÔÓÖÊΪ___________£¬Òª½øÒ»²½Ìá´¿£¬ÏÂÁвÙ×÷ÖбØÐèµÄÊÇ_____________ £¨Ìî×Öĸ£©¡£

A£®Öؽᾧ B£®¹ýÂË C£®ÝÍÈ¡ D£®ÕôÁó

£¨6£©ÊµÑéÖÐÒ²¿ÉÒÔ³·È¥d×°ÁDÖÐÊ¢±ùË®µÄÉÕ±­£¬¸ÄΪ½«Àäˮֱ½Ó¼ÓÈëµ½d×°ÁDµÄÊÔ¹ÜÖУ¬Ôò ´ËʱÀäË®³ýÁËÄÜÆðµ½ÀäÈ´1£¬2-¶þäåÒÒÍéµÄ×÷ÓÃÍ⣬»¹¿ÉÒÔÆðµ½µÄ×÷ÓÃÊÇ____________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑ֪ǦÐîµç³Ø·Åµçʱµç³Ø·´Ó¦ÎªPbO2£«Pb£«4H£«£«=2PbSO4£«2H2O¡£ÈçͼÊÇǦÐîµç³ØµÄ¹¤×÷Ô­ÀíʾÒâͼ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A.KÓëNÁ¬½Óʱ£¬¸Ã×°ÖÃÖеçÄÜת»¯Îª»¯Ñ§ÄÜ

B.KÓëNÁ¬½Óʱ£¬H+Ïò¸º¼«Òƶ¯

C.KÓëMÁ¬½Óʱ£¬aΪµçÔ´µÄ¸º¼«

D.KÓëMÁ¬½Óʱ£¬Ñô¼«¸½½üÈÜÒºµÄpHÖð½¥Ôö´ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³ÏõË᳧´¦ÀíβÆøÖÐNOµÄ·½·¨ÊÇÔÚ´ß»¯¼Á´æÔÚÏ£¬ÓÃH2½«NO»¹Ô­ÎªN2£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪNO(g)£«H2(g)=N2(g)£«H2O(g) ¦¤H£½mkJ¡¤mol£­1£¬ÆäÄÜÁ¿±ä»¯¹ý³ÌÈçͼ£º

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©

A.¹ý³Ì¢Ù¡¢¢Ú¡¢¢Û¡¢¢Ü¶¼ÊÇ·ÅÈȹý³ÌB.m£½£«(a£«b£­c£­d)kJ¡¤mol£­1

C.m£½£­(c£«a£­d£­b)kJ¡¤mol£­1D.m£½£«(c£«d£­a£­b) kJ¡¤mol£­1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂͼÊÇʵÑéÊÒÖÆÂÈÆøµÄ×°Öá£

Ô²µ×ÉÕÆ¿Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________________£¬Éú³É1¸öCl2·Ö×ÓתÒƵç×ÓÊýĿΪ________¸ö£¬±¥ºÍʳÑÎË®µÄ×÷ÓÃÊÇ_____________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓÉÒ»ÖÖÑôÀë×ÓÓëÁ½ÖÖËá¸ùÒõÀë×Ó×é³ÉµÄÑγÆΪ»ìÑΡ£Ïò»ìÑÎNa4S2O3ÖмÓÈë×ãÁ¿Ï¡ÁòËᣬ·¢Éú·´Ó¦£º¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©

A. Na4S2O3µÄË®ÈÜÒºÏÔ¼îÐÔ

B.1mol Na4S2O3Öй²º¬Àë×ÓÊýΪ5NA

C.ÉÏÊö·´Ó¦ÖУ¬Ã¿²úÉú3molS£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿Îª6mol

D.CaOCl2Ò²¿É³ÆΪ»ìÑΣ¬ÏòCaOCl2ÖмÓÈë×ãÁ¿Ï¡ÁòËá»áÓÐCl2²úÉú

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¡°µÍ̼¾­¼Ã¡±Õý³ÉΪ¿Æѧ¼ÒÑо¿µÄÖ÷Òª¿ÎÌâ¡£½«È¼Ãº·ÏÆøÖеÄCO2ת»¯Îª¶þ¼×Ãѵķ´Ó¦Ô­ÀíΪ:

2CO2(g) + 6H2(g) CH3OCH3(g) + 3H2O(g)

(1) ÒÑÖªÒ»¶¨Ìõ¼þÏ£¬¸Ã·´Ó¦ÖÐCO2µÄƽºâת»¯ÂÊËæζȡ¢Í¶ÁϱÈ[n(H2)/n(CO2)]µÄ±ä»¯ÇúÏßÈçÏÂͼ:

¢Ùa¡¢3¡¢bµÄ´óС¹Øϵ___________

¢ÚÇë¸ù¾ÝÏÂͼÖÐÐÅÏ¢ÔÚÏÂͼ(ÓÒ)Öл­³öCO2(g)ºÍH2(g)ת»¯ÎªCH3OCH3(g)ºÍH2O(g)µÄÄÜÁ¿¹ØϵÇúÏß___________¡£

(2)ijζÈÏ£¬½«2.0molCO2(g)ºÍ6.0molH2(g)³äÈëÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬ÉÏÊö·´Ó¦µ½´ïƽºâʱ£¬¸Ä±äѹǿºÍζȣ¬Æ½ºâÌåϵÖÐCH3OCH3(g)µÄÎïÖʵÄÁ¿·ÖÊý±ä»¯Çé¿öÈçͼËùʾ£¬¹ØÓÚζȺÍѹǿµÄ¹ØϵÅжÏÕýÈ·µÄ______

A. P3£¾P2£¬T3£¾T2 B. P1£¾P3£¬T1£¾T3 C. P2£¾P4£¬T4£¾T2 D. P1£¾P4£¬T2£¾T3

(3)ÔÚºãÈÝÃܱÕÈÝÆ÷Àï°´Ìå»ý±ÈΪ1:3³äÈëCO2(g)ºÍH2(g),Ò»¶¨Ìõ¼þÏÂÉÏÊö·´Ó¦´ïµ½Æ½ºâ״̬¡£µ±¸Ä±ä·´Ó¦µÄijһ¸öÌõ¼þºó£¬ÏÂÁб仯ÄÜ˵Ã÷ƽºâÒ»¶¨ÏòÄæ·´Ó¦·½ÏòÒƶ¯µÄÊÇ___________

A. Õý·´Ó¦ËÙÂÊÏÈÔö´óºó¼õС B. Äæ·´Ó¦ËÙÂÊÏÈÔö´óºó¼õС

C. »¯Ñ§Æ½ºâ³£ÊýKÖµÔö´ó D. ·´Ó¦ÎïµÄŨ¶ÈÔö´ó

(4)ÏÂÁÐÒ»¶¨ÄÜ˵Ã÷ÉÏÊö·´Ó¦´ïµ½Æ½ºâµÄÊÇ___________

A.ÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿²»±ä

B.ºãÈÝÌõ¼þÏ£¬ÆøÌåµÄÃܶȲ»±ä

C.¸÷ÎïÖʵÄËÙÂÊÖ®±ÈµÈÓÚϵÊý±È

D.[n(H2)/n(CO2)]²»±ä

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³¼×ÍéȼÁϵç³Ø¹¹ÔìʾÒâͼÈçÏ£¬¹ØÓڸõç³ØµÄ˵·¨²»ÕýÈ·µÄÊÇ

A. µç½âÖÊÈÜÒºÖÐNa+Ïòb¼«Òƶ¯

B. b¼«µÄµç¼«·´Ó¦ÊÇ£ºO2+2H2O+4e-=4OH-

C. a¼«ÊǸº¼«£¬·¢ÉúÑõ»¯·´Ó¦

D. µç×Óͨ¹ýÍâµç·´Óbµç¼«Á÷Ïòaµç¼«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨1£©ÈçͼËùʾ£¬½«ÂÈÆøÒÀ´Îͨ¹ýÊ¢ÓиÉÔïÓÐÉ«²¼ÌõµÄ¹ã¿ÚÆ¿ºÍÊ¢ÓÐʪÈóÓÐÉ«²¼ÌõµÄ¹ã¿ÚÆ¿£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ_______¡£

£¨2£©Îª·ÀÖ¹ÂÈÆøβÆøÎÛȾ¿ÕÆø£¬ÊµÑéÊÒͨ³£ÓÃ________ÈÜÒºÎüÊÕ¶àÓàµÄÂÈÆø£¬Ô­ÀíÊÇ______________________(Óû¯Ñ§·½³Ìʽ±íʾ)¡£¹¤ÒµÉϳ£ÓÃÁ®¼ÛµÄʯ»ÒÈéÎüÊÕ¹¤ÒµÂÈÆøβÆøÖƵÃƯ°×·Û£¬Æ¯°×·ÛµÄÓÐЧ³É·ÖÊÇ________________(Ìѧʽ)£¬Æ¯°×·ÛÈÜÓÚË®ºó£¬ÊÜ¿ÕÆøÖеÄCO2×÷Ó㬲úÉúÓÐƯ°×¡¢É±¾ú×÷ÓõĴÎÂÈËᣬ»¯Ñ§·½³ÌʽΪ_____________________£¬³¤ÆÚ¶ÖÃÓÚ¿ÕÆøÖеÄƯ°×·Û£¬¼ÓÏ¡ÑÎËáºó²úÉúµÄÆøÌåÊÇ________(Ìî×Öĸ£¬ÏÂͬ)¡£

A£®O2B£®Cl2 C£®CO2D£®HClO

£¨3£©Ò»µ©·¢ÉúÂÈÆøй©ºÍ±¬Õ¨Ê¹ʣ¬ÖÜΧȺÖÚÓ¦½ô¼±ÊèÉ¢¡£µ±ÈËÃÇÌÓÀ뱬ըÏÖ³¡Ê±£¬¿ÉÒÔÓýþÓÐÒ»¶¨Å¨¶ÈijÎïÖÊË®ÈÜÒºµÄë½íÎæס±Ç×Ó£¬×îÊÊÒ˲ÉÓõĸÃÎïÖÊÊÇ________¡£

A£®NaOH B£®NaCl

C£®KCl D£®Na2CO3

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸