ÌâÄ¿ÄÚÈÝ


Ô­×ÓÐòÊýÓÉСµ½´óÅÅÁеÄËÄÖÖ¶ÌÖÜÆÚÔªËØX¡¢Y¡¢Z¡¢W£¬ËÄÖÖÔªËØµÄÔ­×ÓÐòÊýÖ®ºÍΪ32£¬ÔÚÖÜÆÚ±íÖÐXÊÇÔ­×Ó°ë¾¶×îСµÄÔªËØ£¬Y¡¢Z×óÓÒÏàÁÚ£¬Z¡¢WλÓÚͬÖ÷×å¡£MÓëXͬÖ÷×壬ÓëWͬÖÜÆÚ¡£

(1)MÔªËØÊÇ________(ÌîÔªËØ·ûºÅ)¡£

(2)Z¡¢WÐÎ³ÉµÄÆøÌ¬Ç⻯ÎïµÄÎȶ¨ÐÔΪ______>________________________________________________________________________

(Ìѧʽ)¡£

(3)M2Z2µÄµç×ÓʽΪ________________£¬Ð´³öM2Z2ÓëË®·´Ó¦µÄÀë×Ó·½³Ìʽ£º________________________¡£

(4)ÓÉX¡¢Y¡¢Z¡¢WËÄÖÖÔªËØÖеÄÈýÖÖ×é³ÉµÄ¡ªÖÖÇ¿Ëᣬ¸ÃÇ¿ËáµÄÏ¡ÈÜÒºÄÜÓëÍ­·´Ó¦£¬Àë×Ó·½³ÌʽΪ__________________________________________¡£

(5)ÓÉX¡¢Y¡¢Z¡¢WËÄÖÖÔªËØ×é³ÉµÄÒ»ÖÖÀë×Ó»¯ºÏÎïA£¬ÒÑÖª£º

¢Ù1 mol AÄÜÓë×ãÁ¿NaOHŨÈÜÒº·´Ó¦Éú³É±ê×¼×´¿öÏÂ44.8 LÆøÌ壻

¢ÚAÄÜÓëÑÎËá·´Ó¦²úÉúÆøÌåB£¬¸ÃÆøÌåÄÜÓëÂÈË®·´Ó¦¡£

ÔòAÊÇ____________(Ìѧʽ)£»Ð´³ö¸ÃÆøÌåBÓëÂÈË®·´Ó¦µÄÀë×Ó·½³Ìʽ£º________________________________________________¡£

(6)ÓÉX¡¢Y¡¢Z¡¢WºÍFeÎåÖÖÔªËØ×é³ÉµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª392µÄ»¯ºÏÎïC£¬1 mol CÖк¬ÓÐ6 mol½á¾§Ë®¡£¶Ô¸Ã»¯ºÏÎïC½øÐÐÒÔÏÂʵÑ飺

a£®È¡CµÄÈÜÒº¼ÓÈë¹ýÁ¿Å¨NaOHÈÜÒº²¢¼ÓÈÈ£¬²úÉú°×É«³ÁµíºÍÎÞÉ«¡¢Óд̼¤ÐÔÆøÎ¶µÄÆøÌå¡£¹ýÒ»¶Îʱ¼ä£¬°×É«³Áµí±äΪ»ÒÂÌÉ«£¬×îÖÕ±äΪºìºÖÉ«£»

b£®ÁíÈ¡CµÄÈÜÒº£¬¼ÓÈë¹ýÁ¿BaCl2ÈÜÒº²úÉú°×É«³Áµí£¬¼ÓÑÎËá³Áµí²»Èܽ⡣

¢Ùд³öCµÄ»¯Ñ§Ê½£º________________________________________________________________________¡£

¢ÚÊÔд³öCÓëM2Z2°´ÎïÖʵÄÁ¿±È1¡Ã2ÔÚÈÜÒºÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________________________________________________________________

________________________________________________________________________¡£


(1)Na

(2)H2O¡¡H2S

(3)Na£«[¡ÃO, ¡ÃO, ¡Ã]2£­Na£«¡¡2Na2O2£«2H2O===4Na£«£«4OH£­£«O2¡ü

(4)3Cu£«2NO£«8H£«===3Cu2£«£«2NO¡ü£«4H2O

(5)(NH4)2SO3¡¡SO2£«Cl2£«2H2O===4H£«£«2Cl£­£«SO

(6)¢Ù(NH4)2Fe(SO4)2¡¤6H2O

¢Ú4(NH4)2Fe(SO4)2¡¤6H2O£«8Na2O2===4Fe(OH)3¡ý£«8NH3¡ü£«3O2¡ü£«8Na2SO4£«22H2O

[½âÎö] ÔªËØÖÜÆÚ±íÖÐÔ­×Ó°ë¾¶×îСµÄÔªËØÊÇÇâÔªËØ£¬ÎªXÔªËØ£¬ÈôZµÄÔ­×ÓÐòÊýΪa£¬Ôò(a£­1)£«a£«(a£«8)£½3a£«7£½31£¬a£½8£¬¹ÊY¡¢Z¡¢W·Ö±ðÊǵªÔªËØ¡¢ÑõÔªËØ¡¢ÁòÔªËØ£»MÔªËØÓëÇâÔªËØÍ¬Ö÷×壬ÓëÁòÔªËØÍ¬ÖÜÆÚ£¬ÔòMÎªÄÆÔªËØ¡£

(4)HNO3¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Äܹ»ºÍÍ­·´Ó¦£¬Ï¡ÏõËáÓëÍ­·´Ó¦Éú³ÉNO¡£

(5)1 mol AÄܹ»ÓëNaOHÈÜÒº·´Ó¦²úÉú44.8 LÆøÌ壬Ôò¸ÃÆøÌåΪNH3£¬AΪï§ÑΣ¬Ôò1 mol Aº¬ÓÐ2 mol NH£»¢ÚµÎ¼ÓÑÎËáÄܹ»Éú³ÉÆøÌ壬¸ÃÆøÌåÄÜÓëÂÈË®·´Ó¦£¬Ôò¸ÃÆøÌåÊÇSO2¡£¶þÑõ»¯Áò¾ßÓÐÇ¿»¹Ô­ÐÔ£¬Äܹ»±»ÂÈÆøÑõ»¯£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪSO2£«Cl2£«2H2O===4H£«£«2Cl£­£«SO¡£½áºÏÉÏÊöÐÅÏ¢¿ÉÖªAΪ(NH4)2SO3¡£

(6)a.CÖмÓÈëŨNaOHÈÜÒºÄܹ»²úÉúÎÞÉ«¡¢Óд̼¤ÐÔÆøÎ¶µÄÆøÌ壬ÔòΪï§ÑΣ»Í¬Ê±Éú³ÉµÄ°×É«³Áµí»á±äΪ»ÒÂÌÉ«£¬×îÖÕ±ä³ÉºìºÖÉ«£¬ÔòCÖк¬ÓÐFe2£«¡£

b£®CÈÜÒºÖмÓÈëBaCl2ÈÜÒº¿ÉÉú³ÉÄÑÈÜÓÚÑÎËáµÄ°×É«³Áµí£¬ÔòCΪÁòËáÑΡ£

¢Ù½áºÏ»¯ºÏÎïCµÄÏà¶ÔÔ­×ÓÖÊÁ¿ºÍ½á¾§Ë®£¬¿ÉÍÆ¶ÏÆä»¯Ñ§Ê½Îª(NH4)2Fe(SO4)2¡¤6H2O¡£

¢Ú¹ýÑõ»¯ÄƾßÓÐÇ¿Ñõ»¯ÐÔ£¬Äܹ»Ñõ»¯Fe2£«ÎªFe3£«£¬ÒòÈÜÒº³Ê¼îÐÔ£¬Í¬Ê±Éú³ÉFe(OH)3¡£¸ù¾Ý¶þÕß±ÈÀý¿ÉÒÔÅ䯽·´Ó¦·½³Ìʽ¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

A¡¢B¡¢D¡¢E¡¢FΪ¶ÌÖÜÆÚÔªËØ£¬·Ç½ðÊôÔªËØA×îÍâ²ãµç×ÓÊýÓëÆäÖÜÆÚÊýÏàͬ£¬BµÄ×îÍâ²ãµç×ÓÊýÊÇÆäËùÔÚÖÜÆÚÊýµÄ2±¶¡£BÔÚDÖгä·ÖȼÉÕÄÜÉú³ÉÆä×î¸ß¼Û»¯ºÏÎïBD2¡£E£«ÓëD2£­¾ßÓÐÏàͬµÄµç×ÓÊý¡£AÔÚFÖÐȼÉÕ£¬²úÎïÈÜÓÚË®µÃµ½Ò»ÖÖÇ¿Ëá¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)AÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ________£¬Ð´³öÒ»ÖÖ¹¤ÒµÖƱ¸µ¥ÖÊFµÄÀë×Ó·½³Ìʽ£º__________________________¡£

(2)B¡¢D¡¢E×é³ÉµÄÒ»ÖÖÑÎÖУ¬EµÄÖÊÁ¿·ÖÊýΪ43%£¬ÆäË×ÃûΪ__________£¬ÆäË®ÈÜÒºÓëFµ¥ÖÊ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________________________________________£»ÔÚ²úÎïÖмÓÈëÉÙÁ¿KI£¬·´Ó¦ºó¼ÓÈëCCl4²¢Õñµ´£¬Óлú²ãÏÔ______É«¡£

(3)ÓÉÕâÐ©ÔªËØ×é³ÉµÄÎïÖÊ£¬Æä×é³ÉºÍ½á¹¹ÐÅÏ¢ÈçÏÂ±í£º

ÎïÖÊ

×é³ÉºÍ½á¹¹ÐÅÏ¢

a

¡¡º¬ÓÐAµÄ¶þÔªÀë×Ó»¯ºÏÎï

b

¡¡º¬ÓзǼ«ÐÔ¹²¼Û¼üµÄ¶þÔªÀë×Ó»¯ºÏÎÇÒÔ­×ÓÊýÖ®±ÈΪ1¡Ã1

c

¡¡»¯Ñ§×é³ÉΪBDF2

d

¡¡Ö»´æÔÚÒ»ÖÖÀàÐÍ×÷ÓÃÁ¦Çҿɵ¼µçµÄµ¥Öʾ§Ìå

aµÄ»¯Ñ§Ê½Îª________£»bµÄ»¯Ñ§Ê½Îª______________£»cµÄµç×ÓʽΪ________£»dµÄ¾§ÌåÀàÐÍÊÇ________¡£

(4)ÓÉAºÍB¡¢DÔªËØ×é³ÉµÄÁ½ÖÖ¶þÔª»¯ºÏÎïÐγÉÒ»ÀàÐÂÄÜÔ´ÎïÖÊ¡£Ò»ÖÖ»¯ºÏÎï·Ö×Óͨ¹ý________¼ü¹¹³É¾ßÓпÕÇ»µÄ¹ÌÌ壻ÁíÒ»ÖÖ»¯ºÏÎï(ÕÓÆøµÄÖ÷Òª³É·Ö)·Ö×Ó½øÈë¸Ã¿ÕÇ»£¬Æä·Ö×ӵĿռä½á¹¹Îª__________¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø