ÌâÄ¿ÄÚÈÝ

(Ò»)ijѧÉúΪÁ˲ⶨ²¿·Ö±äÖʵÄNa2SO3ÑùÆ·µÄ´¿¶È£¬Éè¼ÆÁËÈçÏÂʵÑ飺

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)д³öA×°ÖÃÖв£Á§ÒÇÆ÷µÄÃû³Æ£º¾Æ¾«µÆ¡¢____________¡¢____________¡£

(2)ʵÑ鿪ʼºó£¬Ð´³öBÖз´Ó¦µÄÀë×Ó·½³Ìʽ_____________________________________¡£

(3)CÖеÄÏÖÏóÊÇ________________________£¬E×°ÖõÄ×÷ÓÃÊÇ______________________¡£

(4)°´ÏÂͼËùʾ³ÆÈ¡Ò»¶¨Á¿µÄNa2SO3ÑùÆ··ÅÈëA×°ÖõÄÉÕÆ¿ÖУ¬µÎÈë×ãÁ¿µÄH2SO4ÍêÈ«·´Ó¦¡£È»ºó½«BÖÐÍêÈ«·´Ó¦ºóµÄÈÜÒºÓë×ãÁ¿µÄBaCl2ÈÜÒº·´Ó¦£¬¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔµÃ°×É«³Áµí23.3 g£¬ÔòÔ­ÑùÆ·ÖÐNa2SO3µÄ´¿¶ÈΪ___________(¾«È·µ½0.1%)¡£

(5)ÔÚ¹ýÂ˳ÁµíʱÈô¹ýÂËÒº³öÏÖ»ë×Ç£¬Ôò±ØÐëÒªÖظ´___________²Ù×÷£¬Èô¸ÃѧÉúûÓÐÖظ´¸Ã²Ù×÷Ôò²â¶¨µÄ½á¹û½«___________(Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)¡£

(6)Ҫʹ²â¶¨½á¹û׼ȷ£¬µÚÒ»£¬×°ÖÃÆøÃÜÐÔ±ØÐëÁ¼ºÃ£»µÚ¶þ£¬Ó¦Ïȵãȼ___________´¦¾Æ¾«µÆ(Ìî×°ÖÃ×Öĸ)¡£

(¶þ)ÏÂÁÐÓйػ¯Ñ§ÊµÑéµÄ»ù±¾²Ù×÷¼°°²È«ÖªÊ¶µÄÐðÊö£¬²»ÕýÈ·µÄÊÇ_________(ÌîÐòºÅ)¡£

A.ÓÃÍÐÅÌÌìƽ³ÆÈ¡10.55 g¸ÉÔïµÄNaCl¹ÌÌå

B.¸½×ÅÓÚÊÔ¹ÜÄڱڵı½·Ó£¬¿ÉÓüîҺϴµÓ

C.ÓüîʽµÎ¶¨¹ÜÁ¿È¡20.00 mL 0.100 0 mol¡¤L-1µÄ¸ßÃÌËá¼ØÈÜÒº

D.ÓÃÉøÎö·¨·ÖÀëµí·ÛÖлìÓеÄNaNO3ÔÓÖÊ

E.ÅäÖÆŨÁòËáºÍŨÏõËáµÄ»ìºÏËáʱ£¬½«Å¨ÁòËáÑØÆ÷±ÚÂýÂý¼ÓÈ뵽ŨÏõËáÖУ¬²¢²»¶Ï½Á°è

F.Çиî°×Á×ʱ£¬±ØÐëÓÃÄ÷×Ó¼ÐÈ¡£¬ÖÃÓÚ×ÀÃæÉϵIJ£Á§Æ¬ÉÏ£¬Ð¡ÐÄÓõ¶Çиî

G.ʵÑéʱ£¬²»É÷´ò·­È¼×ŵľƾ«µÆ£¬¿ÉÁ¢¼´ÓÃʪĨ²¼¸ÇÃð»ðÑæ

H.ÓÃÖؽᾧ·¨¿ÉÒÔ³ýÈ¥ÏõËá¼ØÖлìÓеÄÉÙÁ¿ÂÈ»¯ÄÆ

I.ÔÚÖÐѧ¡¶ÁòËáÍ­¾§ÌåÀï½á¾§Ë®º¬Á¿²â¶¨¡·µÄʵÑéÖУ¬³ÆÁ¿²Ù×÷ÖÁÉÙÐèÒªËÄ´Î

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

(Ò»)(1)Ô²µ×ÉÕÆ¿¡¢·ÖҺ©¶·

(2)Cl2+SO2+2H2O4H++2Cl-+

(3)ºìÉ«ÏÊ»¨ÍÊÉ«  ÎüÊÕδ·´Ó¦µÄÓж¾ÆøÌ壬·ÀÖ¹Óж¾ÆøÌå¶Ô´óÆøµÄÎÛȾ

(4)50.8%  (5)¹ýÂË  Æ«µÍ  (6)D

(¶þ)ACFJ

½âÎö£º(Ò»)´ËÌâµÄÉè¼ÆÔ­ÀíÊÇAÖвúÉúµÄSO2ÔÚBÖб»DÖвúÉúµÄCl2Ñõ»¯ÎªH2SO4£¬Í¨¹ý²â¶¨H2SO4µÄÁ¿½ø¶øÇóµÃNa2SO3µÄ´¿¶È¡£

(4)³ÆÁ¿Ê±íÀÂëºÍÒ©Æ··Å´íÁËλÖã¬m(ÑùÆ·)+0.2 g=25.0 g

m(ÑùÆ·)=24.8 g¡£Éú³É°×É«³Áµí23.3 g£¬ÔòÔ­ÑùÆ·ÖÐNa2SO3µÄÖÊÁ¿Îª0.1 mol¡Á126 g¡¤mol-1=12.6 g£¬ËùÒÔNa2SO3µÄ´¿¶ÈΪ¡Á100%=50.8%¡£

(¶þ)ÍÐÅÌÌìƽֻÄܾ«È·µ½Ð¡ÊýµãºóµÚһ룬A´í¡£Á¿È¡¸ßÃÌËá¼ØÈÜÒºÓ¦ÓÃËáʽµÎ¶¨¹Ü£¬C´í¡£°×Á×Ó¦ÔÚË®ÏÂÇиF´í¡£JÏîÓ¦ÖØÐÂÅäÖÆ¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¹ýÑõ»¯ÄÆÊÇÒ»ÖÖµ­»ÆÉ«¹ÌÌ壬ËüÄÜÓë¶þÑõ»¯Ì¼·´Ó¦Éú³ÉÑõÆø£¬ÔÚDZˮͧÖÐÓÃ×÷ÖÆÑõ¼Á£¬¹©ÈËÀàºôÎüÖ®Óã®ËüÓë¶þÑõ»¯Ì¼·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2Na2O2+2CO2¨T2Na2CO3+O2£®Ä³Ñ§ÉúΪÁËÑéÖ¤ÕâһʵÑ飬ÒÔ×ãÁ¿µÄ´óÀíʯ¡¢×ãÁ¿µÄÑÎËáºÍ1.95g¹ýÑõ»¯ÄÆÑùƷΪԭÁÏ£¬ÖÆÈ¡O2£¬Éè¼Æ³öÈçͼ1ʵÑé×°Öãº
¾«Ó¢¼Ò½ÌÍø
£¨1£©AÖÐÖÆÈ¡CO2µÄ×°Öã¬Ó¦´Óͼ2µÄ¢Ù¡¢¢Ú¡¢¢ÛÖÐÑ¡Äĸöͼ£º
 
£¬B×°ÖõÄ×÷ÓÃÊÇ
 
£¬C×°ÖÃÄÚ¿ÉÄܳöÏÖµÄÏÖÏóÊÇ
 
£®ÎªÁ˼ìÑéEÖÐÊÕ¼¯µ½µÄÆøÌ壬ÔÚÈ¡³ö¼¯ÆøÆ¿ºó£¬
 
£®
£¨2£©ÈôEÖеÄʯ»ÒË®³öÏÖ³öÏÖÇá΢°×É«»ë×Ç£¬Çë˵Ã÷Ô­Òò£º
 
£®
£¨3£©ÈôDÖеÄ1.95g¹ýÑõ»¯ÄÆÑùÆ·½Ó½ü·´Ó¦Íê±Ïʱ£¬ÄãÔ¤²âE×°ÖÃÄÚÓкÎÏÖÏó£¿
 
£®
£¨4£©·´Ó¦Íê±Ïʱ£¬Èô²âµÃEÖеļ¯ÆøÆ¿ÊÕ¼¯µ½µÄÆøÌåΪ250mL£¬ÓÖÖªÑõÆøµÄÃܶÈΪ1.43g/L£¬µ±×°ÖõÄÆøÃÜÐÔÁ¼ºÃµÄÇé¿öÏ£¬Êµ¼ÊÊÕ¼¯µ½µÄÑõÆøÌå»ý±ÈÀíÂÛ¼ÆËãÖµ
 
£¨´ó»òС£©£¬Ïà²îÔ¼
 
mL£¨È¡ÕûÊýÖµ£¬ËùÓÃÊý¾Ý¾ùÔÚ±ê×¼×´¿öϲⶨ£©£¬ÕâÊÇÓÉÓÚ
 
£®
£¨5£©ÄãÈÏΪÉÏÊöA-EµÄʵÑé×°ÖÃÖУ¬E²¿·ÖÊÇ·ñ°²È«¡¢ºÏÀí£¿
 
EÊÇ·ñÐèÒª¸ÄΪÏÂÁÐËÄÏîÖеÄÄÄÒ»Ï
 
£®£¨Óüס¢ÒÒ¡¢±û¡¢¶¡»Ø´ð£©
¶ÔÑÀ¸àµÄ̽¾¿ÒªÓõ½Ðí¶à»¯Ñ§ÖªÊ¶£º

(1)ϱíÁгöÁËÈýÖÖÑÀ¸àÖеÄĦ²Á¼Á£¬ÇëÔÚ±íÖÐÌîдÈýÖÖĦ²Á¼ÁËùÊôµÄÎïÖÊÀà±ð£º

 

Á½ÃæÕë¶ùͯÑÀ¸à

ÕäÖéÍõ·À³ôÑÀ¸à

ÖлªÍ¸Ã÷ÑÀ¸à

Ħ²Á¼Á

ÇâÑõ»¯ÂÁ

̼Ëá¸Æ

¶þÑõ»¯¹è

Ħ²Á¼ÁµÄÎïÖÊÀà±ð(Ö¸Ëá¡¢¼î¡¢ÑΡ¢Ñõ»¯Îï)

 

 

 

(2)¸ù¾ÝÄãµÄÍƲ⣬ÑÀ¸àĦ²Á¼ÁµÄÈܽâÐÔÊÇ___________¡£

(3)ÑÀ¸àÖеÄĦ²Á¼Á̼Ëá¸Æ¿ÉÒÔÓÃʯ»ÒʯÀ´ÖƱ¸£¬Ä³Ñ§ÉúÉè¼ÆÁËÒ»ÖÖÖƱ¸Ì¼Ëá¸ÆµÄʵÑé·½°¸£¬ÆäÁ÷³ÌͼΪ

Çëд³öÉÏÊö·½°¸ÖÐÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

¢Ù____________________________________________________________;

¢Ú____________________________________________________________;

¢Û____________________________________________________________¡£

(4)ÇëÄãÈÔÓÃʯ»ÒʯΪԭÁÏ(ÆäËûÊÔ¼Á×ÔÑ¡)£¬Éè¼ÆÁíÒ»ÖÖÖƱ¸Ì¼Ëá¸ÆµÄʵÑé·½°¸¡£·ÂÕÕ(3)Ëùʾ£¬½«ÄãµÄʵÑé·½°¸ÓÃÁ÷³Ìͼ±íʾ³öÀ´£º

ÄãÉè¼ÆµÄ·½°¸ÓŵãÊÇ____________________________________¡£

(5)¼ìÑéÑÀ¸àÖÐÊÇ·ñº¬ÓÐ̼ËáÑεÄʵÑé·½·¨ÊÇ____________________________________¡£

(6)ijѧÉúΪÁ˲ⶨһÖÖÒÔ̼Ëá¸ÆΪĦ²Á¼ÁµÄÑÀ¸àÖÐ̼Ëá¸ÆµÄº¬Á¿£¬ÓÃÉÕ±­³ÆÈ¡ÕâÖÖÑÀ¸à¸àÌå100.0 g£¬ÏòÉÕ±­ÖÐÖð½¥¼ÓÈëÑÎËáÖÁ²»ÔÙÓÐÆøÌå·Å³ö(³ý̼Ëá¸ÆÍ⣬ÕâÖÖÑÀ¸àµÄÆäËûÎïÖʲ»ÄÜÓëÑÎËá·´Ó¦Éú³ÉÆøÌå)£¬¹²ÊÕ¼¯µ½ÆøÌåµÄÖÊÁ¿Îª22 g¡£ÇëÄã¼ÆËãÕâÖÖÑÀ¸àÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý¡£(¿ÉÄÜÓõ½µÄÏà¶ÔÔ­×ÓÖÊÁ¿£ºC¡ª12£»O¡ª16£»Ca¡ª40)

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

¾«Ó¢¼Ò½ÌÍø