ÌâÄ¿ÄÚÈÝ

ij»¯Ñ§ÐËȤС×éÒÔľ̿ºÍŨÏõËáΪÆðʼԭÁÏ£¬Ì½¾¿Ò»Ñõ»¯µªÓë¹ýÑõ»¯ÄÆ·´Ó¦ÖƱ¸ÑÇÏõËáÄÆ¡£Éè¼Æ×°ÖÃÈçÏÂ(ºöÂÔ×°ÖÃÖпÕÆøµÄÓ°Ïì)£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©×é×°ºÃÒÇÆ÷ºó£¬±ØÐë½øÐеÄÒ»Ïî²Ù×÷ÊÇ_________________¡£
£¨2£©×°ÖÃAµÄÊÔ¹ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_______________¡£
£¨3£©ÍƲâBÖпÉÒԹ۲쵽µÄÖ÷ÒªÏÖÏóÊÇ________£»C×°ÖõÄ×÷ÓÃÊÇ________¡£
£¨4£©×°ÖÃDÖгýÉú³ÉNaNO2Í⣬»¹ÓÐÁíÒ»ÖÖ¹Ì̬ÎïÖÊY£¬YµÄ»¯Ñ§Ê½ÊÇ________£»¿ÉÒÔͨ¹ýÊʵ±¸Ä½ø£¬²»²úÉúYÎïÖÊ£¬ÇëÄãÌá³ö¸Ä½ø·½·¨£º______________________¡£
£¨5£©ÒÑÖª£ºÑÇÏõËáÊÇÈõËᣬ²»Îȶ¨£¬ÊÒÎÂÏ´æÔÚ·´Ó¦3HNO2=HNO3£«2NO¡ü£«H2O£»ÔÚËáÐÔÈÜÒºÖУ¬NO2-¿É½«MnO4-»¹Ô­ÎªMn2£«ÇÒÎÞÆøÌåÉú³É¡£
¢Ùд³ö¼ìÑéDÖвúÎïÊÇÑÇÏõËáÄƵķ½·¨£º_________________£»
¢ÚE×°ÖÃÖÐÊÔ¼ÁX¿ÉÒÔÊÇ________¡£

A£®Ï¡ÁòËá B£®ËáÐÔ¸ßÃÌËá¼ØÈÜÒº
C£®Ï¡ÏõËá D£®Ë®

£¨1£©¼ì²é×°ÖõÄÆøÃÜÐÔ
£¨2£©C£«4HNO3(Ũ) CO2£«4NO2¡ü£«2H2O
£¨3£©Í­Æ¬Öð½¥Èܽ⣬ÈÜÒºÖð½¥±äÀ¶£¬²úÉúÎÞÉ«ÆøÅÝ¡¡³ýÈ¥NOÖлìÓеÄCO2
£¨4£©NaOH¡¡ÓÃ×°Óмîʯ»ÒµÄ¸ÉÔï¹Ü´úÌæC×°ÖÃ
£¨5£©¢Ù½«Éú³ÉÎïÖÃÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏ¡ÁòËᣬÈô²úÉúÎÞÉ«ÆøÌå²¢ÔÚÒºÃæÉÏ·½±äΪºì×ØÉ«£¬ÔòDÖвúÎïÊÇÑÇÏõËáÄÆ(»ò½«Éú³ÉÎïÖÃÓÚÊÔ¹ÜÖУ¬¼ÓÈëËáÐÔKMnO4ÈÜÒº£¬ÈôÈÜÒº×ÏÉ«ÍÊÈ¥£¬ÔòDÖвúÎïÊÇÑÇÏõËáÄÆ)¡¡¢ÚB

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¶þÑõ»¯ÃÌÓëŨÑÎËá»ìºÏ¼ÓÈȵõ½ÂÈÆø£¬ÈçͼÊÇÖÆÈ¡²¢Ì½¾¿Cl2»¯Ñ§ÐÔÖʵÄ×°ÖÃͼ¡£

(1)Ô²µ×ÉÕÆ¿Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                     ¡£
(2)ÈôÒªµÃµ½¸ÉÔï´¿¾»µÄÆøÌ壬B¡¢CÖÐÓ¦·Ö±ðÊ¢·ÅµÄÊÔ¼ÁΪ           ¡¢          ¡£
(3)EÖÐÈô×°ÓÐFeCl2ÈÜÒº£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ                         £¬EÖÐÈô×°Óеí·Ûµâ»¯¼ØÈÜÒº£¬Äܹ۲쵽µÄʵÑéÏÖÏóÊÇ                        ¡£
(4)ʵÑéÖз¢ÏÖ£ºÅ¨ÑÎËáÓëMnO2»ìºÏ¼ÓÈÈÉú³ÉÂÈÆø£¬Ï¡ÑÎËáÓëMnO2»ìºÏ¼ÓÈȲ»Éú³ÉÂÈÆø¡£Õë¶ÔÉÏÊöÏÖÏóij»¯Ñ§ÐËȤС×é¶Ô¡°Ó°ÏìÂÈÆøÉú³ÉµÄÔ­Òò¡±½øÐÐÁËÌÖÂÛ£¬²¢Éè¼ÆÁËÒÔÏÂʵÑé·½°¸£º
a£®Ï¡ÑÎËáµÎÈëMnO2ÖУ¬È»ºóͨÈëHClÆøÌå¼ÓÈÈ
b£®Ï¡ÑÎËáµÎÈëMnO2ÖУ¬È»ºó¼ÓÈëNaCl¹ÌÌå¼ÓÈÈ
c£®Ï¡ÑÎËáµÎÈëMnO2ÖУ¬È»ºó¼ÓÈëŨÁòËá¼ÓÈÈ
d£®MnO2ÓëNaClµÄŨÈÜÒº»ìºÏ¼ÓÈÈ
e£®Å¨ÁòËáÓëNaCl¹ÌÌå¡¢MnO2¹ÌÌå¹²ÈÈ
¢ÙʵÑébµÄÄ¿µÄÊÇ                                                     £¬ÊµÑécµÄÄ¿µÄÊÇ                                                  ¡£
¢ÚʵÑéÏÖÏó£ºa¡¢c¡¢eÓлÆÂÌÉ«ÆøÌåÉú³É£¬b¡¢dûÓлÆÂÌÉ«ÆøÌåÉú³É¡£Óɴ˵óöÓ°ÏìÂÈÆøÉú³ÉµÄÔ­ÒòÊÇ                                                             ¡£

¾§Ìå¹èÊÇÒ»ÖÖÖØÒªµÄ·Ç½ðÊô²ÄÁÏ£¬ÖƱ¸´¿¹èµÄÖ÷Òª²½ÖèÈçÏ£º
¢Ù¸ßÎÂÏÂÓÃ̼»¹Ô­¶þÑõ»¯¹èÖƵôֹè
¢Ú´Ö¹èÓë¸ÉÔïHClÆøÌå·´Ó¦ÖƵÃSiHCl3£ºSi£«3HClSiHCl3£«H2
¢ÛSiHCl3Óë¹ýÁ¿H2ÔÚ1 000¡«1 100 ¡æ·´Ó¦ÖƵô¿¹è¡£
ÒÑÖªSiHCl3ÄÜÓëH2OÇ¿ÁÒ·´Ó¦£¬ÔÚ¿ÕÆøÖÐÒ××Ôȼ¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)µÚ¢Ù²½ÖƱ¸´Ö¹èµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ_______________________________¡£
(2)´Ö¹èÓëHCl·´Ó¦ÍêÈ«ºó£¬¾­ÀäÄýµÃµ½µÄSiHCl3(·Ðµã£­33 ¡æ)Öк¬ÓÐÉÙÁ¿SiCl4(·Ðµã57.6 ¡æ)ºÍHCl(·Ðµã£­84.7 ¡æ)£¬Ìá´¿SiHCl3²ÉÓõķ½·¨Îª________¡£
(3)ÓÃSiHCl3Óë¹ýÁ¿H2·´Ó¦ÖƱ¸´¿¹èµÄ×°ÖÃÈçÏÂͼ(ÈÈÔ´¼°¼Ð³Ö×°ÖÃÂÔÈ¥)

¢Ù×°ÖÃBÖеÄÊÔ¼ÁÊÇ________¡£×°ÖÃCÖеÄÉÕÆ¿ÐèÒª¼ÓÈÈ£¬ÆäÄ¿µÄÊÇ
________________________________________________________________¡£
¢Ú·´Ó¦Ò»¶Îʱ¼äºó£¬×°ÖÃDÖй۲쵽µÄÏÖÏóÊÇ________£¬×°ÖÃD²»ÄܲÉÓÃÆÕͨ²£Á§¹ÜµÄÔ­ÒòÊÇ__________________________________¡£
×°ÖÃDÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________________________¡£
¢ÛΪ±£Ö¤ÖƱ¸´¿¹èʵÑéµÄ³É¹¦£¬²Ù×÷µÄ¹Ø¼üÊǼì²éʵÑé×°ÖõÄÆøÃÜÐÔ£¬¿ØÖƺ÷´Ó¦Î¶ÈÒÔ¼°__________________________________________________
¢ÜΪ¼ø¶¨²úÆ·¹èÖÐÊÇ·ñº¬Î¢Á¿Ìúµ¥ÖÊ£¬½«ÊÔÑùÓÃÏ¡ÑÎËáÈܽ⣬ȡÉϲãÇåÒººóÐèÔÙ¼ÓÈëµÄÊÔ¼ÁÊÇ(Ìîд×Öĸ´úºÅ)________¡£
a£®µâË®¡¡b£®ÂÈË®¡¡c£®NaOHÈÜÒº¡¡d£®KSCNÈÜÒº
e£®Na2SO3ÈÜÒº

ij»¯Ñ§¿ÎÍâ»î¶¯Ð¡×éͨ¹ýʵÑéÑо¿NO2µÄÐÔÖÊ¡£
ÒÑÖª:2NO2+2NaOHNaNO3+NaNO2+H2O
ÈÎÎñ1:ÀûÓÃÈçͼËùʾװÖÃ̽¾¿NO2ÄÜ·ñ±»NH3»¹Ô­(K1¡¢K2Ϊֹˮ¼Ð,¼Ð³Ö¹Ì¶¨×°ÖÃÂÔÈ¥)¡£

(1)E×°ÖÃÖÐÖÆÈ¡NO2·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                ¡¡¡£
(2)ÈôNO2Äܹ»±»NH3»¹Ô­,Ô¤Æڹ۲쵽C×°ÖÃÖеÄÏÖÏóÊÇ¡¡                     ¡£
(3)ʵÑé¹ý³ÌÖÐ,δÄܹ۲쵽C×°ÖÃÖеÄÔ¤ÆÚÏÖÏ󡣸ÃС×éͬѧ´Ó·´Ó¦Ô­ÀíµÄ½Ç¶È·ÖÎöÁËÔ­Òò,ÈÏΪ¿ÉÄÜÊÇ:
¢ÙNH3»¹Ô­ÐÔ½ÏÈõ,²»Äܽ«NO2»¹Ô­;
¢ÚÔÚ´ËÌõ¼þÏÂ,NO2µÄת»¯Âʼ«µÍ;
¢Û¡¡                                          ¡£
(4)´ËʵÑé×°ÖôæÔÚÒ»¸öÃ÷ÏÔµÄȱÏÝÊÇ                                 ¡£
ÈÎÎñ2:̽¾¿NO2ÄÜ·ñÓëNa2O2·¢ÉúÑõ»¯»¹Ô­·´Ó¦¡£
(5)ʵÑéÇ°,¸ÃС×éͬѧÌá³öÈýÖÖ¼ÙÉè¡£
¼ÙÉè1:¶þÕß²»·´Ó¦;
¼ÙÉè2:NO2Äܱ»Na2O2Ñõ»¯;
¼ÙÉè3:¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ ¡£
(6)ΪÁËÑéÖ¤¼ÙÉè2,¸ÃС×éͬѧѡÓÃÈÎÎñ1ÖеÄB¡¢D¡¢E×°ÖÃ,½«BÖеÄÒ©Æ·¸ü»»ÎªNa2O2,ÁíÑ¡F×°ÖÃ(ÈçͼËùʾ),ÖØÐÂ×é×°,½øÐÐʵÑé¡£

¢Ù×°ÖõĺÏÀíÁ¬½Ó˳ÐòÊÇ¡¡                                    ¡£
¢ÚʵÑé¹ý³ÌÖÐ,B×°ÖÃÖе­»ÆÉ«·ÛÄ©Öð½¥±ä³É°×É«¡£¾­¼ìÑé,¸Ã°×É«ÎïÖÊΪ´¿¾»Îï,ÇÒÎÞÆäËûÎïÖÊÉú³É¡£ÍƲâB×°ÖÃÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

(1)ij»¯Ñ§ÐËȤС×éµÄͬѧ½øÐÐCl2¡¢NH3µÄÖƱ¸¼°ÐÔÖʼìÑéµÈʵÑéµÄÁ÷³ÌºÍ²¿·Ö×°ÖÃÈçÏ£º

¢ÙÇëÀûÓÃA¡¢G×°ÖÃÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑéÑéÖ¤Cl2¡¢Fe3£«¡¢I2µÄÑõ»¯ÐÔÇ¿ÈõΪCl2>Fe3£«>I2(ʵÑéÖв»¶ÏµØСÐÄÕñµ´G×°ÖÃÖеÄÊÔ¹Ü)¡£Çëд³öAÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º______________________________________________£¬
Çëд³öÊÔ¼ÁMΪ________ÈÜÒº£¬Ö¤Ã÷Ñõ»¯ÐÔΪCl2>Fe3£«>I2µÄʵÑéÏÖÏóÊÇ________________________________________________________________________¡£
¢ÚÒÑÖª3Cl2£«2NH3=6HCl£«N2£¬µ±DµÄÉÕÆ¿ÖгäÂú»ÆÂÌÉ«ÆøÌåºó£¬¹Ø±Õa¡¢c´ò¿ªb£¬DÖеÄÏÖÏóΪ»ÆÂÌÉ«ÆøÌåÏûʧ£¬²úÉú°×ÑÌ£¬·´Ó¦Ò»¶Îʱ¼äºó£¬¹Ø±Õb´ò¿ªc£¬¹Û²ìµ½µÄÏÖÏóΪ________________________________________________________________________¡£
(2)ij·ÏË®Öк¬ÓÐÒ»¶¨Á¿µÄNa£«¡¢SO32¡ª£¬¿ÉÄܺ¬ÓÐCO32¡ª£¬Ä³Ñо¿Ð¡×éÓû²â¶¨ÆäÖÐSO32¡ªµÄŨ¶È£¬Éè¼ÆÈçÏÂʵÑé·½°¸£º

¢Ù´ÓÏÂÁÐÊÔ¼ÁÖÐÑ¡ÔñÊÔ¼ÁXΪ________(ÌîÐòºÅ)£»
A£®0.1 mol/L KMnO4(H2SO4Ëữ)ÈÜÒºB£®0.5 mol/L NaOHÈÜÒº
C£®ÐÂÖÆÂÈË®                       D£®KIÈÜÒº
¢Ú¼ÓÈëÊÔ¼ÁXÉú³ÉSO42¡ªµÄÀë×Ó·½³ÌʽΪ_______________________________________
¢ÛÖ¤Ã÷¸Ã·ÏË®ÖÐÊÇ·ñº¬ÓÐCO32¡ªµÄʵÑé·½°¸Îª_________________________________________________

¹¤ÒµÎ²ÆøÖеªÑõ»¯Îïͨ³£²ÉÓð±´ß»¯ÎüÊÕ·¨£¬ÆäÔ­ÀíÊÇNH3ÓëNOxÔÚ´ß»¯¼Á×÷ÓÃÏ·´Ó¦Éú³ÉÎÞ¶¾µÄÎïÖÊ¡£Ä³Ð£»î¶¯Ð¡×éͬѧ²ÉÓÃÒÔÏÂ×°ÖúͲ½ÖèÄ£Ä⹤ҵÉϵªÑõ»¯ÎïµÄ´¦Àí¹ý³Ì¡£

I£®Ì½¾¿ÖÆÈ¡NH3µÄ·½·¨
£¨1£©ÔÚÉÏÊö×°ÖÃÖУ¬HÄÜ¿ìËÙ¡¢¼ò±ãÖÆÈ¡NH3£¬×°ÖÃÖÐÐèÒªÌí¼ÓµÄ·´Ó¦ÊÔ¼ÁΪ      ¡£
£¨2£©ÎªÌ½¾¿¸üºÃµÄʵÑéЧ¹û£¬»î¶¯Ð¡×éͬѧ²ÉÓÃÉÏÊöC×°ÖÃÀ´ÖÆÈ¡°±Æø£¬ÔÚ¿ØÖÆʵÑéÌõ¼þÏàͬµÄÇé¿öÏ£¬»ñµÃϱíÖÐʵÑéÊý¾Ý¡£

ÊÔ¼Á×éºÏÐòºÅ
¹ÌÌåÊÔ¼Á
NH3Ìå»ý£¨mL£©
a
6.0 g Ca(OH)2£¨¹ýÁ¿£©
5.4 g NH4Cl
1344
b
5.4g (NH4)2SO4
1364
c
6.0 g NaOH£¨¹ýÁ¿£©
5.4 g NH4Cl
1568
d
5.4g (NH4)2SO4
1559
e
6.0 g CaO£¨¹ýÁ¿£©
5.4 g NH4Cl
1753
f
5.4 g (NH4)2SO4
1792
 
·ÖÎö±íÖÐÊý¾Ý£¬ÄãÈÏΪÄÄÖÖ·½°¸ÖÆÈ¡°±ÆøµÄЧ¹û×îºÃ     £¨ÌîÐòºÅ£©£¬´Ó¸Ã·½°¸Ñ¡ÔñµÄÔ­ÁÏ·ÖÎöÖÆÆøЧ¹ûºÃµÄ¿ÉÄÜÔ­ÒòÊÇ                           ¡£
II£®Ä£ÄâβÆø´¦Àí
»î¶¯Ð¡×éͬѧѡÓÃÉÏÊö²¿·Ö×°Ö㬰´ÏÂÁÐ˳ÐòÁ¬½Ó³ÉÄ£ÄâβÆø´¦Àí×°ÖýøÐÐʵÑé¡£
£¨1£©Çë´ÓÉÏÊö×°ÖÃÖÐÑ¡ÔñÄãÈÏΪºÏÀíµÄ½øÐв¹³ä£¨ËùѡװÖò»ÄÜÖظ´£©¡£

£¨2£©AÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ                                  ¡£
£¨3£©D×°ÖõÄ×÷ÓÃÓУºÊ¹ÆøÌå»ìºÏ¾ùÔÈ¡¢µ÷½ÚÆøÁ÷Ëٶȡ¢             ¡£
£¨4£©D×°ÖÃÖеÄÒºÌ廹¿É»»³É        £¨ÌîÐòºÅ£©¡£
a£®H2O   b£®CCl4  c£®Å¨H2SO4    d£®CuSO4ÈÜÒº
£¨5£©¸ÃС×éͬѧËùÉè¼ÆµÄÄ£ÄâβÆø´¦Àí×°ÖÃÖл¹´æÔÚÒ»´¦Ã÷ÏÔµÄȱÏÝÊÇ            ¡£

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

¾«Ó¢¼Ò½ÌÍø