AºÍB¾ùΪÄÆÑεÄË®ÈÜÒº£¬A³ÊÖÐÐÔ£¬B³Ê¼îÐÔ²¢¾ßÓÐÑõ»¯ÐÔ¡£ÏÂÊöΪÏà¹ØʵÑé²½ÖèºÍʵÑéÏÖÏó£º

ÇëÍê³ÉÏÂÁÐÎÊÌ⣺

£¨1£©Ð´³öA¡¢BºÍCµÄ»¯Ñ§Ê½£ºA_____________£¬B_____________£¬C_____________¡£

£¨2£©ÒÀ´Îд³öA¡úDºÍD¡úE£¨EÖк¬ÓÐij+5¼ÛÔªËصĺ¬ÑõËá¸ùÀë×Ó£©µÄÀë×Ó·½³Ìʽ£º______________________________________£¬_______________________________________¡£

£¨3£©½«SO2ÆøÌåͨÈëDÈÜÒº£¬DÈÜÒº±äΪÎÞÉ«£¬Éú³ÉÁ½ÖÖËᡣд³ö·´Ó¦µÄÀë×Ó·½³Ìʽ£º______________________________________________________________________________¡£

£¨4£©Ð´³öÓÉF¡úHµÄ»¯Ñ§·½³Ìʽ£º__________________________________________________¡£

£¨1£©NaI  NaClO  AgI

(2)2I-+ClO-+H2O====I2+Cl-+2OH-

I2+5ClO-+2OH-====2+5Cl-+H2O

(3)I2+SO2+2H2O====2I-++4H+

(4)Cl2+2NaOH====NaCl+NaClO+H2O


½âÎö:

´ËÌâ½âÌâÍ»ÆÆ¿ÚΪ¼¸ÖÖÎïÖʵÄÑÕÉ«£¬ÓÉ´Ë¿É֪Ϊ±Ëص¥Öʼ°Æ仯ºÏÎïÖ®¼äµÄÏ໥ת±ä£¬ÓÖÓÐB³Ê¼îÐÔ²¢¾ßÓÐÑõ»¯ÐÔ¼°HÊǺ¬BµÄÈÜÒº£¬¿ÉÍƳöAΪNaI£¬BΪNaClO£¬CΪAgI£¬DΪI2¡££¨2£©ÒòNaClO¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬¿É°ÑI2Ñõ»¯ÎªNaIO3¡££¨3£©I2¿É°ÑSO2Ñõ»¯Îª¡££¨4£©FHΪCl2ÓëNaOHµÄ·´Ó¦¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2011?ʯ¾°É½Çøһģ£©ÎÒ¹úÖƼҵµÄÏÈÇý--ºîµÂ°ñÏÈÉú£¬1939Äê·¢Ã÷ÁËÖøÃûµÄºîÊÏÖƼ£¬ÆäºËÐÄ·´Ó¦Ô­Àí¿ÉÓÃÈçÏ»¯Ñ§·½³Ìʽ±íʾ£º
NH3+CO2+NaCl+H2O¨TNH4Cl+NaHCO3£¨¾§Ì壩£¬ÒÀ¾Ý´ËÔ­Àí£¬ÓûÖƵÃ̼ËáÇâÄƾ§Ì壬ijУѧÉúÉè¼ÆÁËÈçÏÂʵÑé×°Öã¬ÆäÖÐB×°ÖÃÖеÄÊÔ¹ÜÄÚÊÇÈÜÓа±ºÍÂÈ»¯ÄƵÄÈÜÒº£¬ÇÒ¶þÕß¾ùÒÑ´ïµ½±¥ºÍ£®

£¨1£©A×°ÖÃÖÐËù·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
CaCO3+2H+=Ca2++CO2¡ü+H2O
CaCO3+2H+=Ca2++CO2¡ü+H2O
£®C×°ÖÃÖÐÏ¡ÁòËáµÄ×÷ÓÃΪ
ÎüÊÕ´ÓB×°ÖÃÖеÄÊÔ¹ÜÄÚÒݳöµÄ°±Æø£¬¼õÉÙ¶Ô»·¾³µÄÎÛȾ
ÎüÊÕ´ÓB×°ÖÃÖеÄÊÔ¹ÜÄÚÒݳöµÄ°±Æø£¬¼õÉÙ¶Ô»·¾³µÄÎÛȾ
£®
£¨2£©Ï±íÖÐËùÁгöµÄÊÇÏà¹ØÎïÖÊÔÚ²»Í¬Î¶ÈϵÄÈܽâ¶ÈÊý¾Ý£¨g/100gË®£©
ζÈ
Èܽâ¶È
ÑÎ
0¡æ 10¡æ 20¡æ 30¡æ 40¡æ 50¡æ
NaCl 35.7 35.8 36.0 36.3 36.6 37.0
NaHCO3 6.9 8.1 9.6 11.1 12.7 14.5
NH4Cl 29.4 33.3 37.2 41.4 45.8 50.4
²ÎÕÕ±íÖÐÊý¾Ý£¬Çë·ÖÎöB×°ÖÃÖÐʹÓñùË®µÄÄ¿µÄÊÇ
ζÈÔ½µÍ̼ËáÇâÄÆÈܽâ¶ÈԽС£¬±ãÓÚÎö³ö
ζÈÔ½µÍ̼ËáÇâÄÆÈܽâ¶ÈԽС£¬±ãÓÚÎö³ö
£®
£¨3£©¸ÃУѧÉúÔÚ¼ì²éÍê´ËÌ××°ÖÃÆøÃÜÐÔºó½øÐÐʵÑ飬½á¹ûûÓеõ½Ì¼ËáÇâÄƾ§Ì壬ָѰ½Ìʦָ³öÓ¦ÔÚ
AB
AB
×°ÖÃÖ®¼ä£¨Ìîд×Öĸ£©Á¬½ÓÒ»¸öÊ¢ÓÐ
±¥ºÍ̼ËáÇâÄÆÈÜÒºµÄ
±¥ºÍ̼ËáÇâÄÆÈÜÒºµÄ
µÄÏ´Æø×°Öã¬Æä×÷ÓÃÊÇ
³ýÈ¥¶þÑõ»¯Ì¼ÆøÌåÖеÄÂÈ»¯ÇâÆøÌå
³ýÈ¥¶þÑõ»¯Ì¼ÆøÌåÖеÄÂÈ»¯ÇâÆøÌå
£®
£¨4£©Èô¸ÃУѧÉú½øÐÐʵÑéʱ£¬ËùÓñ¥ºÍʳÑÎË®Öк¬NaClµÄÖÊÁ¿Îª5.85g£¬ÊµÑéºóµÃµ½¸ÉÔïµÄNaHCO3¾§ÌåµÄÖÊÁ¿Îª5.04g£¬ÔòNaHCO3µÄ²úÂÊΪ
60%
60%
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

¹¤ÒµÉ϶Ժ£Ë®×ÊÔ´×ۺϿª·¢ÀûÓõIJ¿·Ö¹¤ÒÕÁ÷³ÌÈçͼ1Ëùʾ£®

£¨1£©´ÖÑÎÖк¬ÓÐCa2+¡¢Mg2+¡¢SO42-µÈÔÓÖÊ£¬´ÖÖƺó¿ÉµÃ±¥ºÍNaClÈÜÒº£¬¾«ÖÆʱͨ³£ÔÚÈÜÒºÖÐÒÀ´ÎÖмÓÈë¹ýÁ¿µÄBaCl2ÈÜÒº¡¢¹ýÁ¿µÄNaOHÈÜÒººÍ¹ýÁ¿µÄNa2CO3ÈÜÒº£¬¹ýÂ˺óÏòÂËÒºÖмÓÈëÑÎËáÖÁÈÜÒº³ÊÖÐÐÔ£®Çëд³ö¼ÓÈëNa2CO3ÈÜÒººóÏà¹Ø»¯Ñ§·´Ó¦µÄÀë×Ó·½³Ìʽ£º
Ba2++CO32-=BaCO3¡ý£»Ca2++CO32-=CaCO3¡ý£»
Ba2++CO32-=BaCO3¡ý£»Ca2++CO32-=CaCO3¡ý£»
£®
£¨2£©±¾¹¤ÒÕÁ÷³ÌÖÐÏȺóÖƵÃBr2¡¢CaSO4¡¢Mg£¨OH£©2£¬ÄÜ·ñ°´Br2¡¢Mg£¨OH£©2¡¢CaSO4µÄ˳ÐòÖƱ¸£¿
·ñ
·ñ
£¨Ìî¡°ÄÜ¡±»ò¡°·ñ¡±£©£¬Ô­ÒòÊÇ
Èç¹ûÏȳÁµíMg£¨OH£©2£¬Ôò³ÁµíÖлá¼ÐÔÓÓÐCaSO4³Áµí£¬²úÆ·²»´¿
Èç¹ûÏȳÁµíMg£¨OH£©2£¬Ôò³ÁµíÖлá¼ÐÔÓÓÐCaSO4³Áµí£¬²úÆ·²»´¿
£®
£¨3£©ÂÈ»¯ÄÆÊÇÖØÒªµÄÂȼҵ»¯¹¤µÄÔ­ÁÏ£®µç½â±¥ºÍʳÑÎË®³£ÓÃÀë×ÓĤµç½â²ÛºÍ¸ôĤµç½â²Û£®Àë×ÓĤºÍ¸ôĤ¾ùÔÊÐíͨ¹ýµÄ·Ö×Ó»òÀë×ÓÊÇ
B
B
£®
A£®Cl-      B£®Na+       C£®OH-        D£®Cl2
£¨4£©±¥ºÍʳÑÎË®µç½âʱÓëµçÔ´Õý¼«ÏàÁ¬µÄµç¼«ÉÏ·¢ÉúµÄ·´Ó¦Îª
Ñõ»¯·´Ó¦£¬2Cl--2e-=Cl2¡ü
Ñõ»¯·´Ó¦£¬2Cl--2e-=Cl2¡ü
£¬ÓëµçÔ´¸º¼«ÏßÁ¬µÄµç¼«¸½½üÈÜÒºpH
±ä´ó
±ä´ó
£¨±ä´ó¡¢²»±ä¡¢±äС£©£®
£¨5£©ÂÈ»¯ÄƵĿÉÓÃÓÚÉú²ú´¿¼î£¬ÎÒ¹ú»¯Ñ§¼ÒºîµÂ°ñ¸Ä¸ï¹úÍâÉú²ú¹¤ÒÕ£¬Éú²úÁ÷³Ì¼òÒª±íʾÈçͼ2£º

¢ÙÉÏÊöÉú²ú´¿¼îµÄ·½·¨³Æ
ÁªºÏÖƼ
ÁªºÏÖƼ
£¬¸±²úÆ·µÄÒ»ÖÖÓÃ;Ϊ
×ö»¯·Ê
×ö»¯·Ê
£®
д³öÉÏÊöÁ÷³ÌÖÐXÎïÖʵķÖ×Óʽ
CO2
CO2
£®Ê¹Ô­ÁÏÂÈ»¯ÄƵÄÀûÓÃÂÊ´Ó70%Ìá¸ßµ½90%ÒÔÉÏ£¬Ö÷ÒªÊÇÉè¼ÆÁË
¢ñ
¢ñ
£¨ÌîÉÏÊöÁ÷³ÌÖеıàºÅ£©µÄÑ­»·£®´Ó³Áµí³ØÖÐÈ¡³ö³ÁµíµÄ²Ù×÷ÊÇ
¹ýÂË
¹ýÂË
£®
¢ÚºÏ³É°±Ô­ÁÏÆøÖеªÆøÖƱ¸µÄ·½·¨Ö®Ò»Îª
Һ̬¿ÕÆøÒÀ¾Ý·Ðµã·ÖÀë
Һ̬¿ÕÆøÒÀ¾Ý·Ðµã·ÖÀë
£¬ÁíÒ»Ô­ÁÏÆøÇâÆøµÄÖÆÈ¡»¯Ñ§·½³ÌʽΪ
C+H2O
 ¸ßΠ
.
 
CO+H2£¬CO+H2O
 ¸ßΠ
.
 
CO2+H2
C+H2O
 ¸ßΠ
.
 
CO+H2£¬CO+H2O
 ¸ßΠ
.
 
CO2+H2
£®
¢Û³Áµí³ØÖз¢ÉúµÄ»¯Ñ§·´Ó¦·½³ÌʽÊÇ
NH3+CO2+NaCl+H2O=NH4Cl+NaHCO3¡ý
NH3+CO2+NaCl+H2O=NH4Cl+NaHCO3¡ý
£®ÒªÊµÏָ÷´Ó¦£¬ÄãÈÏΪӦ¸ÃÈçºÎ²Ù×÷£º
Ïò°±»¯µÄ±¥ºÍÂÈ»¯ÄÆÈÜÒºÖÐͨÈë¶þÑõ»¯Ì¼ÆøÌåµÃµ½Ì¼ËáÇâÄƾ§Ìå
Ïò°±»¯µÄ±¥ºÍÂÈ»¯ÄÆÈÜÒºÖÐͨÈë¶þÑõ»¯Ì¼ÆøÌåµÃµ½Ì¼ËáÇâÄƾ§Ìå
£»
¢ÜΪ¼ìÑé²úƷ̼ËáÄÆÖÐÊÇ·ñº¬ÓÐÂÈ»¯ÄÆ£¬¿ÉÈ¡ÉÙÁ¿ÊÔÑùÈÜÓÚË®ºó£¬ÔٵμÓ
ÓÃÏõËáËữµÄÏõËáÒø£¬¹Û²ì²úÉú°×É«³Áµí
ÓÃÏõËáËữµÄÏõËáÒø£¬¹Û²ì²úÉú°×É«³Áµí
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

¢ñÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÃèÊöÕýÈ·µÄÊÇ
 
£¨ÌîÐòºÅ£©£»
¢ÙʵÑéÊÒÅäÖÆÂÈ»¯ÑÇÌúÈÜҺʱ£¬½«ÂÈ»¯ÑÇÌúÏÈÈܽâÔÚÑÎËáÖУ¬È»ºóÓÃÕôÁóˮϡÊͲ¢¼ÓÈëÉÙÁ¿Ìú·Û£®
¢ÚÅäÖÆÒ»¶¨Å¨¶ÈµÄÈÜҺʱ£¬¸©ÊÓÈÝÁ¿Æ¿µÄ¿ÌÏߣ¬»áʹÅäÖƵÄŨ¶ÈÆ«¸ß£»ÊµÑéÊҲⶨÖкÍÈÈʱ£¬¹ýÔç¶ÁÊý»áʹ²â¶¨½á¹ûÆ«µÍ£®
¢Û½«Fe2£¨SO4£©3ÈÜÒº¼ÓÈÈÕô·¢ÖÁ¸É²¢×ÆÉÕ£¬×îºóµÃµ½ºì×ØÉ«·ÛÄ©
¢ÜʵÑéÊÒÓÃͭƬÓëÏ¡ÏõËá·´Ó¦²úÉúµÄÆøÌå¿ÉÓÃÅÅË®·¨ÊÕ¼¯
¢ÝÊÔ¹ÜÖмÓÈëÉÙÁ¿µí·Û£¬ÔÙ¼ÓÈëÒ»¶¨Á¿Ï¡ÁòËᣬ¼ÓÈÈ3-4·ÖÖÓ£¬È»ºó¼ÓÈëÒø°±ÈÜÒº£¬Æ¬¿Ìºó¹Ü±ÚÉÏÓС°Òø¾µ¡±³öÏÖ
¢ÞÏò°±Ë®ÖеμÓAl2£¨SO4£©3ÈÜÒººÍÏòAl2£¨SO4£©3ÈÜÒºÖеμӰ±Ë®ÏÖÏóÏàͬ
¢ß±½ÓëäåË®ÔÚÌú·ÛµÄ´ß»¯×÷ÓÃÏÂÖƱ¸äå±½
¢à·Ö±ðÏòÌå»ýºÍpH¾ùÏàͬµÄÑÎËáºÍ´×ËáÖеμӵÈŨ¶ÈµÄÇâÑõ»¯ÄÆÈÜÒº£¬ÍêÈ«ÖкÍʱÏûºÄµÄÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ýÒ»Ñù¶à
¢ò·ú»¯ÄÆÊÇÒ»ÖÖÖØÒªµÄ·úÑΣ¬Ö÷ÒªÓÃÓÚÅ©×÷Îïɱ¾ú¡¢É±³æ¡¢Ä¾²ÄµÄ·À¸¯£®ÊµÑéÊÒ¿Éͨ¹ýÏÂͼËùʾµÄÁ÷³ÌÒÔ·ú¹èËᣨH2SiF6£©µÈÎïÖÊΪԭÁÏÖÆÈ¡·ú»¯ÄÆ£¬²¢µÃµ½¸±²úÆ·ÂÈ»¯ï§£»
¾«Ó¢¼Ò½ÌÍø
ÒÑÖª£º20¡æʱÂÈ»¯ï§µÄÈܽâ¶ÈΪ37.2g£¬·ú»¯ÄƵÄÈܽâ¶ÈΪ4g£®Na2SiF6΢ÈÜÓÚË®£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÉÏÊöÁ÷³Ì¢ÙÖвúÉú»ë×ǵÄÔ­ÒòÊÇÉú³ÉÁË
 
£¨Ñ¡ÔñÐòºÅ£©£»
A£®NH4F   B£®H2SiO3C£®£¨NH4£©2SiF6 D£®£¨NH4£©2CO3
£¨2£©Ð´³öÁ÷³Ì¢ÚÖÐÏà¹ØµÄ»¯Ñ§·½³Ìʽ£º
 
£»
£¨3£©²Ù×÷IºÍ²Ù×÷
 
ÊÇÏàͬµÄ£¬²Ù×÷IIºÍ²Ù×÷VÊÇÏàͬµÄ£¬Æä²Ù×÷Ãû³ÆΪ
 
£»
£¨4£©Á÷³Ì¢ÙÖÐNH4HCO3±ØÐë¹ýÁ¿£¬ÆäÔ­ÒòÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ººþÄÏʦ´ó¸½ÖÐ2006£­2007ѧÄê¶ÈÉÏѧÆÚ¸ßÈýÔ¿¼ÊÔ¾í(Èý)¡¢»¯Ñ§ÊÔÌâ ÌâÐÍ£º022

ÓÐA¡¢BÁ½ÖÖÄÆÑΣ¬·Ö±ðΪÎÞÑõËáÑκͺ¬ÑõËáÑΣ¬¸÷È¡ÊÊÁ¿»ìºÏºó¼ÓÈÈ£¬Éú³ÉÆøÌåCºÍÖÖÑÎD£®C¼«Ò×ÈÜÓÚË®£¬ÆäŨÈÜÒºÓëMnO2y¹²ÈÈÉú³ÉÆøÌåE£®BºÍDÑÎÈÜÒº¾ù¿ÉÓëBaCl2ÈÜÒº·´Ó¦Éú³É²»ÈÜÓÚÏ¡ÏõËáµÄ°×É«³ÁµíF£®

(1)

ÊÔÍƶÏA¡¢F¸÷ΪºÎÖÖÎïÖÊ£¬Ð´³öÆ仯ѧʽ£ºA£º________¡¢F£º________£®

(2)

д³öÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

A£«B________£®

B£«BaCl2¡ú________£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

Çë´Ó¸ø³öµÄ3µÀ»¯Ñ§ÌâÖÐÈÎÑ¡Ò»Ìâ×ö´ð

1.£Û»¯Ñ§¡ª¡ªÑ¡ÐÞ»¯Ñ§Óë¼¼Êõ£Ý

Çë»Ø´ðÂȼҵÖеÄÈçÏÂÎÊÌ⣺

(1)ÂÈÆø¡¢ÉÕ¼îÊǵç½âʳÑÎˮʱ°´Õչ̶¨µÄ±ÈÂÊk(ÖÊÁ¿±È)Éú³ÉµÄ²úÆ·¡£ÀíÂÛÉÏk__________(ÒªÇó¼ÆËã±í´ïʽºÍ½á¹û)£»

(2)Ô­ÁÏ´ÖÑÎÖг£º¬ÓÐÄàɳºÍCa2+¡¢Mg2+¡¢Fe3+¡¢µÈÔÓÖÊ£¬±ØÐ뾫Öƺó²ÅÄܹ©µç½âʹÓ᣾«ÖÆʱ£¬´ÖÑÎÈÜÓÚË®¹ýÂ˺󣬻¹Òª¼ÓÈëµÄÊÔ¼Á·Ö±ðΪ¢ÙNa2CO3¡¢¢ÚHCl(ÑÎËá)¡¢¢ÛBaCl2£¬Õâ3ÖÖÊÔ¼ÁÌí¼ÓµÄºÏÀí˳ÐòÊÇ_________(ÌîÐòºÅ)£»

(3)ÂȼҵÊǸߺÄÄܲúÒµ£¬Ò»ÖÖ½«µç½â³ØÓëȼÁϵç³ØÏà×éºÏµÄй¤ÒÕ¿ÉÒÔ½Ú(µç)ÄÜ30%ÒÔÉÏ¡£ÔÚÕâÖÖ¹¤ÒÕÉè¼ÆÖУ¬Ïà¹ØÎïÁϵĴ«ÊäÓëת»¯¹ØϵÈçÏÂͼËùʾ£¬ÆäÖеĵ缫δ±ê³ö£¬ËùÓõÄÀë×ÓĤ¶¼Ö»ÔÊÐíÑôÀë×Óͨ¹ý¡£

¢ÙͼÖÐX¡¢Y·Ö±ðÊÇ_________¡¢_________(Ìѧʽ)£¬·ÖÎö±È½ÏͼʾÖÐÇâÑõ»¯ÄÆÖÊÁ¿·ÖÊýa%Óëb%µÄ´óС_________£»

¢Ú·Ö±ðд³öȼÁϵç³ØBÖÐÕý¼«¡¢¸º¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦

Õý¼«£º_________£»¸º¼«£º_________£»

¢ÛÕâÑùÉè¼ÆµÄÖ÷Òª½Ú(µç)ÄÜÖ®´¦ÔÚÓÚ(д³ö2´¦)

___________________________¡¢___________________________¡£

2.£Û»¯Ñ§¡ª¡ªÑ¡ÐÞÎïÖʽṹÓëÐÔÖÊ£Ý

ÒÑÖªX¡¢YºÍZÈýÖÖÔªËصÄÔ­×ÓÐòÊýÖ®ºÍµÈÓÚ42¡£XÔªËØÔ­×ÓµÄ4p¹ìµÀÉÏÓÐ3¸öδ³É¶Ôµç×Ó£¬YÔªËØÔ­×ÓµÄ×îÍâ²ã2p¹ìµÀÉÏÓÐ2¸öδ³É¶Ôµç×Ó¡£X¸úY¿ÉÐγɻ¯ºÏÎïX2Y3£¬ZÔªËØ¿ÉÒÔÐγɸºÒ»¼ÛÀë×Ó¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)XÔªËØÔ­×Ó»ù̬ʱµÄµç×ÓÅŲ¼Ê½Îª_________£¬¸ÃÔªËصķûºÅΪ_________£»

(2)YÔªËØÔ­×ӵļ۲ãµç×ӵĹìµÀ±íʾʽΪ_________£¬¸ÃÔªËصÄÃû³ÆÊÇ_________£»

(3)XÓëZ¿ÉÐγɻ¯ºÏÎïXZ3£¬¸Ã»¯ºÏÎïµÄ¿Õ¼ä¹¹ÐÍΪ________________________£»

(4)ÒÑÖª»¯ºÏÎïX2Y3ÔÚÏ¡ÁòËáÈÜÒºÖпɱ»½ðÊôп»¹Ô­ÎªXZ3£¬²úÎﻹÓÐZnSO4ºÍH2O£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ____________________________________£»

(5)±È½ÏXµÄÇ⻯ÎïÓëͬ×åµÚ¶þ¡¢ÈýÖÜÆÚÔªËØËùÐγɵÄÇ⻯ÎïÎȶ¨ÐÔ¡¢·Ðµã¸ßµÍ²¢ËµÃ÷ÀíÓÉ____________________________________¡£

3.£Û»¯Ñ§¡ª¡ªÑ¡ÐÞÓлú»¯Ñ§»ù´¡£Ý

A¡«J¾ùΪÓлú»¯ºÏÎËüÃÇÖ®¼äµÄת»¯ÈçÏÂͼËùʾ£º

 
ʵÑé±íÃ÷£º

¢ÙD¼ÈÄÜ·¢ÉúÒø¾µ·´Ó¦£¬ÓÖÄÜÓë½ðÊôÄÆ·´Ó¦·Å³öÇâÆø£»

¢ÚºË´Å¹²ÕñÇâÆ×±íÃ÷F·Ö×ÓÖÐÓÐÈýÖÖÇ⣬ÇÒÆä·åÃæ»ýÖ®±ÈΪ1¡Ã1¡Ã1£»

¢ÛGÄÜʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«£»

¢Ü1 mol JÓë×ãÁ¿½ðÊôÄÆ·´Ó¦¿É·Å³ö22.4 LÇâÆø(±ê×¼×´¿ö)¡£

Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺

(1)AµÄ½á¹¹¼òʽΪ_________(²»¿¼ÂÇÁ¢ÌåÒì¹¹)£¬ÓÉAÉú³ÉBµÄ·´Ó¦ÀàÐÍÊÇ_________·´Ó¦£»

(2)DµÄ½á¹¹¼òʽΪ___________________________£»

(3)ÓÉEÉú³ÉFµÄ»¯Ñ§·½³ÌʽΪ____________£¬EÖеĹÙÄÜÍÅÓÐ____________(ÌîÃû³Æ)£¬ÓëE¾ßÓÐÏàͬ¹ÙÄÜÍŵÄEµÄͬ·ÖÒì¹¹Ì廹ÓÐ____________(д³ö½á¹¹¼òʽ£¬²»¿¼ÂÇÁ¢ÌåÒì¹¹)£»

(4)GµÄ½á¹¹¼òʽΪ____________________________________£»

(5)ÓÉIÉú³ÉJµÄ»¯Ñ§·½³ÌʽΪ________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸