ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÈçͼËùʾµÄͼ½â±íʾ¹¹³Éϸ°ûµÄÔªËØ¡¢»¯ºÏÎa¡¢b¡¢c¡¢d´ú±í²»Í¬µÄС·Ö×ÓÎïÖÊ£¬A¡¢B¡¢C´ú±í²»Í¬µÄ´ó·Ö×ÓÎïÖÊ£¬Çë·ÖÎö»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÎïÖÊaÊÇ_________£¬ÔÚ¶¯Îïϸ°ûÄÚ£¬ÓëÎïÖÊA×÷ÓÃ×îÏà½üµÄÎïÖÊÊÇ_________¡£ÈôÎïÖÊAÔÚ¶¯Îï¡¢Ö²Îïϸ°ûÖоù¿Éº¬ÓУ¬²¢ÇÒ×÷Ϊϸ°ûÄÚ×îÀíÏëµÄ´¢ÄÜÎïÖÊ£¬²»½öº¬ÄÜÁ¿¶à¶øÇÒÌå»ý½ÏС£¬ÔòAÊÇ______________¡£

£¨2£©ÎïÖÊbÊÇ____________£¬ÈôijÖÖB·Ö×ÓÓÉn¸öb·Ö×Ó£¨Æ½¾ùÏà¶Ô·Ö×ÓÖÊÁ¿Îªm£©×é³ÉµÄ2ÌõÁ´×é³É£¬Ôò¸ÃB·Ö×ÓµÄÏà¶Ô·Ö×ÓÖÊÁ¿´óԼΪ____________¡£

£¨3£©ÎïÖÊcÔÚÈËÌåϸ°ûÖй²ÓÐ_______ÖÖ£¬·Ö×ÓÖÐ___________µÄ²»Í¬¾ö¶¨ÁËcµÄÖÖÀ಻ͬ¡£

£¨4£©ÎïÖÊdÊÇ____________£¬dºÍ__________¡¢Î¬ÉúËØD¶¼ÊôÓڹ̴¼ÀàÎïÖÊ¡£

¡¾´ð°¸¡¿ÆÏÌÑÌÇ ÌÇÔ­ Ö¬·¾ °±»ùËá mn£­18£¨n£­2£© 4 º¬µª¼î»ù ÐÛÐÔ¼¤ËØ µ¨¹Ì´¼

¡¾½âÎö¡¿

ÔĶÁÌâ¸ÉºÍÌâͼ¿ÉÖª£¬¸ÃÌâµÄ֪ʶµãÊÇÌÇÀàµÄ·ÖÀà¡¢·Ö²¼ºÍ¹¦ÄÜ£¬Ö¬ÖʵķÖÀàºÍ¹¦ÄÜ£¬µ°°×ÖʵĻù±¾×é³Éµ¥Î»¡¢½á¹¹ºÍ¹¦ÄÜ£¬ºËËáµÄ·ÖÀà¡¢·Ö²¼½á¹¹ºÍ¹¦ÄÜ£¬Ïȸù¾ÝÌâͼÊáÀíÏà¹Ø֪ʶµã£¬È»ºó½áºÏÌâ¸ÉÐÅÏ¢½øÐнâ´ð¡£Í¼ÖÐAÊǵí·Û£¬aÊÇÆÏÌÑÌÇ£¬BÊǵ°°×ÖÊ£¬bÊÇ°±»ùËᣬCÊÇRNA£¬cÊǺËÌǺËÜÕËᣬdÊÇÐÔ¼¤ËØ¡£

£¨1£©Ö²Îïϸ°ûµÄ´ó·Ö×Ó´¢ÄÜÎïÖÊÊǵí·Û»ù±¾×é³Éµ¥Î»ÊÇÆÏÌÑÌÇ£¬¶¯Îïϸ°ûÖд¢´æÄÜÁ¿µÄ¶àÌÇÊÇÌÇÔ­£»Ö¬·¾ÊǶ¯Ö²Îïϸ°û¶¼º¬ÓеÄÎïÖÊ£¬ÏàͬÖÊÁ¿µÄÖ¬·¾º¬ÓеÄÄÜÁ¿´óÓÚ¶àÌÇ£¬ÊǶ¯Ö²Îïϸ°û¹²ÓеÄÁ¼ºÃµÄ´¢ÄÜÎïÖÊ¡£
£¨2£©bÊǵ°°×ÖʵĻù±¾×é³Éµ¥Î»°±»ùËᣬ¶àëÄÁ´µÄÏà¶Ô·Ö×ÓÖÊÁ¿=°±»ùËá¸öÊý¡Á°±»ùËáƽ¾ù·Ö×ÓÖÊÁ¿-Ë®·Ö×ÓÊý£¨°±»ùËá¸öÊý-ëÄÁ´Êý£©¡ÁË®µÄÏà¶Ô·Ö×ÓÖÊÁ¿£¬ÔòB·Ö×ÓµÄÏà¶Ô·Ö×ÓÖÊÁ¿´óԼΪmn£­18£¨n£­2£©¡£

£¨3£©cÊǺËÌǺËÜÕËᣬºËÌǺËÜÕËẬÓеļî»ùÊÇA¡¢U¡¢C¡¢G£¬¸ù¾Ýº¬µª¼î»ùµÄ²»Í¬ÐγÉ4ÖÖºËÌǺËÜÕËá¡£

£¨4£©´ÓdµÄ¹¦ÄÜ¿ÉÒÔ¿´³ö£¬dÊÇÐÛÐÔ¼¤ËØ£»¹Ì´¼ÀàÎïÖÊ°üÀ¨ÐÔ¼¤ËØ¡¢Î¬ÉúËØDºÍµ¨¹Ì´¼¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ºì·¯ÄÆ(Na2Cr2O7¡¤2H2O)¹ã·ºÓÃ×÷Ç¿Ñõ»¯¼Á¡¢÷·¸ï¼Á¡£ÒÔ¸õ¿óʯ(Ö÷Òª³É·ÖΪCr2O3,»¹º¬ÓÐFeO¡¢Al2O3¡¢SiO2µÈÔÓÖÊ)ΪԭÁÏÖÆÈ¡ºì·¯ÄƵÄÁ÷³ÌÈçÏÂ:

ÒÑÖª:¢ÙCrO42-ÓëCr2O72-´æÔÚÈçÏÂƽºâ:2CrO42-+2H+Cr2O72-+H2O£¬µ±pH<3ʱ£¬ÒÔCr2O72-ΪÖ÷£¬µ±pH>9ʱ£¬ÒÔCrO42-ΪÖ÷¡£

¢ÚÄÆÑÎÔÚ²»Í¬Î¶ÈϵÄÈܽâ¶È(g):

ζÈ(¡æ)

10

20

30

40

50

60

70

80

Na2SO4

9.0

19.4

40.8

48.8

46.7

45.3

44.1

43.7

Na2Cr2O7

170.2

180.1

196.7

220.5

248.4

283.1

323.8

385.4

×¢32.38¡æÒÔÉÏ£¬Óë±¥ºÍÈÜҺƽºâµÄ¹ÌÏàΪÎÞË®Na2SO4,ÒÔÏÂÔòΪNa2SO4¡¤10H2O¡£

¢ÛCr3+ÍêÈ«³ÁµíʱpHΪ6.8£¬Cr(OH)3¿ªÊ¼ÈܽâʱpHΪ12¡£

Çë»Ø´ðÏÂÁÐÎÊÌâ:

(1)ìÑÉÕ¸õ¿óʯʱ£¬Éú³ÉNa2CrO4µÄ»¯Ñ§·´Ó¦·½³ÌʽΪ_________

(2)ÂËÔü¢òµÄ³É·ÖÊÇ__________(Ìѧʽ)¡£

(3)ÓÐÈËÈÏΪ¹¤ÒÕÁ÷³ÌÖС°ÓÃÏ¡ÁòËáµ÷pH¡±¸ÄΪ¡°Í¨Èë¹ýÁ¿CO2¡±£¬²»Ðèµ÷½ÚpHͬÑù¿ÉÒԴﵽʵÑéЧ¹û£¬ÀíÓÉÊÇ___________¡£

(4)Ïòºì·¯ÄÆÈÜÒºÖмÓÈëÊÊÁ¿KCl,Õô·¢Å¨Ëõ£¬ÀäÈ´½á¾§£¬¹ýÂ˵õ½K2Cr2O7¹ÌÌå¡£ÀäÈ´µ½________(ÌîÐòºÅ)µÃµ½µÄK2Cr2O7¹ÌÌå²úÆ·×î¶à¡£

A.80 B.60 C.40 D.10

(5)¸Ã¹¤ÒÕËùµÃ¸±²úÆ·Ö÷ҪΪÎÞË®ÁòËáÄƲ¢»ìÓÐÉÙÁ¿ÖظõËáÄÆ£¬ÇëÉè¼Æ´Ó¸±²úÆ·»ñµÃâÏõ(Na2SO4¡¤10H2O)µÄʵÑé·½°¸:½«¸Ã¸±²úÆ·°´¹ÌÒºÖÊÁ¿±È100:230ÈÜÓÚÈÈË®£¬¼ÓÈëÉÔ¹ýÁ¿µÄNa2SO3ÈÜÒº£¬½Á°è,____£¬¹ýÂË£¬Ï´µÓ£¬µÍθÉÔï¡£(ʵÑéÖÐÐëʹÓõÄÊÔ¼Á:Ï¡H2SO4¡¢NaOHÈÜÒº)¡£

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

¾«Ó¢¼Ò½ÌÍø