ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÖظõËá¼Ø(K2Cr2O7)ÊÇÒ»ÖÖ³£¼ûµÄÇ¿Ñõ»¯¼Á£¬ÄÜÑõ»¯ÁòËáÑÇÌú¡¢ÑÎËáµÈÎïÖÊ¡£Ä³ÐËȤС×éÄ£ÄâÆóÒµ´¦Àíº¬¸õ·ÏË®£¨Ö÷Òªº¬Cr2O72-ºÍCr3+)£¬Í¬Ê±»ñµÃÖظõËá¼Ø¾§ÌåµÄÁ÷³ÌÈçÏ£º

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©²Ù×÷IÊÇ______£¬²ÐÔüµÄÖ÷Òª³É·ÖÊÇ______¡£

£¨2£©µ÷½Ú³ØÖз¢ÉúµÄÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________¡£

£¨3£©²Ù×÷¢ó__________£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©ÓÃÑÎËáµ÷½ÚÈÜÒºpH£¬Ô­ÒòÊÇ___________¡£

£¨4£©´ÓNa2Cr2O7ÈÜÒºÖлñµÃK2Cr2O7¾§ÌåµÄ²Ù×÷ÒÀ´ÎÊǼÓÈëÊÊÁ¿KCl¹ÌÌ壬½Á°è¡¢Èܽ⣬ÔÚˮԡÉϼÓÈÈŨËõÖÁ________ʱֹͣ¼ÓÈÈ¡£½ÓÏÂÀ´»ñµÃK2Cr2O7¾§ÌåÐèÒªµÄһϵÁвÙ×÷ÖУ¬ÏÂÁÐÒÇÆ÷¿ÉÄÜ»áÓõ½µÄÊÇ__________£¨Ìî±êºÅ£©¡£

£¨5£©Îª¼ì²â´¦Àíºó·ÏË®ÊÇ·ñ´ïµ½Åŷűê×¼£¬Ä³Í¬Ñ§½øÐÐÁËÈçÏÂʵÑ飺ȡ100mL´¦ÀíºóµÄ·ÏÒºÑùÆ·ÓÚ׶ÐÎÆ¿ÖУ¬ÓÃŨ´×Ëáµ÷½ÚpH=5²¢¼ÓÈëÊÊÁ¿¹ÌÌ忹»µÑªËᣬʹCr2O72-Íêȫת»¯ÎªCr2+£¬ÔÙÓÃcmol¡¤ L-1µÄEDTA£¨ÓÃH4Y±íʾ£©±ê×¼ÈÜÒº½øÐе樣¬Æä·´Ó¦Ô­ÀíΪ£ºCr3++Y4-=CrY-£¬ÈôʵÑéÏûºÄEDTA±ê×¼ÈÜÒºVmL£¬Ôò´¦ÀíºóµÄ·ÏÒºÖк¬¸õÔªËØŨ¶ÈΪ______mg¡¤L-1£¨Óú¬c¡¢VµÄʽ×Ó±íʾ£©¡£

¡¾´ð°¸¡¿ ¹ýÂË Fe(OH)3 6Fe2++Cr2O72-+14H+=6Fe3++2Cr3++7H2O ²»ÄÜ ÔÚËáÐÔÈÜÒºÖÐCr2O72-¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ÄÜÑõ»¯ÑÎËá ÈÜÒº±íÃæ³öÏÖ¾§Ä¤ AD 520cV

¡¾½âÎö¡¿±¾Ì⿼²é»¯Ñ§¹¤ÒÕÁ÷³Ì£¬£¨1£©²Ù×÷IµÃµ½²ÐÔüºÍÂËÒº£¬Òò´Ë²Ù×÷IÊǹýÂË£¬ÖظõËá¼Ø(K2Cr2O7)ÊÇÒ»ÖÖ³£¼ûµÄÇ¿Ñõ»¯¼Á£¬ÄÜÑõ»¯ÁòËáÑÇÌú£¬µ÷½Ú³ØÖÐCr2O72£­°ÑFe2£«Ñõ»¯³ÉFe3£«£¬±¾Éí±»»¹Ô­³ÉCr3£«£¬·´Ó¦²ÛÖмÓÈëNaOHÈÜÒº£¬Fe3£«ºÍOH£­·´Ó¦Éú³ÉFe(OH)3³Áµí£¬Òò´Ë²ÐÔüµÄ³É·ÖÊÇFe(OH)3£»£¨2£©¸ù¾Ý(1)µÄ·ÖÎö£¬Àë×Ó·´Ó¦·½³ÌΪCr2O72£­£«Fe2£«¡úFe3£«£«Cr3£«£¬¸ù¾Ý»¯ºÏ¼ÛÉý½µ·¨½øÐÐÅäƽ£¬¼´Cr2O72£­£«6Fe2£«¡ú6Fe3£«£«2Cr3£«£¬¼ÓÈëFeSO4µÄͬʱ£¬»¹¼ÓÈëÁËH2SO4£¬Òò´ËÀë×Ó·½³ÌʽΪ£º6Fe2£«+Cr2O72£­+14H£«=6Fe3£«+2Cr3£«+7H2O £»£¨3£©ÖظõËá¼Ø(K2Cr2O7)ÊÇÒ»ÖÖ³£¼ûµÄÇ¿Ñõ»¯¼Á£¬ÄÜÑõ»¯ÁòËáÑÇÌú¡¢ÑÎËᣬNa2CrO4ÈÜÒº´æÔÚ2CrO42£­£«2H£«Cr2O72£­£«H2O£¬¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ÄÜ°ÑÑÎËáÑõ»¯£¬Òò´Ë²»ÄÜÓÃÑÎËáµ÷½ÚpH£»£¨4£©Na2Cr2O7ÖƱ¸K2Cr2O7£¬ËµÃ÷K2Cr2O7µÄÈܽâ¶ÈСÓÚNa2Cr2O7£¬Òò´Ëµ±¼ÓÈȳöÏÖ¾§Ä¤Ê±£¬Í£Ö¹¼ÓÈÈ£»½ÓÏÂÀ´µÄ²Ù×÷ÊÇÀäÈ´¡¢¹ýÂË£¬Òò´ËÐèÒªµÄÒÇÆ÷ÊÇAºÍD£»£¨5£©¸ù¾ÝÔªËØÊغãºÍ·´Ó¦Ô­Àí£¬½¨Á¢¹ØϵʽΪCr2O72£­¡«2Cr3£«¡«2Y4£­¡«2H4Y£¬Òò´Ë¸õÔªËصÄÖÊÁ¿ÎªV¡Á10£­3¡Ác¡Á52¡Á103mg=52Vcmg£¬ÔòŨ¶ÈΪ52Vc/(100¡Á10£­3)mg¡¤L£­1=520Vc mg¡¤L£­1¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿1£¬2-¶þäåÒÒÍé¿É×÷¿¹±¬¼ÁµÄÌí¼Ó¼Á¡£ÈçͼΪʵÑéÊÒÖƱ¸1£¬2-¶þäåÒÒÍéµÄ×°ÁDͼ£¬ ͼÖзÖÒºÖƶ·ºÍÉÕÆ¿aÖзֱð×°ÓÐŨH2SO4ºÍÎÞË®ÒÒ´¼£¬d×°ÁDÊÔ¹ÜÖÐ×°ÓÐÒºäå¡£

¼ºÖª£ºCH3CH2OHCH2=CH2¡ü+H2O£»2CH3CH2OHCH3CH2OCH2CH3+H2O

Ïà¹ØÊý¾ÝÁбíÈçÏ£º

ÒÒ´¼

1£¬2-¶þäåÒÒÍé

ÒÒÃÑ

äå

״̬

ÎÞÉ«ÒºÌå

ÎÞÉ«ÒºÌå

ÎÞÉ«ÒºÌå

ºì×ØÉ«ÒºÌå

ÃܶÈ/g¡¤cm-3

0.79

2.18

0.71

3.10

·Ðµã/¡æ

78.5

131.4

34.6

58.8

ÈÛµã/¡æ

-114.3

9.79

- 116.2

-7.2

Ë®ÈÜÐÔ

»ìÈÜ

ÄÑÈÜ

΢ÈÜ

¿ÉÈÜ

£¨1£©ÊµÑéÖÐӦѸËÙ½«Î¶ÈÉý…lµ½170¡æ×óÓÒµÄÔ­ÒòÊÇ______________________________¡£

£¨2£©°²È«Æ¿bÔÚʵÑéÖÐÓжàÖØ×÷Óá£ÆäÒ»¿ÉÒÔ¼ì²éʵÑé½øÐÐÖÐd×°ÁDÖе¼¹ÜÊÇ·ñ·¢Éú¶ÂÈû£¬

Çëд³ö·¢Éú¶ÂÈûʱƿbÖеÄÏÖÏ󣺢Ù_______________________________£»Èç¹ûʵÑéʱd×°ÁDÖе¼¹Ü¶ÂÈû£¬ÄãÈÏΪ¿ÉÄܵÄÔ­ÒòÊÇ¢Ú_______________________________________________£»°²È«Æ¿b»¹¿ÉÒÔÆðµ½µÄ×÷ÓÃÊÇ¢Û__________________¡£

£¨3£©ÈÝÆ÷c¡¢eÖж¼Ê¢ÓÐNaOHÈÜÒº£¬cÖÐNaOHÈÜÒºµÄ×÷ÓÃÊÇ________________________________¡£

______________¡¢______________£¨Ð´³öÁ½Ìõ¼´¿É£©¡£

£¨5£©³ýÈ¥²úÎïÖÐÉÙÁ¿Î´·´Ó¦µÄBr2ºó£¬»¹º¬ÓеÄÖ÷ÒªÔÓÖÊΪ___________£¬Òª½øÒ»²½Ìá´¿£¬ÏÂÁвÙ×÷ÖбØÐèµÄÊÇ_____________ £¨Ìî×Öĸ£©¡£

A£®Öؽᾧ B£®¹ýÂË C£®ÝÍÈ¡ D£®ÕôÁó

£¨6£©ÊµÑéÖÐÒ²¿ÉÒÔ³·È¥d×°ÁDÖÐÊ¢±ùË®µÄÉÕ±­£¬¸ÄΪ½«Àäˮֱ½Ó¼ÓÈëµ½d×°ÁDµÄÊÔ¹ÜÖУ¬Ôò ´ËʱÀäË®³ýÁËÄÜÆðµ½ÀäÈ´1£¬2-¶þäåÒÒÍéµÄ×÷ÓÃÍ⣬»¹¿ÉÒÔÆðµ½µÄ×÷ÓÃÊÇ____________________________¡£

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

¾«Ó¢¼Ò½ÌÍø