ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿K¡¢Ka¡¢KW·Ö±ð±íʾ»¯Ñ§Æ½ºâ³£Êý¡¢µçÀë³£ÊýºÍË®µÄÀë×Ó»ý³£Êý£¬ÏÂÁÐÅжÏÕýÈ·µÄÊÇ£¨ £©

A. ÔÚ500¡æ¡¢20 MPaÌõ¼þÏ£¬ÔÚ5 LÃܱÕÈÝÆ÷ÖнøÐкϳɰ±µÄ·´Ó¦£¬Ê¹Óô߻¯¼ÁºóKÔö´ó

B. ÊÒÎÂÏÂKa(HCN) < Ka(CH3COOH)£¬ËµÃ÷CH3COOHµçÀë²úÉúµÄc(H+)Ò»¶¨±ÈHCNµçÀë²úÉúµÄc(H+)´ó

C. 25¡æʱ£¬pH¾ùΪ4µÄÑÎËáºÍNH4IÈÜÒºÖÐKW²»ÏàµÈ

D. 2SO2£«O22SO3´ïƽºâºó£¬¸Ä±äijһÌõ¼þʱK²»±ä£¬SO2µÄת»¯ÂÊ¿ÉÄÜÔö´ó¡¢¼õС»ò²»±ä

¡¾´ð°¸¡¿D

¡¾½âÎö¡¿A£®»¯Ñ§Æ½ºâ³£ÊýÖ»ÓëζÈÓйأ¬Î¶Ȳ»±ä»¯Ñ§Æ½ºâ³£Êý²»±ä£¬Ê¹Óô߻¯¼Á¼Ó¿ì·´Ó¦ËÙÂÊ£¬Æ½ºâ²»Òƶ¯£¬»¯Ñ§Æ½ºâ³£Êý²»±ä£¬A´íÎó£»B£®ÏàͬζÈÏÂCH3COOH¡¢HCNµÄµçÀë³öµÄÇâÀë×ÓŨ¶ÈÓëÆðʼŨ¶ÈÓйأ¬B´íÎó£»C£®Ë®µÄÀë×Ó»ýÊÊÓÃÓÚËá¡¢¼î¡¢ÑÎÈÜÒº£¬Ò»¶¨Î¶ÈÏ£¬Ë®µÄÀë×Ó»ýÊdz£Êý£¬25¡æʱ£¬ÑÎËáºÍNH4I(aq)ÖÐKWÏàµÈ£¬C´íÎó£»D£®¸Ä±äѹǿƽºâ·¢ÉúÒƶ¯£¬SO2µÄת»¯ÂÊ¿ÉÄÜÔö´ó¡¢¼õС£¬Ê¹Óô߻¯¼Áƽºâ²»Òƶ¯£¬DÕýÈ·£»´ð°¸Ñ¡D¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿KMnO4ÊÇÖÐѧ»¯Ñ§³£¼ûµÄÊÔ¼Á¡£

¢ñ£®Ä³Ð¡×éÄâÓÃËáÐÔKMnO4ÈÜÒºÓëH2C2O4£¨Èõµç½âÖÊ£©µÄ·´Ó¦À´Ì½¾¿¡°Íâ½çÌõ¼þ¶Ô»¯Ñ§·´Ó¦ËÙÂʵÄÓ°Ï족£¬²¢Éè¼ÆÁËÈçϵķ½°¸¼Ç¼ʵÑé½á¹û£¨ºöÂÔÈÜÒº»ìºÏÌå»ý±ä»¯£©¡£ÏÞÑ¡ÊÔ¼ÁºÍÒÇÆ÷£º0.20mol£¯L H2C2O4ÈÜÒº¡¢0.010mol£¯L KMnO4ËáÐÔÈÜÒº¡¢ÕôÁóË®¡¢ÊԹܡ¢Á¿Í²¡¢Ãë±í¡¢ºãÎÂˮԡ²Û

ÎïÀíÁ¿

񅧏

V£¨0.20 mol£¯L

H2C2O4ÈÜÒº£©£¯mL

¼×

V£¨0.010mol£¯L

KMnO4ÈÜÒº£©£¯mL

M£¨MnSO4¹ÌÌ壩£¯g

T£¯¡æ

ÒÒ

¢Ù

2.0

0

4.0

0

50

¢Ú

2.0

0

4.0

0

25

¢Û

1.0

a

4.0

0

25

¢Ü

2.0

0

4.0

0.1

25

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Íê³ÉÉÏÊö·´Ó¦Ô­ÀíµÄÀë×Ó·´Ó¦·½³Ìʽ________

£¨2£©ÉÏÊöʵÑé¢Ù¢ÚÊÇ̽¾¿________¶Ô»¯Ñ§·´Ó¦ËÙÂʵÄÓ°Ï죻ÈôÉÏÊöʵÑé¢Ú¢ÛÊÇ̽¾¿Å¨¶È¶Ô»¯Ñ§·´Ó¦ËÙÂʵÄÓ°Ï죬Ôò±í¸ñÖС°¼×¡±Ó¦Ìîд________£¬aΪ________£»

ÒÒÊÇʵÑéÐèÒª²âÁ¿µÄÎïÀíÁ¿£¬Ôò±í¸ñÖС°ÒÒ¡±Ó¦Ìîд________¡£

ÉÏÊöʵÑé¢Ú¢ÜÊÇ̽¾¿________¶Ô»¯Ñ§·´Ó¦ËÙÂʵÄÓ°Ïì¡£

¢ò£®Ä³»¯Ñ§Ð¡×éÀûÓÃÏÂͼװÖýøÐÐʵÑéÖ¤Ã÷Ñõ»¯ÐÔ£ºKMnO4£¾Cl2£¾Br2¡£ÏÞÑ¡ÊÔ¼Á£ºKBrÈÜÒº¡¢KMnO4ÈÜÒº¡¢Å¨ÑÎËᡢŨÁòËá

£¨3£©×°ÖÃa¡¢dÖÐÊ¢·ÅµÄÊÔ¼Á·Ö±ðÊÇ£º________¡¢________£»

£¨4£©ÊµÑéÖй۲쵽µÄÏÖÏóΪ________________________________£»

´ËʵÑé×°ÖõIJ»×ãÖ®´¦ÊÇ____________________¡£

¡¾ÌâÄ¿¡¿¢ñ.³ÁµíµÄÉú³É¡¢ÈܽâºÍת»¯ÔÚÎÞ»úÎïÖƱ¸ºÍÌá´¿ÒÔ¼°¿ÆÑеÈÁìÓòÓй㷺ӦÓá£

£¨1£©ÒÑÖª25¡æʱ£¬Ksp(BaSO4)£½1¡Á10£­10£¬½«BaSO4µÄÐü×ÇÒº¹ýÂË£¬ÂËÒºÖÐc(Ba2£«)£½_______mol¡¤L£­1¡£È¡100 mLÂËÒºÓë100 mL 2 mol¡¤L£­1µÄNa2SO4ÈÜÒº»ìºÏ£¬»ìºÏÒºÖÐc(Ba2£«)£½___________ mol¡¤L£­1¡£

£¨2£©³¤ÆÚʹÓõĹø¯ÐèÒª¶¨ÆÚ³ýË®¹¸£¬·ñÔò»á½µµÍȼÁϵÄÀûÓÃÂÊ¡£Ë®¹¸Öк¬ÓеÄCaSO4£¬¿ÉÏÈÓÃNa2CO3ÈÜÒº´¦Àí£¬Ê¹Ö®×ª»¯ÎªÊèËÉ¡¢Ò×ÈÜÓÚËáµÄCaCO3£¬¶øºóÓÃËá³ýÈ¥¡£

¢ÙCaSO4ת»¯ÎªCaCO3µÄÀë×Ó·½³ÌʽΪ£º_____________________________________________________

¢ÚÇë·ÖÎöCaSO4ת»¯ÎªCaCO3µÄÔ­Àí£º______________________________________________________

¢ò.ÔÚ25 ¡æʱ£¬HSCN¡¢HClO¡¢H2CO3µÄµçÀë³£ÊýÈçÏÂ±í£º

HClO

HSCN

H2CO3

K£½3.2¡Á10£­8

K£½0.13

K1£½4.2¡Á10£­7

K2£½5.6¡Á10£­11

£¨1£©1 mol¡¤L£­1µÄKSCNÈÜÒºÖУ¬ËùÓÐÀë×ÓµÄŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ_________________________________

£¨2£©ÏòNa2CO3ÈÜÒºÖмÓÈë¹ýÁ¿HClOÈÜÒº£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________________________

£¨3£©25 ¡æʱ£¬ÎªÖ¤Ã÷HClOΪÈõËᣬijѧϰС×éµÄͬѧÉè¼ÆÁËÒÔÏÂÈýÖÖʵÑé·½°¸¡£ÏÂÁÐÈýÖÖ·½°¸ÖУ¬ÄãÈÏΪÄܹ»´ïµ½ÊµÑéÄ¿µÄµÄÊÇ______________________(ÌîÏÂÁи÷ÏîÖÐÐòºÅ)¡£

a£®ÓÃpH¼Æ²âÁ¿0.1mol¡¤L£­1NaClOÈÜÒºµÄpH£¬Èô²âµÃpH>7£¬¿ÉÖ¤Ã÷HClOΪÈõËá

b£®ÓÃpHÊÔÖ½²âÁ¿0.01 mol¡¤L£­1 HClOÈÜÒºµÄpH£¬Èô²âµÃpH>2£¬¿ÉÖ¤Ã÷HClOΪÈõËá

c£®ÓÃÒÇÆ÷²âÁ¿Å¨¶È¾ùΪ0.1 mol¡¤L£­1µÄHClOÈÜÒººÍÑÎËáµÄµ¼µçÐÔ£¬Èô²âµÃHClOÈÜÒºµÄµ¼µçÐÔÈõÓÚÑÎËᣬ¿ÉÖ¤Ã÷HClOΪÈõËá

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

¾«Ó¢¼Ò½ÌÍø