£¨10·Ö£©ÊµÑéÊÒÐèÒª0.10 mol/L NaOHÈÜÒº475mLºÍ0.40 mol/L¡£ÁòËáÈÜÒº500 mL¡£¸ù¾ÝÕâÁ½ÖÖÈÜÒºµÄÅäÖÆÇé¿ö»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈçͼËùʾµÄÒÇÆ÷ÖÐÅäÖÆÈÜÒº¿Ï¶¨²»ÐèÒªµÄÊÇ      £¨ÌîÐòºÅ£©£¬ÅäÖÆÉÏÊöÈÜÒº»¹ÐèÓõ½µÄ²£Á§ÒÇÆ÷ÊÇ                      £¨ÌîÒÇÆ÷Ãû³Æ£©¡£
£¨2£©ÈËÃdz£½«ÅäÖƹý³Ì¼òÊöΪÒÔϸ÷²½Ö裺
A£®ÀäÈ´B£®³ÆÁ¿C£®Ï´µÓD£®¶¨ÈÝ E£®Èܽâ F£®Ò¡ÔÈ¡£ G£®ÒÆÒº
ÕýÈ·µÄ²Ù×÷˳ÐòÓ¦ÊÇ                  £¨Ìî¸÷²½ÖèÐòºÅ£©¡£
£¨3£©¼ÆËã¸ÃʵÑéÓÃÍÐÅÌÌìƽ³ÆÈ¡NaOHµÄÖÊÁ¿Îª      g¡£ÔÚʵÑéÖÐÆäËû²Ù×÷¾ùÕýÈ·£¬Èô¶¨ÈÝʱ¸©ÊÓÊӿ̶ÈÏߣ¬¡¢ÔòËùµÃÈÜҺŨ¶È       0.lmol/L£¨Ìî¡°´óÏ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£¬ÏÂͬ£©¡£ÈôÒÆҺʱ£¬ÈÝÁ¿Æ¿ÄÚÒÑÓÐÉÙÐíÕôÁóË®£¬ÔòËùµÃÈÜҺŨ¶È   0.lmol£¯L¡£  
£¨4£©¸ù¾Ý¼ÆËãµÃÖª£¬ËùÐèÖÊÁ¿·ÖÊýA 98%¡¢ÃܶÈΪ1184g/cm3µÄŨÁòËáµÄÌå»ýΪ____mL£¨±£ÁôһλСÊý£©¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨12·Ö£©£®ÏÖÓк¬NaCl¡¢Na2SO4ºÍNaNO3µÄ»ìºÏÎѡÔñÊʵ±µÄÊÔ¼Á½«Æäת»¯ÎªÏàÓ¦µÄ³Áµí»ò¹ÌÌ壬´Ó¶øµÃµ½½ÏΪ´¿¾»µÄNaNO3¡£ÊµÑéÁ÷³Ì¿ÉÓÃÏÂͼ±íʾ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öʵÑéÁ÷³ÌÖÐÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£º
ÊÔ¼ÁX£º      £¬³ÁµíB£º           ¡£
£¨2£©ÊµÑéÁ÷³ÌÖмÓÈë¹ýÁ¿µÄNa2CO3µÄÄ¿µÄÊÇ             ¡£
£¨3£©ÊµÑé·½°¸µÃµ½µÄÈÜÒº3Öп϶¨º¬ÓР     £¨Ìѧʽ£©ÔÓÖÊ£»ÎªÁ˽â¾öÕâ¸öÎÊÌ⣬¿ÉÒÔÏòÈÜÒº3ÖмÓÈëÊÊÁ¿µÄ    £¬Çëд³öËù·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ_______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

(15·Ö)ij»¯Ñ§ÐËȤС×éÐè100 mLijÎïÖʵÄÁ¿Å¨¶ÈµÄFeSO4ÈÜÒº×÷±ê×¼ÊÔ¼Á£¬ÏÖÓÐÍâ¹ÛÑÕÉ«·¢»ÆµÄÁòËáÑÇÌú¾§Ì塢ŨÁòËᡢϡÑÎËá¡¢KMnO4ÈÜÒº¡¢äåË®¡¢KSCNÈÜÒº¡¢Ê¯Èï¡¢·Ó̪¼°ÖÐѧ»¯Ñ§³£Óû¯Ñ§ÒÇÆ÷£¬ÊµÑé¹ý³ÌÈçÏ£º
¢ñ.ÅäÖÆFeSO4ÈÜÒº
(1)ÏÂÁйØÓÚFeSO4ÈÜÒºÅäÖƲÙ×÷²½ÖèµÄºÏÀí˳ÐòΪ________(Ìî×Öĸ)¡£
A£®ÔÚÊ¢ÊÊÁ¿Ë®µÄÉÕ±­ÖеμÓÉÙÁ¿Å¨H2SO4ºó½Á°è¾ùÔȲ¢ÀäÈ´µ½ÊÒÎÂ
B£®³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄFeSO4¾§ÌåÑùÆ·
C£®½«ÑùÆ·ÈÜÓÚÒÑÅäÖƵÄÏ¡H2SO4ÖУ¬Óò£Á§°ô½Á°èÖÁÑùÆ·³ä·ÖÈܽâ
D£®¹ýÂ˺󣬽«ÂËҺתÒƵ½ÈÝÁ¿Æ¿ÖУ¬¼ÓˮϡÊÍÖÁÖ¸¶¨¿Ì¶È
E£®ÍùÁòËáÑÇÌúÈÜÒºÖмÓÈëÉÔ¹ýÁ¿µÄÌú·Û£¬³ä·Ö½Á°è£¬¾²ÖÃÒ»¶Îʱ¼äÖÁ²»ÔÙÓÐÆøÌåð³öΪֹ
(2)»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù½«ÑùÆ·ÈÜÓÚÏ¡H2SO4£¬¶ø²»Ö±½ÓÈÜÓÚË®µÄÀíÓÉÊÇ_____________________________________________________________________¡£
¢ÚÔÚÅäÖÆÁòËáÑÇÌúÈÜҺʱ£¬Ðè¼ÓÈëÌú·ÛµÄÀíÓÉÊÇ
________________________________________________________________________¡£
¢ò.±ê¶¨FeSO4ÈÜÒºµÄŨ¶È
(1)ÓÃÒÆÒº¹ÜÁ¿È¡20.00 mL FeSO4ÈÜÒº·ÅÈë׶ÐÎÆ¿ÖУ¬ÓÃ0.10 mol¡¤L£­1µÄËáÐÔKMnO4ÈÜÒºµÎÖÁÖյ㣬ºÄÈ¥KMnO4ÈÜÒº20.00 mL£¬ÈôÉú³ÉÎïÖÐMnÔªËØÈ«²¿³Ê£«2¼Û£¬µÎ¶¨·´Ó¦µÄÀë×Ó·½³ÌʽΪ______                                 __£¬¾Ý´Ë¿É²âµÃFeSO4ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ______  __mol¡¤L£­1¡£
(2)µÎ¶¨Ê±Ñ¡ÓÃ________(A.ʯÈï¡¡B£®·Ó̪¡¡C£®²»ÓÃָʾ¼Á£¬Ìî×Öĸ)Ϊָʾ¼Á£¬ÀíÓÉÊÇ____________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨14·Ö£©ÓÒͼΪʵÑéÊÒijŨÑÎËáÊÔ¼ÁÆ¿ÉϵıêÇ©£¬ÊÔ¸ù¾ÝÓйØÊý¾Ý»Ø´ðÏÂÁÐÎÊÌâ

(1)¸ÃŨÑÎËáÖÐHClµÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________mol/L¡£
(2)ÅäÖÆ250mL 0.1mol/LµÄÑÎËáÈÜÒº  
Ó¦Á¿È¡Å¨ÑÎËáÌå»ý/mL
ӦѡÓÃÈÝÁ¿Æ¿µÄ¹æ¸ñ/mL
 
 
A£®ÓÃÁ¿Í²Á¿È¡ËùÐèµÄŨÑÎËáµÄÌå»ý£¬Ñز£Á§°ôµ¹ÈëÉÕ±­ÖУ¬ÔÙ¼ÓÈëÕôÁóË®£¬Óò£Á§°ôÂýÂý½Á¶¯£¬Ê¹Æä»ìºÏ¾ùÔÈ¡£
B£®½«ÒÑÀäÈ´µÄÑÎËáÑز£Á§°ô×¢ÈëÈÝÁ¿Æ¿ÖС£
C£®ÓÃÕôÁóˮϴµÓ                       2¡ª3´Î£¬Ï´µÓÒº¾ù×¢ÈëÈÝÁ¿Æ¿£¬Õñµ´¡£
D£®¼ÌÐøÍùÈÝÁ¿Æ¿ÄÚСÐļÓË®£¬Ö±µ½ÒºÃæ½Ó½ü¿Ì¶ÈÏß1~2cm´¦£¬¸ÄÓà              ¼ÓË®£¬Ê¹ÈÜÒº°¼ÃæÇ¡ºÃÓë¿Ì¶ÈÏàÇС£
E£®½«ÈÝÁ¿Æ¿¸Ç½ô£¬Õñµ´£¬Ò¡ÔÈ¡£
(3) ÈôʵÑéÖгöÏÖÏÂÁÐÏÖÏó¶ÔËùÅäÈÜҺŨ¶ÈÓÐʲôӰÏ죿(ÌîÆ«¸ß¡¢Æ«µÍ¡¢ÎÞÓ°Ïì)
¢Ù½«ÉÕ±­ÖÐÈÜҺתÒƵ½ÈÝÁ¿Æ¿Ö®Ç°£¬ÈÝÁ¿Æ¿ÖÐÓÐÉÙÁ¿ÕôÁóË®               £»
¢ÚÏòÈÝÁ¿Æ¿ÖÐתÒÆÈÜҺʱ²»É÷ÓÐÒºµÎµôÔÚÈÝÁ¿Æ¿ÍâÃ棬ÔòŨ¶È             £»
¢Û¶¨ÈÝʱ¸©ÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß                  ¡£
(4)ʵÑéÊÒÀïѸËÙÖƱ¸ÉÙÁ¿ÂÈÆø¿ÉÀûÓÃÒÔÏ·´Ó¦£º2 KMnO4+16 HCl ="==" 2 KCl + 2 MnCl2 + 5 Cl2¡ü + 8 H2O ´Ë·´Ó¦²»ÐèÒª¼ÓÈÈ£¬³£ÎÂϾͿÉÒÔѸËÙ½øÐС£
¢ÙÓá°Ë«ÏßÇÅ·¨¡± ±êÃ÷µç×ÓתÒƵķ½ÏòºÍÊýÄ¿£º
2 KMnO4+16 HCl ="==" 2 KCl + 2 MnCl2 + 5 Cl2¡ü + 8 H2O
¢ÚÓøÃŨÑÎËáÖƵÃÁ˱ê¿öÏÂ560mlCl2,Ôò±»Ñõ»¯µÄHClΪ    mol£¬ÐèÒªKMnO4µÄÖÊÁ¿      g

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

(12·Ö)ij»¯Ñ§Ñо¿ÐÔѧϰС×é̽ÌÖFe3+ºÍSO32-Ö®¼ä·¢ÉúÔõÑùµÄ·´Ó¦£¬ÇëÄãÒ»Æð²ÎÓ벢ЭÖúËûÃÇÍê³ÉʵÑé¡£
¢ÅÌá³ö²ÂÏ룺
¼×ͬѧÈÏΪ·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽΪ               £»
ÒÒͬѧÈÏΪ·¢ÉúË«Ë®½â·´Ó¦£¬Æä·´Ó¦·½³ÌʽΪ2Fe3++3SO32-+6H2O=2Fe(OH)3(½ºÌå)+3H2SO3£»
¢ÆʵÑéÑéÖ¤£º
±ûͬѧÉè¼ÆÁËÏÂÁÐʵÑéÀ´Ì½¾¿·´Ó¦µÄ¿ÉÄÜÐÔ¡£
¢ÙΪÁ˼ìÑéËùÓÃNa2SO3ÊÇ·ñ±äÖÊ£¬Ó¦Ñ¡ÓõÄÊÔ¼ÁÊÇ                   ¡£
¢ÚÈ¡5mLFeCl3ÈÜÒºÓÚÊÔ¹ÜÖУ¬ÖðµÎ¼ÓÈëNa2SO3ÈÜÒºÖÁ¹ýÁ¿£¬¹Û²ìµ½ÈÜÒºÑÕÉ«ÓÉ»ÆÉ«±äΪºì×ØÉ«£¨ÎÞÆøÅݲúÉú£¬Ò²ÎÞ³ÁµíÉú³É£©¡£
¢Û½«¢ÚÈÜÒº·Ö³ÉÁ½µÈ·Ý£¬ÆäÖÐÒ»·Ý¼ÓÈëÏ¡ÑÎËáÖÁ¹ýÁ¿£¬ÔÙ¼ÓÈëBaCl2Ï¡ÈÜÒº£¬Óа×É«³ÁµíÉú³É£»ÁíÒ»·ÝµÎÈ뼸µÎKSCNÈÜÒº£¬ÈÜÒº±ä³ÉѪºìÉ«¡£
¢ÇµÃ³ö½áÂÛ£º
¢Ù¸ù¾Ý±ûͬѧµÄʵÑéµÃ³öµÄ½áÂÛÊÇ£º                               £»
¢ÚʵÑé¢ÛÖÐÈÜÒº±ä³ÉѪºìÉ«µÄÀë×Ó·½³ÌʽΪ                        ¡£
¢ÈÍØչ̽¾¿£º
¢Ù¶¡Í¬Ñ§ÔÚFeCl3ÈÜÒºÖмÓÈëNa2CO3ÈÜÒº£¬¹Û²ìµ½ºìºÖÉ«³Áµí²¢ÇÒ²úÉúÎÞÉ«ÆøÌ壬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                                   ¡£
¢Ú´ÓÐÎʽÉÏ¿´£¬Na2CO3ºÍNa2SO3ÏàËÆ£¬µ«ÊÇ´ÓÉÏÊöʵÑéÖпÉÒÔ¿´µ½£¬¶þÕßµÄË®ÈÜÒºÓëÂÈ»¯ÌúÈÜÒº·´Ó¦µÄÏÖÏó²î±ðºÜ´ó£¬Æä¿ÉÄܵÄÔ­Òò³ýSO32-Ë®½âÄÜÁ¦½ÏCO32- £¨Ìî¡°´ó¡±»ò¡°Ð¡¡±£©Í⣬»¹ÓР                                   

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁмÒÍ¥»¯Ñ§Ð¡ÊµÑéÄÜ´ïµ½Ô¤ÆÚÄ¿µÄµÄÊÇ£º
A£®ÓÃÃ×ÌÀ¼ìÑéʳÓüӵâÑÎ(º¬KIO3)Öк¬Óеâ
B£®Óõâ¾Æ¼ìÑéÆûÓÍÖÐÊÇ·ñº¬Óв»±¥ºÍÌþ
C£®Ìá´¿µ°°×ÖÊ¿ÉÒÔÔÚµ°°×ÖÊÈÜÒºÖмӱ¥ºÍCuSO4ÈÜÒº£¬µ°°×ÖÊÎö³ö£¬È»ºó°Ñ³ÁµíÈÜÓÚÕôÁóË®ÖÐ
D£®µí·ÛÈÜÒºÖмÓÈëÁòËá¼ÓÈÈË®½â£¬ÀäÈ´£¬¼ÓÒø°±ÈÜÒº×öÒø¾µ·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÂÌÉ«£º»·¾³±£»¤Ò²ÊÇÎÒÃǵÄÉú»îÀíÄî¡£ÏÂÁÐʵÑé·ûºÏ¡°ÂÌÉ«»·±£¡±Ë¼ÏëµÄÊÇ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ij»¯Ñ§Ê½È¤Ð¡×éÉè¼ÆÁËÏÂÁÐËĸöʵÑé×°Öã¬ÊÔͼÔÚ¶Ìʱ¼äÄÚͨ¹ý¹Û²ìʵÑéÏÖÏó˵Ã÷CO2ÓëNaOHÈÜÒº·¢ÉúÁË·´Ó¦¡£ÆäÖÐÎÞ·¨´ïµ½ÊµÑéÄ¿µÄµÄÊÇ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁгýÔӵķ½·¨ÕýÈ·µÄÊÇ£º                     (     )
A£®³ýÈ¥N2ÖеÄÉÙÁ¿O2£ºÍ¨¹ý×ÆÈȵÄCuO·ÛÄ©£¬ÊÕ¼¯ÆøÌå
B£®³ýÈ¥CO2ÖеÄÉÙÁ¿HCl£ºÍ¨ÈëNa2CO3ÈÜÒº£¬ÊÕ¼¯ÆøÌå
C£®³ýÈ¥NaClÈÜÒºÖÐÉÙÁ¿CaCl2£º¼ÓÈëÊÊÁ¿Na2CO3£¬¹ýÂË
D£®³ýÈ¥KClÈÜÒºÖÐÉÙÁ¿MgCl2£º¼ÓÈëÊÊÁ¿NaOHÈÜÒº£¬¹ýÂË

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸