B£® [ʵÑ黯ѧ]
´ÎÁòËáÇâÄƼ×È©(NaHSO2¡¤HCHO¡¤2H2O)ÔÚӡȾ¡¢Ò½Ò©ÒÔ¼°Ô­×ÓÄܹ¤ÒµÖÐÓ¦Óù㷺¡£ÒÔNa2SO3¡¢SO2¡¢HCHO ºÍп·ÛΪԭÁÏÖƱ¸´ÎÁòËáÇâÄƼ×È©µÄʵÑé²½ÖèÈçÏ£º
²½Öè1£ºÔÚÉÕÆ¿ÖÐ(×°ÖÃÈçͼ Ëùʾ) ¼ÓÈëÒ»¶¨Á¿Na2SO3ºÍË®£¬½Á°èÈܽ⣬»ºÂýͨÈëSO2£¬ÖÁÈÜÒºpH ԼΪ4£¬ÖƵÃNaHSO3ÈÜÒº¡£²½Öè2£º½«×°ÖÃA Öе¼Æø¹Ü»»³ÉÏðƤÈû¡£ÏòÉÕÆ¿ÖмÓÈëÉÔ¹ýÁ¿µÄп·ÛºÍÒ»¶¨Á¿¼×È©ÈÜÒº£¬ÔÚ80 ~ 90ÒæÏ£¬·´Ó¦Ô¼3h£¬ÀäÈ´ÖÁÊÒΣ¬³éÂË¡£²½Öè3£º½«ÂËÒºÕæ¿ÕÕô·¢Å¨Ëõ£¬ÀäÈ´½á¾§¡£

(1)×°ÖÃB µÄÉÕ±­ÖÐÓ¦¼ÓÈëµÄÈÜÒºÊÇ                     ¡£
(2)¢Ù²½Öè2 ÖУ¬·´Ó¦Éú³ÉµÄZn(OH)2»á¸²¸ÇÔÚп·Û±íÃæ×èÖ¹·´Ó¦½øÐУ¬·ÀÖ¹¸ÃÏÖÏó·¢ÉúµÄ´ëÊ©ÊÇ                            
    ¡£¢ÚÀäÄý¹ÜÖлØÁ÷µÄÖ÷ÒªÎïÖʳýH2O Í⻹ÓР                    (Ìѧʽ)¡£
(3)¢Ù³éÂË×°ÖÃËù°üº¬µÄÒÇÆ÷³ý¼õѹϵͳÍ⻹ÓР           ¡¢            (ÌîÒÇÆ÷Ãû³Æ)¡£¢ÚÂËÔüµÄÖ÷Òª³É·ÖÓР              ¡¢             (Ìѧʽ)¡£
(4)´ÎÁòËáÇâÄƼ×È©¾ßÓÐÇ¿»¹Ô­ÐÔ£¬ÇÒÔÚ120ÒæÒÔÉÏ·¢Éú·Ö½â¡£²½Öè3 Öв»ÔÚ³¨¿ÚÈÝÆ÷ÖÐÕô·¢Å¨ËõµÄÔ­ÒòÊÇ                     ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

½ñÓÐÏÂÁÐÆøÌ壺H2¡¢Cl2¡¢HCl¡¢NH3¡¢NO¡¢H2S¡¢SO2£¬ÓÃÈçͼËùʾµÄ×°ÖýøÐÐʵÑ飬Ìî¿ÕÏÂÁпհףº

£¨1£©ÉÕÆ¿¸ÉÔïʱ£¬´ÓA¿Ú½øÆø¿ÉÊÕ¼¯µÄÆøÌåÊÇ________£¬´ÓB¿Ú½øÆø¿ÉÊÕ¼¯µÄÆøÌåÊÇ______________¡£
£¨2£©ÉÕÆ¿ÖгäÂúˮʱ£¬¿ÉÓÃÀ´²âÁ¿___________µÈÆøÌåµÄÌå»ý¡£
£¨3£©µ±ÉÕÆ¿ÖÐ×°ÈëÏ´Òº£¬ÓÃÓÚÏ´Æøʱ£¬ÆøÌåÓ¦´Ó________¿Ú½øÈëÉÕÆ¿¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

(13·Ö)ÒÒËáÒìÎìõ¥ÊÇ×é³ÉÃÛ·äÐÅÏ¢ËØÖʵijɷÖÖ®Ò»£¬¾ßÓÐÏ㽶µÄÏãζ£¬ÊµÑéÊÒÖƱ¸ÒÒËáÒìÎìõ¥µÄ·´Ó¦×°ÖÃʾÒâͼºÍÓйØÊý¾ÝÈçÏ£º

ʵÑé²½Ö裺
ÔÚAÖмÓÈë4.4 gµÄÒìÎì´¼£¬6.0 gµÄÒÒËá¡¢ÊýµÎŨÁòËáºÍ2¡«3ƬËé´ÉƬ£¬¿ªÊ¼»ºÂý¼ÓÈÈA£¬»ØÁ÷50·ÖÖÓ£¬·´Ó¦ÒºÀäÖÁÊÒκ󣬵¹Èë·ÖҺ©¶·ÖУ¬·Ö±ðÓÃÉÙÁ¿Ë®£¬±¥ºÍ̼ËáÇâÄÆÈÜÒººÍˮϴµÓ£¬·Ö³öµÄ²úÎï¼ÓÈëÉÙÁ¿ÎÞË®ÁòËáþ¹ÌÌ壬¾²ÖÃƬ¿Ì£¬¹ýÂ˳ýÈ¥ÁòËáþ¹ÌÌ壬£¬½øÐÐÕôÁó´¿»¯£¬ÊÕ¼¯140¡«143 ¡æÁó·Ö£¬µÃÒÒËáÒìÎìõ¥3.9 g¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©×°ÖÃBµÄÃû³ÆÊÇ£º                    
£¨2£©ÔÚÏ´µÓ²Ù×÷ÖУ¬µÚÒ»´ÎˮϴµÄÖ÷ҪĿµÄÊÇ£º               £» µÚ¶þ´ÎˮϴµÄÖ÷ҪĿµÄÊÇ£º                       ¡£
£¨3£©ÔÚÏ´µÓ¡¢·ÖÒº²Ù×÷ÖУ¬Ó¦³ä·ÖÕñµ´£¬È»ºó¾²Ö㬴ý·Ö²ãºó       £¨Ìî±êºÅ£©£¬
A£®Ö±½Ó½«ÒÒËáÒìÎìõ¥´Ó·ÖҺ©¶·ÉÏ¿Úµ¹³ö  
B£®Ö±½Ó½«ÒÒËáÒìÎìõ¥´Ó·ÖҺ©¶·Ï¿ڷųö
C£®ÏȽ«Ë®²ã´Ó·ÖҺ©¶·µÄÏ¿ڷųö£¬ÔÙ½«ÒÒËáÒìÎìõ¥´ÓÏ¿ڷųö
D£®ÏȽ«Ë®²ã´Ó·ÖҺ©¶·µÄÏ¿ڷųö£¬ÔÙ½«ÒÒËáÒìÎìõ¥´ÓÉϿڷųö
£¨4£©±¾ÊµÑéÖмÓÈë¹ýÁ¿ÒÒËáµÄÄ¿µÄÊÇ£º                             
£¨5£©ÊµÑéÖмÓÈëÉÙÁ¿ÎÞË®ÁòËáþµÄÄ¿µÄÊÇ£º                                        
£¨6£©ÔÚÕôÁó²Ù×÷ÖУ¬ÒÇÆ÷Ñ¡Ôñ¼°°²×°¶¼ÕýÈ·µÄÊÇ£º                   (Ìî±êºÅ)

£¨7£©±¾ÊµÑéµÄ²úÂÊÊÇ£º        
A£®30¨G             B£®40¨G               C£®50¨G              D£®60¨G
£¨8£©ÔÚ½øÐÐÕôÁó²Ù×÷ʱ£¬Èô´Ó130 ¡æ¿ªÊ¼ÊÕ¼¯Áó·Ö£¬²úÂÊÆ«      (Ìî¸ß»òÕßµÍ)Ô­ÒòÊÇ             

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

£¨13·Ö£©´×ËáÑǸõË®ºÏÎï([Cr(CH3COO)2)]2¡¤2H2O£¬ÉîºìÉ«¾§Ì壩ÊÇÒ»ÖÖÑõÆøÎüÊÕ¼Á£¬Í¨³£ÒÔ¶þ¾ÛÌå·Ö×Ó´æÔÚ£¬²»ÈÜÓÚÀäË®ºÍÃÑ£¬Î¢ÈÜÓÚ´¼£¬Ò×ÈÜÓÚÑÎËᡣʵÑéÊÒÖƱ¸´×ËáÑǸõË®ºÏÎïµÄ×°ÖÃÈçͼËùʾ£¬Éæ¼°µÄ»¯Ñ§·½³ÌʽÈçÏ£º

Zn(s) + 2HCl(aq) = ZnCl2(aq) + H2(g)
2CrCl3(aq) + Zn(s)= 2CrCl2 (aq) + ZnCl2(aq)
2Cr2+(aq) + 4CH3COO-(aq) + 2H2O(l) = [Cr(CH3COO)2]2¡¤2H2O (s)
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¼ì²éÐé¿òÄÚ×°ÖÃÆøÃÜÐԵķ½·¨ÊÇ     ¡£
£¨2£©´×ËáÄÆÈÜÒºÓ¦·ÅÔÚ×°Öà    ÖУ¨ÌîдװÖñàºÅ£¬ÏÂͬ£©£»ÑÎËáÓ¦·ÅÔÚ×°Öà    ÖУ»
×°ÖÃ4µÄ×÷ÓÃÊÇ     ¡£
£¨3£©±¾ÊµÑéÖÐËùÓÐÅäÖÆÈÜÒºµÄË®ÐèÖó·Ð£¬ÆäÔ­ÒòÊÇ     ¡£
£¨4£©½«Éú³ÉµÄCrCl2ÈÜÒºÓëCH3COONaÈÜÒº»ìºÏʱµÄ²Ù×÷ÊÇ     ·§ÃÅA¡¢     ·§ÃÅB £¨Ìî¡°´ò¿ª¡±»ò¡°¹Ø±Õ¡±£©¡£
£¨5£©±¾ÊµÑéÖÐпÁ£Ðë¹ýÁ¿£¬ÆäÔ­ÒòÊÇ     ¡£                                             
£¨6£©ÎªÏ´µÓ[Cr(CH3COO)2)]2¡¤2H2O²úÆ·£¬ÏÂÁз½·¨ÖÐ×îÊʺϵÄÊÇ     ¡£        
A£®ÏÈÓÃÑÎËáÏ´£¬ºóÓÃÀäˮϴ        B£®ÏÈÓÃÀäˮϴ£¬ºóÓÃÒÒ´¼Ï´
C£®ÏÈÓÃÀäˮϴ£¬ºóÓÃÒÒÃÑÏ´        D£®ÏÈÓÃÒÒ´¼Ï´µÓ£¬ºóÓÃÒÒÃÑÏ´

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

£¨12·Ö£©ÌìȻˮÊÇÈËÀàÒûÓÃË®µÄÖ÷ÒªÀ´Ô´¡£´ÓÌìȻˮ»ñµÃ¿ÉÒÔÒûÓõÄˮһ°ãÐë¾­¹ý³Á½µÐü¸¡Îɱ¾úÏû¶¾µÈ²½Öè¡£
£¨1£©³Á½µÐü¸¡ÎïÒªÔÚË®ÖмÓÈëÐõÄý¼Á£¬È罫ÂÁÑμÓÈëË®ÖÐÄÜ´ïµ½¾»Ë®Ä¿µÄ£¬
Ô­ÒòÊÇ              £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
£¨2£©ÂÈÆø¿ÉÓÃÓÚ×ÔÀ´Ë®É±¾úÏû¶¾¼Á£¬½áºÏÀë×Ó·½³ÌʽºÍÎÄ×ÖÀíÓÉ            ¡£
£¨3£©ÐÂÐÍË®´¦Àí¼Á¸ßÌúËá¼Ø (K2FeO4)¾ßÓÐÇ¿µÄÑõ»¯×÷ÓúÍÐõÄý×÷Ó᣹¤ÒµÉÏ¿Éͨ¹ýÒÔÏÂÁ÷³ÌÖƱ¸¸ßÌúËá¼Ø£º

²éÔÄ×ÊÁÏ:¸ßÌúËáÑÎÔÚÖÐÐÔ»òËáÐÔÈÜÒºÖлáÖ𽥷ֽ⣬ÔÚ¼îÐÔÈÜÒºÖÐÎȶ¨¡£
Íê³É¡°Ñõ»¯¡±¹ý³ÌÖеÄÀë×Ó·½³Ìʽ
¡õFe3+ + ¡õClO- +¡õ      ="¡õ" FeO42- + ¡õCl- + ¡õ   
¡°×ª»¯¡±¹ý³ÌÖÐʵÏÖÓÉNa2FeO4ÖƵÃK2FeO4£¬ÊÇÀûÓöþÕß        ÐԵIJ»Í¬¡£
¢Û½áºÏ×ÊÁÏÍê³É´ÖK2FeO4¾§ÌåµÄÌá´¿£º½«´Ö²úÆ·Óà      Èܽ⣬ȻºóÔÙ¼ÓÈë±¥ºÍKOHÈÜÒº¡¢ÀäÈ´½á¾§¡¢¹ýÂË¡£
¢Ü¸ßÌúËá¼ØµÄÓ¦Óû¹ÔÚ²»¶ÏÀ©Õ¹ÖС£Èç¿ÉÖƳɸßÌúµç³Ø£¬ µç³Ø·´Ó¦Îª£º
3Zn + 2K2FeO4 + 8H2O   3Zn(OH)2 + 2Fe(OH)3 + 4KOH 
·Åµçʱ£¬Õý¼«·´Ó¦Îª£º            ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ij»¯Ñ§ÊµÑéС×éµÄͬѧΪ̽¾¿ºÍ±È½ÏSO2ºÍÂÈË®µÄƯ°×ÐÔ£¬Éè¼ÆÁËÈçϵÄ
ʵÑé×°Öá£

£¨1£©ÊµÑéÊÒÓÃ×°ÖÃAÖƱ¸SO2ijͬѧÔÚʵÑéʱ·¢ÏÖ´ò¿ªAµÄ·ÖҺ©¶·»îÈûºó£¬Â©¶·ÖÐÒºÌåδ
Á÷Ï£¬ÄãÈÏΪԭÒò¿ÉÄÜÊÇ£º                                
£¨2£©ÊµÑéÊÒÓÃ×°ÖÃEÖƱ¸Cl2Æä·´Ó¦µÄ»¯Ñ§»¯Ñ§·½³ÌʽΪ£ºMnO2+4HCl£¨Å¨£©=C12¡ü+MnCl2+2H2O
ŨÑÎËáµÄ×÷ÓÃΪ£º                             
£¨3£©·´Ó¦¿ªÊ¼Ò»¶Îʱ¼äºó£¬¹Û²ìµ½ B¡¢DÁ½¸öÊÔ¹ÜÖеÄÆ·ºìÈÜÒº³öÏÖµÄÏÖÏóÊÇ£º
B£º                     D                                    ¡£
¢ÚֹͣͨÆøºó£¬ÔÙ¸øB¡¢DÁ½¸öÊԹֱܷð¼ÓÈÈ£¬Á½¸öÊÔ¹ÜÖеÄÏÖÏó·Ö±ðΪ
B£º                     D                               ¡£
£¨4£©ÁíÒ»¸öʵÑéС×éµÄͬѧÈÏΪSO2ºÍÂÈË®¶¼ÓÐƯ°×ÐÔ£®¶þÕß»ìºÏºóµÄƯ°×ÐԿ϶¨»á¸üÇ¿¡£ËûÃǽ«ÖƵõÄSO2ºÍCI2°´1£º1ͬʱͨÈ뵽ƷºìÈÜÒºÖУ¬½á¹û·¢ÏÖÍÊɫЧ¹û²¢²»ÏñÏëÏóµÄÄÇÑù¡£ÇëÄã·ÖÎö¸ÃÏÖÏóµÄÔ­Òò£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£º                         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

CaCO3¹ã·º´æÔÚÓÚ×ÔÈ»½ç£¬ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£´óÀíʯÖ÷Òª³É·ÖΪCaCO3£¬ÁíÍâÓÐÉÙÁ¿µÄº¬Áò»¯ºÏÎï(ÈçFeSµÈ)¡£ÊµÑéÊÒÓôóÀíʯºÍÏ¡ÑÎËá·´Ó¦ÖƱ¸CO2ÆøÌå¡£ÏÂÁÐ×°ÖÿÉÓÃÓÚCO2ÆøÌåµÄÌá´¿ºÍ¸ÉÔï¡£

Íê³ÉÏÂÁÐÌî¿Õ£º
(1)ÓÃŨÑÎËáÅäÖÆ1¡Ã1(Ìå»ý±È)µÄÏ¡ÑÎËá(Ô¼6 mol¡¤L-1)£¬Ó¦Ñ¡ÓõÄÒÇÆ÷ÊÇ_____¡£
a.ÉÕ±­¡¡¡¡¡¡b.²£Á§°ô¡¡¡¡¡¡c.Á¿Í²¡¡¡¡¡¡d.ÈÝÁ¿Æ¿
(2)ÉÏÊö×°ÖÃÖУ¬AÊÇ_____ÈÜÒº£¬NaHCO3ÈÜÒº¿ÉÒÔÎüÊÕ_____¡£
(3)ÉÏÊö×°ÖÃÖУ¬BÎïÖÊÊÇ_____¡£°ÑÕâ¸öʵÑéµÃµ½µÄÆøÌåÊÕ¼¯ÆðÀ´£¬ÓÃÀ´²â¶¨CO2µÄ·Ö×ÓÁ¿£¬Èç¹ûBÎïÖÊʧЧ£¬²â¶¨½á¹û_____(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°²»ÊÜÓ°Ï족)¡£
(4)Ò»´ÎÐÔ·¹ºÐÖÐʯÀ¯(¸ß¼¶ÍéÌþ)ºÍCaCO3ÔÚʳÎïÖеÄÈܳöÁ¿ÊÇÆÀ¼Û·¹ºÐÖÊÁ¿µÄÖ¸±êÖ®Ò»£¬²â¶¨ÈܳöÁ¿µÄÖ÷ҪʵÑé²½ÖèÉè¼ÆÈçÏ£º
¼ôËé¡¢³ÆÖØ¡ú½þÅÝÈܽâ¡ú¹ýÂË¡ú²ÐÔüºæ¸É¡úÀäÈ´¡¢³ÆÖØ¡úºãÖØ£¬ÎªÁ˽«Ê¯À¯Èܳö£¬Ó¦Ñ¡ÓõÄÊÔ¼ÁÊÇ_____£¬Ì¼Ëá¸ÆÈܳö£¬Ó¦Ñ¡ÓõÄÊÔ¼ÁÊÇ_____¡£
a.ÂÈ»¯ÄÆÈÜÒº¡¡¡¡¡¡¡¡b.Ï¡´×Ëá
c.Ï¡ÁòËá¡¡¡¡¡¡¡¡¡¡¡¡d.Õý¼ºÍé
(5)ÔÚÈܳöÁ¿²â¶¨ÊµÑéÖУ¬ÎªÁË»ñµÃʯÀ¯ºÍ̼Ëá¸ÆµÄ×î´óÈܳöÁ¿£¬Ó¦ÏÈÈܳö_____£¬Ô­ÒòÊÇ_______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

K3[Fe(C2O4)3]¡¤3H2O[Èý²ÝËáºÏÌú(¢ó)Ëá¼Ø¾§Ìå]Ò×ÈÜÓÚË®,ÄÑÈÜÓÚÒÒ´¼,¿É×÷ΪÓлú·´Ó¦µÄ´ß»¯¼Á¡£ÊµÑéÊÒ¿ÉÓÃÌúмΪԭÁÏÖƱ¸,Ïà¹Ø·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ:Fe+H2SO4FeSO4+H2¡ü
FeSO4+H2C2O4+2H2OFeC2O4¡¤2H2O¡ý+H2SO4
2FeC2O4¡¤2H2O+H2O2+H2C2O4+3K2C2O42K3[Fe(C2O4)3]+6H2O
2Mn+5C2+16H+2Mn2++10CO2¡ü+8H2O
»Ø´ðÏÂÁÐÎÊÌâ:
(1)ÌúмÖг£º¬ÁòÔªËØ,Òò¶øÔÚÖƱ¸FeSO4ʱ»á²úÉúÓж¾µÄH2SÆøÌå,¸ÃÆøÌå¿ÉÓÃÇâÑõ»¯ÄÆÈÜÒºÎüÊÕ¡£ÏÂÁÐÎüÊÕ×°ÖÃÕýÈ·µÄÊÇ¡¡¡¡¡¡¡¡¡¡¡£ 

(2)ÔÚ½«Fe2+Ñõ»¯µÄ¹ý³ÌÖÐ,Ðè¿ØÖÆÈÜҺζȲ»¸ßÓÚ40 ¡æ,ÀíÓÉÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡;µÃµ½K3[Fe(C2O4)3]ÈÜÒººó,¼ÓÈëÒÒ´¼µÄÀíÓÉÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
(3)¾§ÌåÖÐËùº¬½á¾§Ë®¿Éͨ¹ýÖØÁ¿·ÖÎö·¨²â¶¨,Ö÷Òª²½ÖèÓÐ:¢Ù³ÆÁ¿,¢ÚÖÃÓÚºæÏäÖÐÍѽᾧˮ,¢ÛÀäÈ´,¢Ü³ÆÁ¿,¢Ý¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡(ÐðÊö´Ë²½²Ù×÷),¢Þ¼ÆËã¡£²½Öè¢ÛÈôδÔÚ¸ÉÔïÆ÷ÖнøÐÐ,²âµÃµÄ¾§ÌåÖÐËùº¬½á¾§Ë®º¬Á¿¡¡¡¡¡¡¡¡(Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족);²½Öè¢ÝµÄÄ¿µÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
(4)¾§ÌåÖÐC2º¬Á¿µÄ²â¶¨¿ÉÓÃËáÐÔKMnO4±ê×¼ÈÜÒºµÎ¶¨¡£³ÆÈ¡Èý²ÝËáºÏÌú(¢ó)Ëá¼Ø¾§Ìåm gÈÜÓÚË®Åä³É250 mLÈÜÒº,È¡³ö20.00 mL·ÅÈë׶ÐÎÆ¿ÖÐ,ÓÃ0.010 0 mol¡¤L-1ËữµÄ¸ßÃÌËá¼ØÈÜÒº½øÐеζ¨¡£
¢ÙÏÂÁвÙ×÷¼°Ëµ·¨ÕýÈ·µÄÊÇ¡¡¡¡¡¡¡£ 
A.µÎ¶¨¹ÜÓÃÕôÁóˮϴµÓºó,¼´¿É×°Èë±ê×¼ÈÜÒº
B.×°Èë±ê×¼ÈÜÒººó,°ÑµÎ¶¨¹Ü¼ÐÔڵζ¨¹Ü¼ÐÉÏ,ÇáÇáת¶¯»îÈû,·Å³öÉÙÁ¿±ê×¼Òº,ʹ¼â×ì³äÂúÒºÌå
C.½Ó½üÖÕµãʱ,ÐèÓÃÕôÁóË®³åÏ´Æ¿±ÚºÍµÎ¶¨¹Ü¼â¶ËÐü¹ÒµÄÒºµÎ
¢ÚÓÐͬѧÈÏΪ¸ÃµÎ¶¨¹ý³Ì²»ÐèҪָʾ¼Á,ÄÇôµÎ¶¨ÖÕµãµÄÏÖÏóΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡,Èô´ïµ½µÎ¶¨ÖÕµãÏûºÄ¸ßÃÌËá¼ØÈÜÒºV mL,ÄÇô¾§ÌåÖÐËùº¬C2µÄÖÊÁ¿·ÖÊýΪ¡¡¡¡¡¡¡¡(Óú¬V¡¢mµÄʽ×Ó±íʾ)¡£ 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ÒÑÖªÓÃP2O5×÷´ß»¯¼Á£¬¼ÓÈÈÒÒ´¼¿ÉÖƱ¸ÒÒÏ©£¬·´Ó¦Î¶ÈΪ80¡æ~210¡æ¡£Ä³Ñо¿ÐÔС×éÉè¼ÆÁËÈçϵÄ×°ÖÃÖƱ¸²¢¼ìÑé²úÉúµÄÒÒÏ©ÆøÌ壨¼Ð³ÖºÍ¼ÓÈÈÒÇÆ÷ÂÔÈ¥£©¡£
    
£¨1£©ÒÇÆ÷aµÄÃû³ÆΪ_________________
£¨2£©Óû¯Ñ§·´Ó¦·½³Ìʽ±íʾÉÏÊöÖƱ¸ÒÒÏ©µÄÔ­Àí______________________________¡£
£¨3£©ÒÑÖªP2O5ÊÇÒ»ÖÖËáÐÔ¸ÉÔï¼Á£¬ÎüË®·Å´óÁ¿ÈÈ£¬ÔÚʵÑé¹ý³ÌÖÐP2O5ÓëÒÒ´¼ÄÜ·¢Éú×÷Óã¬Òò·´Ó¦ÓÃÁ¿µÄ²»Í¬£¬»áÉú³É²»Í¬µÄÁ×Ëáõ¥£¬ËüÃǾùΪÒ×ÈÜÓÚË®µÄÎïÖÊ£¬·Ðµã½ÏµÍ¡£Ð´³öÒÒ´¼ºÍÁ×Ëá·´Ó¦Éú³ÉÁ×Ëá¶þÒÒõ¥µÄ»¯Ñ§·½³Ìʽ£¨Á×ËáÓýṹʽ±íʾΪ£©__________________________________¡£
£¨4£©Ä³Í¬Ñ§ÈÏΪÓÃÉÏÊö×°ÖÃÑéÖ¤²úÉúÁËÒÒÏ©²»¹»ÑÏÃÜ£¬ÀíÓÉÊÇ___________________¡£
£¨5£©Ä³Í¬Ñ§²éÎÄÏ×µÃÖª£º40%µÄÒÒÏ©Àû£¨·Ö×ÓʽΪC2H6ClO3P£©ÈÜÒººÍNaOH¹ÌÌå»ìºÍ¿É¿ìËÙ²úÉúË®¹û´ßÊìÓõÄÒÒÏ©£¬ÇëÔÚÉÏÊöÐéÏß¿òÄÚ»­³öÓÃÒÒÏ©ÀûÈÜÒººÍNaOH¹ÌÌåÖÆÈ¡ÒÒÏ©µÄ×°Öüòͼ(¼Ð³ÖÒÇÆ÷ÂÔ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸