ij¿ÎÍâ»î¶¯Ð¡×éΪÁ˼ìÑéÄÆÓëË®·´Ó¦µÄ²úÎÉè¼ÆÈçÏÂͼװÖÃ(¼Ð³Ö×°ÖÃÊ¡ÂÔ)£¬Ê×ÏÈÔÚUÐιÜÄÚ¼ÓÈëÉÙÁ¿ÃºÓͺͼ¸Á£Äƿ飬ÔÙ´ÓUÐιܸ߶˼ÓÈëË®(º¬ÓзÓ̪)£¬¸Ï³ö¿ÕÆø£¬Ò»»á¶ù¼ÓÈÈÍ­Ë¿¡£

¸ù¾Ý·´Ó¦Öй۲쵽µÄÏÖÏ󣬻شðÏÂÁÐÎÊÌ⣺

(1)½ðÊôÄƵı仯ÏÖÏó______________________________¡£

(2)UÐιÜÖÐÈÜÒºµÄÑÕÉ«    ____________________£¬

˵Ã÷ÓÐ________Éú³É¡£

(3)Í­Ë¿µÄ±ä»¯ÏÖÏó    ______________________________£¬

˵Ã÷ÓÐ________Éú³É£»ÈôÈ¥µôºóÃæµÄ×°Ö㬼òÊö¼ìÑéÆøÌåµÄ·½·¨_____________________________________________________________¡£

(4)Èôa gÄÆÓëb mLË®ÍêÈ«·´Ó¦£¬Ôò¸ÃÈÜÒºµÄÈÜÖʵÄÖÊÁ¿·ÖÊýÊÇ________________¡£


¡¾½âÎö¡¿¡¡´ËÌâ×°ÖÃÊÇÄÆÓëË®·´Ó¦ÊµÑéµÄ¸Ä½ø×°Öá£ÓÉÓÚ¸Ã×°ÖÃΪÃܱÕÌõ¼þÇÒÔö¼ÓÁËúÓÍ£¬·´Ó¦ÏÖÏó¸ü¼Ó±ãÓڹ۲죬Éú³ÉµÄÎïÖʱãÓÚ¼ìÑ飬ÄÆÓëË®µÄ·´Ó¦ÌåÏÖÄƵĶàÖÖÐÔÖÊ£¬ÈçÑÕÉ«¡¢×´Ì¬¼°ÈÛµãµÍºÍÃܶÈСµÄÌص㡣ÄÆλÓÚË®ºÍúÓͽçÃæÉÏ£¬ÓëË®·¢Éú·´Ó¦£º2Na£«2H2O===2NaOH£«H2¡ü£¬Éú³ÉµÄH2·¢Éú·´Ó¦£ºH2£«CuOCu£«H2O¡£ÄÆÓëË®ÍêÈ«·´Ó¦µÃNaOHÈÜÒº£º

Ôòw(NaOH)£½¡Á100%

£½¡Á100%¡£

¡¾´ð°¸¡¿¡¡(1)ÈÛ³ÉÒø°×ɫСÇò£¬ÔÚË®ºÍúÓ͵ĽçÃæÉÏ×÷ÉÏÏÂÌø¶¯ÇÒÖð½¥±äС

(2)´ÓÉϵ½ÏÂÖð½¥ÓÉÎÞÉ«±äΪºìÉ«¡¡NaOH

(3)ÓɺÚÉ«±äºìÉ«¡¡H2¡¡ÔÚµ¼¹Ü¿Ú´¦µãȼÆøÌ壬Óе­À¶É«»ðÑæ²úÉú

(4)¡Á100%


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÑÖª8 g AÄÜÓë32 g BÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³É22 g CºÍÒ»¶¨Á¿D£¬ÏÖ½«16 g AÓë70 g BµÄ»ìºÏÎï³ä·Ö·´Ó¦ºó£¬Éú³É2 mol DºÍÒ»¶¨Á¿µÄC£¬ÔòDµÄĦ¶ûÖÊÁ¿Îª¶àÉÙ£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ʵÑéÊÒÖƵõÄNa2S2O3´Ö¾§ÌåÖÐÍùÍùº¬ÓÐÉÙÁ¿ÔÓÖÊ¡£ÎªÁ˲ⶨ´Ö²úÆ·ÖÐNa2S2O3¡¤5H2OµÄº¬Á¿£¬Ò»°ã²ÉÓÃÔÚËáÐÔÌõ¼þÏÂÓÃËáÐÔKMnO4±ê×¼ÒºµÎ¶¨µÄ·½·¨£¨¼Ù¶¨´Ö²úÆ·ÖÐÔÓÖÊÓëËáÐÔKMnO4ÈÜÒº²»·´Ó¦£©¡£³ÆÈ¡1.28 gµÄ´ÖÑùÆ·ÈÜÓÚË®£¬ÓÃ0.40 mol/LµÄËáÐÔ KMnO4ÈÜÒºµÎ¶¨£¬µ±ÈÜÒºÖÐS2O32¡ªÈ«²¿±»Ñõ»¯Ê±£¬ÏûºÄKMnO4ÈÜÒºÌå»ý20.00 mL¡£

¢Ù µÎ¶¨ÖÕµãʱµÄÏÖÏóÊÇ________________________________________________________

¢Ú д³ö¸ÃµÎ¶¨·´Ó¦µÄÀë×Ó·½³Ìʽ__________________________________________________

¢ÛÈôµÎ¶¨Ê±Õñµ´²»³ä·Ö£¬¸Õ¿´µ½ÈÜÒº¾Ö²¿±äÉ«¾ÍÍ£Ö¹µÎ¶¨£¬Ôò»áʹÑùÆ·ÖÐNa2S2O3¡¤5H2OµÄÖÊÁ¿·ÖÊýµÄ²â¶¨½á¹û________(Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»±ä¡±)

¢Ü¾­¼ÆË㣬²úÆ·ÖÐNa2S2O3¡¤5H2OµÄÖÊÁ¿·ÖÊýΪ ________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ȡһС¿é½ðÊôÄÆ·ÅÔÚȼÉÕ³×ÖмÓÈÈ£¬ÏÂÁÐʵÑéÏÖÏóÃèÊöÕýÈ·µÄÊÇ(¡¡¡¡)

¢Ù½ðÊôÄÆÏÈÈÛ»¯¡¡¢ÚÔÚ¿ÕÆøÖÐȼÉÕ£¬²úÉú»ÆÉ«»ðÑæ

¢ÛȼÉÕºóµÃ°×É«¹ÌÌå¡¡¢ÜȼÉÕºóÉú³Éµ­»ÆÉ«¹ÌÌåÎïÖÊ

A£®¢Ù¢Ú                         B£®¢Ù¢Ú¢Û

C£®¢Û¢Ü                         D£®¢Ù¢Ú¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


½ðÊôÄƶÖÃÔÚ¿ÕÆøÖУ¬Æä±íÃæ²»¿ÉÄÜÉú³ÉµÄÎïÖÊÊÇ(¡¡¡¡)

A£®Na2O                         B£®NaOH

C£®Na2CO3                        D£®NaHCO3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Óû³ýÈ¥Cl2ÖеÄÉÙÁ¿HClÆøÌ壬¿ÉÑ¡ÓÃ(¡¡¡¡)

A£®NaOHÈÜÒº                     B£®AgNO3ÈÜÒº

C£®±¥ºÍʳÑÎË®                   D£®Ê¯»ÒË®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÖÓÐA¡¢B¡¢CÈýÖÖÆøÌ壬AÊÇÃܶÈ×îСµÄÆøÌ壬BÔÚͨ³£Çé¿öϳʻÆÂÌÉ«£¬´¿¾»µÄA¿ÉÒÔÔÚBÖа²¾²µØȼÉÕÉú³ÉC¡£°ÑÆøÌåBͨÈëµ½ÊÊÁ¿Ê¯»ÒÈéÖпÉÒԵõ½°×É«»ë×ÇÎïD¡£Çë¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)д³öÏÂÁи÷·´Ó¦µÄ»¯Ñ§·½³Ìʽ¡£

¢Ù´¿¾»µÄAÔÚBÖа²¾²µØȼÉÕÉú³ÉC£º__________________________£»

¢Ú½«ÆøÌåBͨÈ뵽ˮÖУº_______________________________________£»

¢Û½«ÆøÌåBͨÈëµ½NaOHÈÜÒºÖУº______________________________£»

¢Ü½«ÆøÌåBͨÈëµ½ÊÊÁ¿Ê¯»ÒÈéÖУº________________________________¡£

(2)°ÑÈýÖÖÆøÌå·Ö±ðͨÈëËáÐÔÏõËáÒøÈÜÒºÖУ¬³öÏÖ°×É«³ÁµíµÄÆøÌåÊÇ£º________(ÓÃ×Öĸ±íʾ)¡£

(3)°×É«»ë×ÇÎïDÒò¾ßÓÐƯ°×ÐÔÓÖ³ÆΪ________£¬¸ÃÎïÖÊÔÚ¿ÕÆøÖÐÈÝÒ×±äÖʵÄÔ­ÒòΪ(Óû¯Ñ§·½³Ìʽ±íʾ)________________ ___________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐËÄÖÖÔªËØÖУ¬µÚÒ»µçÀëÄÜÓÉ´óµ½Ð¡Ë³ÐòÕýÈ·µÄÊÇ

¢Ù£®Ô­×Óº¬ÓÐδ³É¶Ôµç×Ó×î¶àµÄµÚ¶þÖÜÆÚÔªËØ 

¢Ú£®µç×ÓÅŲ¼Îª1s2µÄÔªËØ

¢Û£®ÖÜÆÚ±íÖе縺ÐÔ×îÇ¿µÄÔªËØ           

¢Ü£®Ô­×Ó×îÍâ²ãµç×ÓÅŲ¼Îª3s23p4µÄÔªËØ

A.¢Ú¢Û¢Ù¢Ü         B.¢Û¢Ù¢Ü¢Ú         C.¢Ù¢Û¢Ü¢Ú        D.ÎÞ·¨±È½Ï

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÑÖª£º4NH3(g)+5O2(g)=4NO(g)+6H2O(g)£¬Èô·´Ó¦ËÙÂÊ·Ö±ðÓÃv(NH3)¡¢v(O2)¡¢v(NO)¡¢v(H2O) [µ¥Î»£ºmol/(L¡¤s)]±íʾ£¬ÔòÕýÈ·µÄ¹ØϵÊÇ£º(   )

A.4/5v(NH3)=v(O2)     B.5/6 v(O2)=v(H2O)    C.2/3 v(NH3)=v(H2O)    D.4/5 v(O2)=v(NO)

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸