ijÓлúÎïA·Ö×ÓʽΪCxHyOz,15 g AÍêȫȼÉÕÉú³É22 g CO2ºÍ9 g H2O¡£

(1)¸ÃÓлúÎïµÄ×î¼òʽÊÇ____________¡£

(2)ÈôAÊÇÒ»ÖÖÎÞÉ«¾ßÓÐÇ¿ÁҴ̼¤ÐÔÆøζµÄÆøÌ壬¾ßÓл¹Ô­ÐÔ£¬ÔòÆä½á¹¹¼òʽÊÇ__________¡£

(3)ÈôAºÍNa2CO3»ìºÏÓÐÆøÌå·Å³ö£¬ºÍ´¼ÄÜ·¢Éúõ¥»¯·´Ó¦£¬ÔòAµÄ½á¹¹¼òʽΪ______________¡£

(4)ÈôAÊÇÒ×»Ó·¢ÓÐË®¹ûÏãζµÄÒºÌ壬ÄÜ·¢ÉúË®½â·´Ó¦£¬ÔòÆä½á¹¹¼òʽΪ____________¡£

(5)ÈôÆä·Ö×ӽṹÖк¬ÓÐ6¸ö̼ԭ×Ó£¬¾ßÓжàÔª´¼ºÍÈ©µÄÐÔÖÊ£¬ÔòÆä½á¹¹¼òʽΪ____________¡£


½âÎö¡¡±¾Ìâ¿É°´Èçϲ½Öè½â´ð£º¢Ù¸ù¾ÝȼÉÕ·½³Ìʽ¿ÉÇóµÃ×î¼òʽΪCH2O£¬¢Ú¸ù¾Ý¸ø³öµÄÐÔÖÊ¿ÉÍƵÃÓлúÎïµÄ½á¹¹¼òʽ¡£

´ð°¸¡¡(1)CH2O¡¡(2)HCHO¡¡(3)CH3COOH

(4)HCOOCH3¡¡(5)CH2OH(CHOH)4CHO


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁи÷×éÀë×ÓÔÚͬһÈÜÒºÖÐÒ»¶¨ÄÜ´óÁ¿¹²´æµÄÊÇ(¡¡¡¡)

A.º¬´óÁ¿Ba2+µÄÈÜÒºÖÐ:Cl-¡¢K+¡¢S¡¢C

B.º¬´óÁ¿H+µÄÈÜÒºÖÐ:Mg2+¡¢Na+¡¢C¡¢S

C.º¬´óÁ¿OH-µÄÈÜÒºÖÐ:K+¡¢N¡¢S¡¢Cu2+

D.º¬´óÁ¿Na+µÄÈÜÒºÖÐ:H+¡¢K+¡¢S¡¢N

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


NA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ

A£®1 L 0.2mol¡¤CH3COONaÈÜÒºÖк¬ÓÐ0.2NA¸öCH3COO¡ª

B£®±ê×¼×´¿öÏ£¬11.2LCl2ÈÜÓÚË®£¬×ªÒƵĵç×ÓÊýΪNA

C£®³£Î³£Ñ¹Ï£¬23g NO2ºÍN2O4µÄ»ìºÏÆøÌ庬ÓеÄÔ­×ÓÊýΪ1.5NA

D£®100 mL 18.4mol¡¤Å¨ÁòËáÓë×ãÁ¿Í­¼ÓÈÈ·´Ó¦£¬Éú³ÉSO2µÄ·Ö×ÓÊýΪ0.92NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


»¯Ñ§·´Ó¦Ô­ÀíÔÚ¿ÆÑк͹¤Å©ÒµÉú²úÖÐÓй㷺ӦÓá£

(1)ij»¯Ñ§ÐËȤС×é½øÐй¤ÒµºÏ³É°±µÄÄ£ÄâÑо¿£¬·´Ó¦µÄ·½³ÌʽΪN2(g)+3H2(g)2NH3(g) ¡£ÔÚlLÃܱÕÈÝÆ÷ÖмÓÈë0.1 mol N2ºÍ0.3mol H2£¬ÊµÑé¢Ù¡¢¢Ú¡¢¢ÛÖÐc(N2)Ëæʱ¼äµÄ±ä»¯ÈçÏÂͼËùʾ£º

ʵÑé¢Ú´Ó³õʼµ½Æ½ºâµÄ¹ý³ÌÖУ¬¸Ã·´Ó¦µÄƽ¾ù·´Ó¦ËÙÂÊv(NH3)=__________________£»ÓëʵÑé¢ÙÏà±È£¬ÊµÑé¢ÚºÍʵÑé¢ÛËù¸Ä±äµÄʵÑéÌõ¼þ·Ö±ðΪÏÂÁÐÑ¡ÏîÖеÄ__________¡¢__________(Ìî×Öĸ±àºÅ)¡£

a£®Ôö´óѹǿ  b£®¼õСѹǿ  C£®Éý¸ßζÈd£®½µµÍζȠ e£®Ê¹Óô߻¯¼Á

(2)ÒÑÖªNO2ÓëN2O4¿ÉÒÔÏ໥ת»¯£º¡£

¢ÙT¡æʱ£¬½«0.40 mol NO2ÆøÌå³äÈëÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬´ïµ½Æ½ºâºó£¬²âµÃÈÝÆ÷ÖÐc(N2O4)=0.05 mol¡¤L£¬Ôò¸Ã·´Ó¦µÄƽºâ³£ÊýK=_______________£»

¢ÚÒÑÖªN2O4ÔڽϸßζÈÏÂÄÑÒÔÎȶ¨´æÔÚ£¬Ò×ת»¯ÎªNO2£¬ÈôÉý¸ßζȣ¬ÉÏÊö·´Ó¦µÄƽºâ³£ÊýK½«_____________(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)¡£

¢ÛÏò¾øÈÈÃܱÕÈÝÆ÷ÖÐͨÈëÒ»¶¨Á¿µÄNO2£¬Ä³Ê±¼ä¶ÎÄÚÕý·´Ó¦ËÙÂÊËæʱÎʵı仯ÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ__________(Ìî×Öĸ±àºÅ)¡£

A£®·´Ó¦ÔÚcµã´ïµ½Æ½ºâ״̬   

B£®·´Ó¦ÎïŨ¶È£ºaµãСÓÚbµã

C£®Ê±£¬NO2µÄת»¯ÂÊ£ºa¡«b¶ÎСÓÚ b¡«c¶Î

(3)25¡æʱ£¬½«amol¡¤LµÄ°±Ë®Óëb mol¡¤LÒ»1ÑÎËáµÈÌå»ý»ìºÏ£¬·´Ó¦ºóÈÜҺǡºÃÏÔÖÐÐÔ£¬Ôòa___________b(Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±)£»ÓÃa¡¢b±íʾNH3H2OµÄµçÀëƽºâ³£Êý=________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐ˵·¨ÖдíÎóµÄÊÇ

(¡¡¡¡ )

A£®Í¨³£ÕáÌǺ͵í·Û¶¼²»ÏÔ»¹Ô­ÐÔ

B£®ÓôóÃ×ÄðµÄ¾ÆÔÚÒ»¶¨Ìõ¼þÏÂÃÜ·â±£´æ£¬Ê±¼äÔ½³¤Ô½Ïã´¼

C£®ÏËάËØ¡¢ÕáÌÇ¡¢ÆÏÌÑÌǺÍÖ¬·¾ÔÚÒ»¶¨Ìõ¼þ϶¼¿É·¢ÉúË®½â·´Ó¦

D£®ÏËάËØ·Ö×ÓÊÇÓÉÆÏÌÑÌǵ¥Ôª×é³ÉµÄ£¬¿ÉÒÔ±íÏÖ³öһЩ¶àÔª´¼µÄÐÔÖÊ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÑ֪ij°±»ùËáµÄÏà¶Ô·Ö×ÓÖÊÁ¿Ð¡ÓÚ200£¬ÇÒÑõµÄÖÊÁ¿·ÖÊýԼΪ0.5£¬ÔòÆä·Ö×ÓÖÐ̼µÄ¸öÊý×î¶àΪ

(¡¡¡¡ )

A£®5¸ö                       B£®6¸ö

C£®7¸ö                       D£®8¸ö

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


×Ïɼ´¼ÊÇÒ»ÖÖÐÂÐÍ¿¹°©Ò©£¬Æä·Ö×ÓʽΪC47H51NO14£¬ËüÊÇÓÉÈçϵÄAËáºÍB´¼Éú³ÉµÄÒ»ÖÖõ¥¡£

B£®R¡ªOH(RÊÇÒ»¸öº¬C¡¢H¡¢OµÄ»ùÍÅ)

(1)A¿ÉÔÚÎÞ»úËá´ß»¯ÏÂË®½â£¬Æä·´Ó¦·½³ÌʽÊÇ

__________________________________________________________________

________________________________________________________________¡£

(2)AË®½âËùµÃµÄ°±»ùËá²»ÊÇÌìÈ»µ°°×ÖʵÄË®½â²úÎÒòΪ°±»ù²»ÔÚ(ÌîÏ£À°×Öĸ)________λ¡£

(3)д³öROHµÄ·Ö×Óʽ£º__________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÎïÖÊÖзֱðµÎÈë´×Ëᣬ»á²úÉúÏàͬÆøÌåµÄÊÇ(¡¡¡¡)

¢Ù´óÀíʯ£»¢ÚÖÓÈéʯ£»¢ÛË®¹¸£»¢Ü±´¿Ç£»¢Ýµ°¿Ç

A£®¢Ù¢Ú                              B£®¢Ü¢Ý

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

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


½«NH4HCO3ÔÚÊÔ¹ÜÖмÓÈÈ,ʹ·Å³öµÄÆøÌåÒÀ´Îͨ¹ýÊ¢ÓÐ×ãÁ¿¹ýÑõ»¯ÄƵĸÉÔï¹Ü¡¢×ãÁ¿Å¨ÁòËáµÄÏ´ÆøÆ¿,×îºóµÃµ½µÄÆøÌåÊÇ(¡¡¡¡)

A.NH3¡¡¡¡¡¡     B.O2¡¡¡¡    ¡¡C.H2O¡¡¡¡¡¡       D.CO2

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸