ijͬѧҪÅäÖÆ0.2mol/LNa2SO4ÈÜÒº500mL£¬ÇëÍê³ÉÏÂÁÐÎÊÌ⣺
£¨1£©ÐèÒªNa2SO4¹ÌÌåµÄÖÊÁ¿Îª____________g
£¨2£©ÅäÖÆʱ£¬ÕýÈ·µÄ²Ù×÷˳ÐòÊÇ£¨ÓÃ×Öĸ±íʾ£¬Ã¿¸ö×ÖĸֻÄÜÓÃÒ»´Î£©______________
A£®ÓÃ30mLˮϴµÓÉÕ±­2~3´Î£¬Ï´µÓÒº¾ù×¢ÈëÈÝÁ¿Æ¿£¬Õñµ´
B£®ÓÃÍÐÅÌÌìƽ׼ȷ³ÆÁ¿Na2SO4¹ÌÌ壬ÔÚÉÕ±­ÖÐÓÃÉÙÁ¿Ë®Èܽâ
C£®½«ÈÝÁ¿Æ¿¸Ç½ô£¬Õñµ´£¬Ò¡ÔÈ
D£®½«ÒÑÀäÈ´µÄÈÜҺתÒƵ½ÈÝÁ¿Æ¿ÖÐ
E.ÖðµÎ¼ÓË®£¬Ê¹ÈÜÒº°¼ÒºÃæÇ¡ºÃÓë¿Ì¶ÈÏàÇÐ
F.¼ÌÐøÏòÈÝÁ¿Æ¿ÄÚСÐļÓË®£¬Ö±µ½ÒºÃæ½Ó½ü¿Ì¶È2~3cm´¦
£¨3£©²Ù×÷AÖУ¬½«Ï´µÓÒº¶¼ÒÆÈëÈÝÁ¿Æ¿£¬ÆäÄ¿µÄÊÇ____________________________________
(1)14.2g    (2)BDAFEC  £¨3£©°ÑÈ«²¿Na2SO4תÒƵ½ÈÝÁ¿Æ¿ÖÐÒÔ¼õСʵÑéÎó²î

ÊÔÌâ·ÖÎö£º£¨1£©ÐèÒªNa2SO4¹ÌÌåµÄÖÊÁ¿Îª0.2mol/L¡Á0.5L¡Á142g/mol£½14.2g¡£
£¨2£©¸ù¾ÝÅäÖƵÄÔ­ÀíÒÔ¼°ÊµÑé²Ù×÷ÒªÇó¿ÉÖª£¬ÕýÈ·µÄ²Ù×÷˳ÐòÊÇBDAFEC¡£
£¨3£©½«Ï´µÓÒºÒ²×¢Èëµ½ÈÝÁ¿Æ¿ÖеÄÄ¿µÄÊǼõСʵÑéÎó²î¡£
µãÆÀ£º¸ÃÌâÊÇ»ù´¡ÐÔʵÑéÌâµÄ¿¼²é£¬Ö÷ÒªÊÇ¿¼²éѧÉú¶ÔÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÅäÖƵÄÁ˽âÕÆÎÕÇé¿ö£¬ÒâÔÚÅàÑøѧÉú¹æ·¶ÑϽ÷µÄʵÑéÉè¼ÆÒÔ¼°¶¯ÊÖ²Ù×÷ÄÜÁ¦£¬Ò²ÓÐÁ˵÷¶¯Ñ§ÉúµÄѧϰÐËȤ¡£¸ÃÌâµÄ¹Ø¼üÊÇÃ÷ȷʵÑéÔ­Àí£¬È»ºó½áºÏÌâÒâÁé»îÔËÓü´¿É¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡ ¡¡£©
A£®º¬ÓÐNA¸öº¤Ô­×ӵĺ¤ÆøÔÚ±ê×¼×´¿öϵÄÌå»ýԼΪ11.2L
B£®25¡æ£¬1.01¡Á105 Pa, 44g CO2Öк¬ÓеÄÔ­×ÓÊýΪ3NA
C£®1 mol Cl2Óë×ãÁ¿Fe·´Ó¦£¬×ªÒƵĵç×ÓÊýΪ3NA
D£®±ê×¼×´¿öÏ£¬11.2LH2O º¬ÓеķÖ×ÓÊýΪ0.5NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÏ¡ÑÎËáʱ£¬×îºóÓýºÍ·µÎ¹ÜÏòÈÝÁ¿Æ¿µÎ¼ÓÕôÁóË®£¬ÓÉÓÚ²»É÷£¬ÒºÃ泬¹ýÁËÈÝÁ¿Æ¿¾±ÉϵĿ̶ȣ¬ÕâʱӦ²ÉÈ¡µÄ´ëÊ©ÊÇ£¨   £©
A£®ÓýºÍ·µÎ¹ÜÎü³ö³¬¹ýµÄÄÇÒ»²¿·Ö
B£®°ÑÈÜÒºµ¹³öÒ»²¿·ÖÔÚÕô·¢ÃóÖУ¬Õô·¢Ò»²¿·ÖË®ºó°ÑÕô·¢ÃóÖÐÊ£ÓàµÄÈÜÒºÈÔ×¢ÈëÈÝÁ¿Æ¿£¬ÖØмÓË®ÖÁ¿Ì¶È
C£®¸ù¾Ý³¬³öµÄÕôÁóË®Á¿£¬¾­¹ý¼ÆË㣬²¹¼ÓŨÑÎËá
D£®°ÑÈÝÁ¿Æ¿ÖеÄÈÜҺȫ²¿ÆúÈ¥£¬ÖØÐÂÅäÖÆ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

NA´ú±í°¢·ü¼ÓµÂ³£Êý£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®±ê×¼×´¿öÏ£¬22.4LCCl4º¬ÓÐNA¸öCCl4·Ö×Ó
B  2gÇâÆøËùº¬Ô­×ÓÊýĿΪNA
C ÔÚ³£Î³£Ñ¹Ï£¬11.2LµªÆøËùº¬µÄÔ­×ÓÊýĿΪNA
D  17g°±ÆøËùº¬µç×ÓÊýĿΪ10NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐÐðÊöÖÐÕýÈ·µÄÊÇ       
A£®0.1mol¡¤L£­1Ï¡ÁòËá100mLÖк¬ÓÐÁòËá¸ùÀë×Ó¸öÊýΪ0.1NA
B£®³£Î³£Ñ¹Ï£¬ÑõÆøºÍ³ôÑõµÄ»ìºÏÎï16gÖк¬ÓÐNA¸öÑõÔ­×Ó
C£®7.8 g Na2O2º¬ÓеÄÒõÀë×ÓÊýĿΪ0.2 NA
D£®NO2Óë×ãÁ¿µÄË®ÍêÈ«·´Ó¦£¬ÈôÓÐ1molNOÉú³É£¬ÔòתÒƵç×ÓÊýΪNA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©ÔÚ±ê×¼×´¿öÏ£¬COºÍCO2»ìºÏÆøÌåµÄÖÊÁ¿Îª36 g£¬Ìå»ýΪ22.4 L£¬ÔòCOËùÕ¼µÄÌå»ýΪ       L£¬ÖÊÁ¿Îª        g£¬»ìºÏÆøÌåÖÐCO2µÄ·Ö×ÓÊýΪ         ¡£
£¨2£©3.01¡Á1023¸öOH£­µÄÎïÖʵÄÁ¿Îª       mol£¬ÖÊÁ¿Îª       g£¬ÕâЩOH£­µÄÎïÖʵÄÁ¿Óë±ê×¼×´¿öÏÂÌå»ýΪ      LµÄNH3µÄÎïÖʵÄÁ¿ÏàµÈ£¬Óë      g Na£«º¬ÓеÄÀë×ÓÊýÏàͬ¡£
£¨3£©±ê×¼×´¿öÏ£¬33.6 LµÄNH3Ëù¾ßÓеÄÎïÖʵÄÁ¿Îª_________mol£¬½«ÆäÈܽâÓÚË®Åä³É1 LµÄÈÜÒº£¬ÔòÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________mol/L¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

a gþÂÁºÏ½ðͶÈëxmL£¬2mol/LµÄÑÎËáÖУ¬½ðÊôÍêÈ«Èܽ⣬ÔÙ¼ÓÈëymL£¬1mol/L NaOHÈÜÒº£¬³Áµí´ïµ½×î´óÖµ£¬ÖÊÁ¿Îª£¨a+1.7£©g£¬ÔòÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨   £©
A£®Ã¾ÂÁºÏ½ðÓëÑÎËᷴӦתÒƵç×ÓÊýΪ0.1NA
B£®³ÁµíΪMg£¨OH£©2ºÍAl£¨OH£©3»ìºÏÎï
C£®x =2y
D£®aµÄÈ¡Öµ·¶Î§Îª0.9g <a<1.2g

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂÁÐËùº¬·Ö×ÓÊýÓɶൽÉÙµÄÅÅÁÐ˳ÐòÊÇ                ,
A£®±ê×¼×´¿öÏÂ33.6 L H2B£®Ëùº¬µç×ÓµÄÎïÖʵÄÁ¿Îª4 molµÄH2
C£®20 ¡æʱ45 g H2OD£®³£ÎÂÏ£¬16 g O2Óë14 g N2µÄ»ìºÏÆøÌå
E£®º¬Ô­×Ó×ÜÊýԼΪ1.204¡Á1024µÄNH3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

Èô20 gÃܶÈΪd g£¯cm3µÄÏõËá¸ÆÈÜÒºÀﺬ1 g Ca2£«£¬ÔòNO3£­µÄŨ¶ÈÊÇ
A£®d/400 mol¡¤L£­1B£®20/d mol¡¤L£­1C£®2.5d mol¡¤L£­1D£®1.25d mol¡¤L£­1

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸