È¡50.0 mL Na2CO3ºÍNa2SO4µÄ»ìºÏÈÜÒº£¬¼ÓÈë¹ýÁ¿BaCl2ÈÜÒººóµÃµ½14.51 g°×É«³Áµí£¬ÓùýÁ¿Ï¡ÏõËá´¦Àíºó³ÁµíÁ¿¼õÉÙµ½4.66 g £¬²¢ÓÐÆøÌå·Å³ö¡£

(1)ÓÃÏ¡ÏõËá´¦ÀíºóµÄ²»ÈÜÎïΪ______________(Ìѧʽ)£¬Ð´³öÕû¸ö¹ý³ÌÖÐËù·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º__________________£¬__________________£¬__________________¡£

(2)Ô­»ìºÏÈÜÒºÖÐNa2CO3ºÍNa2SO4µÄÎïÖʵÄÁ¿Å¨¶È·Ö±ðÊÇ________¡¢________¡£

(3)²úÉúµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ________¡£


½âÎö¡¡(2)×îºóÊ£Óà³ÁµíΪBaSO4

n(Na2SO4)£½n(BaSO4)£½£½0.02mol

c(Na2SO4)£½£½0.4mol¡¤L£­1

n(Na2CO3)£½n(BaCO3)£½£½0.05mol

c(Na2CO3)£½£½1mol¡¤L£­1

(3)V(CO2)£½0.05mol¡Á22.4L¡¤mol£­1£½1.12L¡£

´ð°¸¡¡(1)BaSO4

CO£«Ba2£«===BaCO3¡ý

SO£«Ba2£«===BaSO4¡ý

BaCO3£«2H£«===Ba2£«£«H2O£«CO2¡ü

(2)0.4 mol¡¤L£­1

1£®0 mol¡¤L£­1

(3)1.12 L


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Äܹ»²â¶¨ÓлúÎï·Ö×ÓÖл¯Ñ§¼üºÍ¹ÙÄÜÍŵÄÊÇ£¨     £©

A¡¢ÖÊÆ×          B¡¢ºìÍâ¹âÆ×        C¡¢×ÏÍâ¹âÆ×     D¡¢ºË´Å¹²ÕñÆ×

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÀûÓÃÉú»îÖг£¼ûµÄ²ÄÁÏ¿ÉÒÔ½øÐкܶà¿ÆѧʵÑ飬ÉõÖÁÖÆ×÷³öһЩÓÐʵ¼ÊÓ¦ÓüÛÖµµÄ×°ÖÃÀ´£¬

Èç·ÏÂÁ¹ÞºÍ̼°ô£¬Ê³ÑÎË®µÈ²ÄÁÏÖÆ×÷¿ÉÓÃÓÚÇý¶¯Íæ¾ßµÄµç³Ø¡£ÉÏÊöµç³Ø¹¤×÷ʱ£¬

ÓйØ˵·¨ÕýÈ·µÄÊÇ                      £¨    £©

A£®ÂÁ¹Þ½«Öð½¥±»¸¯Ê´

B£®Ì¼°ôÉÏ·¢ÉúµÄ·´Ó¦Îª£ºO2£«4e£­===2O2£­

C£®Ì¼°ôÓ¦ÓëÍæ¾ßµç»úµÄ¸º¼«ÏàÁ¬

D£®¸Ãµç³Ø¹¤×÷Ò»¶Îʱ¼äºó̼°ôµÄÖÊÁ¿»á¼õÇá

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ΪÁ˼ìÑéNa2SO3ÖÐÊÇ·ñ»ìÓÐNa2SO4£¬Ó¦Ñ¡ÓõÄÊÔ¼ÁÊÇ(¡¡¡¡)

A£®BaCl2ÈÜÒº                            B£®BaCl2ÈÜÒººÍÏ¡ÑÎËá

C£®BaCl2ÈÜÒººÍÏ¡ÏõËá                    D£®BaCl2ÈÜÒººÍÏ¡ÁòËá

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ijͬѧΪ¼ìÑéÈÜÒºÖÐÊÇ·ñº¬Óг£¼ûµÄËÄÖÖÎÞ»úÀë×Ó£¬½øÐÐÁËÏÂͼËùʾµÄʵÑé²Ù×÷¡£ÆäÖмìÑé¹ý³ÌÖвúÉúµÄÆøÌåÄÜʹºìɫʯÈïÊÔÖ½±äÀ¶¡£ÓɸÃʵÑéÄܵõ½µÄÕýÈ·½áÂÛÊÇ(¡¡¡¡)

A£®Ô­ÈÜÒºÖÐÒ»¶¨º¬ÓÐSO

B£®Ô­ÈÜÒºÖÐÒ»¶¨º¬ÓÐNHÀë×Ó

C£®Ô­ÈÜÒºÖÐÒ»¶¨º¬ÓÐCl£­Àë×Ó

D£®Ô­ÈÜÒºÖÐÒ»¶¨º¬ÓÐFe3£«Àë×Ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


°Ñ0.05 mol NaOH¹ÌÌå·Ö±ð¼ÓÈëÏÂÁÐ100 mLÈÜÒºÖУ¬ÈÜÒºµÄµ¼µçÐÔ»ù±¾²»±äµÄÊÇ

A£®×ÔÀ´Ë®                        B£®0.5 mol/LÑÎËá  

C£®0.5 mol/L´×Ëá                D£®0.5 mol/L mol·L¡¥1°±Ë®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔÚζÈÏàͬ£¬Ñ¹Ç¿·Ö±ðΪp1¡¢p2Ìõ¼þÏ£¬A(g)£«2B(g)nC(g)µÄ·´Ó¦ÌåϵÖÐ,CµÄÌå»ý·ÖÊý(C£¥)Ëæʱ¼ä(t)±ä»¯µÄÇúÏßÈçͼËùʾ£®

ÏÂÁнáÂÛÕýÈ·µÄÊÇ      

A£®p1 > p2  n < 3                 B£®p1 < p2  n > 3

C£®p1 < p2 n £½ 3                 D£®p1 > p2 n > 3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ( )

A£®±ê×¼×´¿öÏ£¬22.4LCCl4º¬ÓеķÖ×ÓÊýΪNA

B£®³£Î³£Ñ¹Ï£¬17gNH3 Ëùº¬µÄÔ­×ÓÊýĿΪ4NA

C£®1 mol Na2O2ÓëH2OÍêÈ«·´Ó¦£¬×ªÒÆ2NA¸öµç×Ó

D£®0.1mol/LNa2CO3ÈÜÒºÖк¬ÓеÄNa+ÊýĿΪ0.2NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


̼ºÍµªµÄÐí¶à»¯ºÏÎïÔÚ¹¤Å©ÒµÉú²úºÍÉú»îÖÐÓÐÖØÒªµÄ×÷Óá£

£¨1£©ÒÔCO2ÓëNH3ΪԭÁϿɺϳɻ¯·ÊÄòËØ[»¯Ñ§Ê½ÎªCO(NH2)2]¡£ÒÑÖª£º

¢Ù2NH3(g)£«CO2(g) === NH2CO2NH4(s)       ∆H£½-159.5 kJ¡¤mol£­1

¢ÚNH2CO2NH4(s) === CO(NH2)2(s)£«H2O(g)   ∆H£½+116.5 kJ¡¤mol£­1

¢ÛH2O(l) === H2O(g)                      ∆H£½+44.0 kJ¡¤mol£­1

Ôò·´Ó¦2NH3(g)£«CO2(g) === CO(NH2)2(s)£«H2O(l)µÄ∆H£½           kJ¡¤mol£­1

£¨2£©ÓûîÐÔÌ¿»¹Ô­·¨¿ÉÒÔ´¦ÀíµªÑõ»¯ÎijÑо¿Ð¡×éÏòijÃܱÕÈÝÆ÷¼ÓÈëÒ»¶¨Á¿µÄ»îÐÔÌ¿ºÍNO£¬·¢Éú·´Ó¦C(s)£«2NO(g)N2(g)£«CO2 (g)    ∆H£½Q kJ¡¤mol£­1¡£

ÔÚT1¡æʱ£¬·´Ó¦½øÐе½Í¬Ê±¼ä²âµÃ¸÷ÎïÖʵÄŨ¶ÈÈçÏ£»30minºó£¬Ö»¸Ä±äijһÌõ¼þ£¬·´Ó¦ÖØдﵽƽºâ£¬¸÷ÎïÖÊŨ¶ÈÈçͼËù¡£

¢Ù0~10minÄÚ£¬NOµÄƽ¾ù·´Ó¦ËÙÂÊv(NO)£½                                   £¬

T1¡æʱ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK£½                           £»

¢Ú30minºó£¬Ö»¸Ä±äijһÌõ¼þ£¬·´Ó¦ÖØдﵽƽºâ£¬¸÷ÎïÖÊŨ¶ÈÈçͼËùʾ¡£¸ù¾ÝÉϱíÖеÄÊý¾ÝÅжϸıäµÄÌõ¼þ¿ÉÄÜÊÇ          £¨Ìî×Öĸ±àºÅ£©¡£

a£®¼ÓÈëÒ»¶¨Á¿µÄ»îÐÔÌ¿            b£®Í¨ÈëÒ»¶¨Á¿µÄNO 

c£®Êʵ±ËõСÈÝÆ÷µÄÌå»ý            d£®¼ÓÈëºÏÊʵĴ߻¯¼Á       e£®Éý¸ßζÈ

¢Û30minºó£¬Èç¹û½«Î¶ÈÉý¸ßÖÁT2¡æ£¬´ïµ½Æ½ºâʱ£¬ÈÝÆ÷ÖÐNO¡¢N2¡¢CO2µÄŨ¶ÈÖ®±ÈΪ5:3:3£¬Ôò Q         0£¨Ìî¡°£¾¡±»ò¡°£¼¡±£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸