½«32.64gÍ­Óë140mLÒ»¶¨Å¨¶ÈµÄÏõËá·´Ó¦£¬Í­ÍêÈ«Èܽâ²úÉúµÄNOºÍNO2»ìºÏÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ11.2L¡£Çë»Ø´ð£º
£¨1£©NOµÄÌå»ýΪ       L£¬NO2µÄÌå»ýΪ         L¡£
£¨2£©´ý²úÉúµÄÆøÌåÈ«²¿Êͷźó£¬ÏòÈÜÒº¼ÓÈëV mL amol¡¤L£­1µÄNaOHÈÜÒº£¬Ç¡ºÃʹÈÜÒºÖеÄCu2£«È«²¿×ª»¯³É³Áµí£¬ÔòÔ­ÏõËáÈÜÒºµÄŨ¶ÈΪ          mol/L¡£
£¨3£©ÓûʹͭÓëÏõËá·´Ó¦Éú³ÉµÄÆøÌåÔÚNaOHÈÜÒºÖÐÈ«²¿×ª»¯ÎªNaNO3£¬ÖÁÉÙÐèÒª30%µÄË«ÑõË®          g¡£

£¨1£©5.824 L   5.376 L £¨2£©(10-3 va +0.5)/0.14 £¨3£©57.8 g 

½âÎöÊÔÌâ·ÖÎö£º£¨1£©Éè·´Ó¦ÖвúÉúNOºÍNO2µÄÌå»ý·Ö±ðΪx¡¢y(¾ÝµÃʧµç×ÓÊغãÓÐ)
x+y=11.2
(x/22.4)¡Á3+(y/22.4) ¡Á1=(32.64/64)¡Á2 
½âµÃx="5.824" L   y="5.376" L
£¨2£©¸ù¾Ý·´Ó¦ºóµÄÈÜÒºÖÐÖ»ÓÐÒ»ÖÖÈÜÖÊNaNO3 ,ÉèÔ­ÏõËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪC
£¨ÓɵªÔªËØÊغ㣩Ôò£º 0.14¡ÁC= 10-3 ¡Áv¡Áa + 11.2/22.4
µÃÔ­ÏõËáµÄŨ¶ÈΪ£º C= (10-3 va +0.5)/0.14
£¨3£©´Ó·´Ó¦µÄʼ̬ºÍ×îÖÕ״̬¿´,Í­ÔÚ·´Ó¦ÖÐʧȥµç×Ó,Ë«ÑõË®ÔÚ·´Ó¦Öеõ½µç×Ó,ÐèÒª30%µÄË«ÑõË®µÄÖÊÁ¿Îªm Ôò(¾ÝµÃʧµç×ÓÊغãÓÐ)£º(32.64/64)¡Á2=¡²(30%¡Ám)/34¡³¡Á2 ½âµÃ:m="57.8" g 
¿¼µã£ºÑõ»¯»¹Ô­·´Ó¦¹æÂÉ£¬Ñõ»¯»¹Ô­·´Ó¦·½³ÌʽµÄ¼ÆËã¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ëæ×ÅÄÜÔ´ÎÊÌâµÄ½øÒ»²½Í»³ö£¬ÀûÓÃÈÈ»¯Ñ§Ñ­»·ÖÆÇâµÄÑо¿Êܵ½Ðí¶à·¢´ï¹ú¼ÒµÄÇàíù¡£×î½üµÄÑо¿·¢ÏÖ£¬¸´ºÏÑõ»¯ÎïÌúËáÃÌ(MnFe2O4)Ò²¿ÉÒÔÓÃÓÚÈÈ»¯Ñ§Ñ­»··Ö½âË®ÖÆÇ⣬MnFe2O4µÄÖƱ¸Á÷³ÌÈçÏ£º
 
(1)Ô­ÁÏFe(NO3)nÖÐn£½________£¬Í¶ÈëÔ­ÁÏFe(NO3)nºÍMn(NO3)2µÄÎïÖʵÄÁ¿Ö®±ÈӦΪ________¡£
(2)²½Öè¶þÖС°Á¬Ðø½Á°è¡±µÄÄ¿µÄÊÇ__________________________________________
²½ÖèÈýÖÐÏ´µÓ¸É¾»µÄ±ê×¼ÊÇ________________________________________________
(3)ÀûÓÃMnFe2O4ÈÈ»¯Ñ§Ñ­»·ÖÆÇâµÄ·´Ó¦¿É±íʾΪ£º
MnFe2O4MnFe2O4£­x£«O2¡ü£»
MnFe2O4£­x£«xH2OMnFe2O4£«xH2¡ü
ÇëÈÏÕæ·ÖÎöÉÏÊöÁ½¸ö·´Ó¦²¢»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÈôMnFe2O4£­xÖÐx£½0.8£¬ÔòMnFe2O4£­xÖÐFe2£«Õ¼È«²¿ÌúÔªËصİٷÖÂÊΪ________¡£
¢Ú¸ÃÈÈ»¯Ñ§Ñ­»·ÖÆÇâ·¨µÄÓŵãÓÐ_____________________¡¢________________________ (´ðÁ½µã¼´¿É)¡£
¸ÃÈÈ»¯Ñ§Ñ­»··¨ÖÆÇâÉÐÓв»×ãÖ®´¦£¬½øÒ»²½¸Ä½øµÄÑо¿·½ÏòÊÇ___________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉÏÀûÓý¹Ì¿ÔÚʯ»ÒÒ¤ÖÐȼÉÕ·ÅÈÈ£¬Ê¹Ê¯»Òʯ·Ö½âÉú²úCO2¡£Ö÷Òª·´Ó¦ÈçÏ£º
C+O2¡úCO2 ¢Ù£¬        CaCO3¡úCO2¡ü+CaO ¢Ú
£¨1£©º¬Ì¼Ëá¸Æ95%µÄʯ»Òʯ2.0 t°´¢ÚÍêÈ«·Ö½â£¨ÉèÔÓÖʲ»·Ö½â£©£¬¿ÉµÃ±ê×¼×´¿öÏÂCO2µÄÌå»ýΪ_________________m3¡£
£¨2£©´¿¾»µÄCaCO3ºÍ½¹Ì¿°´¢Ù¢ÚÍêÈ«·´Ó¦£¬µ±Ò¤ÄÚÅä±ÈÂÊ=2.2ʱ£¬Ò¤ÆøÖÐCO2µÄ×î´óÌå»ý·ÖÊýΪ¶àÉÙ£¿£¨Éè¿ÕÆøÖ»º¬N2ÓëO2£¬ÇÒÌå»ý±ÈΪ4¡Ã1£¬ÏÂͬ£©
£¨3£©Ä³´ÎÒ¤Æø³É·ÖÈçÏ£ºO2 0.2%£¬CO 0.2%£¬CO2 41.6%£¬ÆäÓàΪN2¡£Ôò´Ë´ÎÒ¤ÄÚÅä±ÈÂÊΪºÎÖµ£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

½«11.2gÌúͶÈë200mLijŨ¶ÈµÄÑÎËáÖУ¬ÌúºÍÑÎËáÇ¡ºÃÍêÈ«·´Ó¦¡£Çó£º
£¨1£©ËùÓÃÑÎËáÖÐHClµÄÎïÖʵÄÁ¿Å¨¶È
£¨2£©·´Ó¦ÖÐÉú³ÉµÄH2ÔÚ±ê×¼×´¿öϵÄÌå»ý
£¨3£©ÔÚ·´Ó¦ºóµÄÈÜÒºÖÐͨÈëCl2£¬Ð´³öËù·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ²¢ÓÃË«ÏßÇűê³öµç×ÓתÒƵķ½ÏòºÍÊýÄ¿£º
                                                       

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©½«agÂÈ»¯¸ÆÈÜÓÚ1.8LË®ÖУ¬Ç¡ºÃʹ¸ÆÀë×ÓÊýÓëË®·Ö×ÓÊýÖ®±ÈΪ1:100£¬ÔòaֵΪ               ¡£
£¨2£©ÔÚ·´Ó¦2A+B=3C+2DÖУ¬ÒÑÖª3.4gAÓë3.2gBÍêÈ«·´Ó¦£¬Éú³É4.8gC£¬ÓÖÖªµÀDµÄʽÁ¿Îª18£¬ÔòBµÄʽÁ¿ÊÇ           
£¨3£©25.4g ij¶þ¼Û½ðÊôÂÈ»¯Îï(ACl2)Öк¬ÓÐ0.4mol Cl£­£¬ÔòACl2µÄĦ¶ûÖÊÁ¿ÊÇ         £»AµÄÏà¶ÔÔ­×ÓÖÊÁ¿ÊÇ      £»ACl2µÄ»¯Ñ§Ê½ÊÇ        ¡£
£¨4£© ij»ìºÏÎïÓÉNa2SO4¡¢Al2£¨SO4£©3×é³É£¬ÒÑÖªNa¡¢AlÁ½ÔªËصÄÖÊÁ¿Ö®±ÈΪ23: 9£¬ÔòNa2SO4ºÍAl2£¨SO4£©3ÎïÖʵÄÁ¿Ö®±ÈΪ             £¬º¬1.00mol SO42¨CµÄ¸Ã»ìºÏÎïµÄÖÊÁ¿Îª             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ʵÑéÊÒÓÃÂÈ»¯ÄƹÌÌåÅäÖÆ1£®00 mol/LµÄNaClÈÜÒº0£®5 L£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Çëд³ö¸ÃʵÑéµÄʵÑé²½Ö裺
¢Ù       £¬¢Ú       £¬¢Û       £¬¢Ü        £¬¢Ý        £¬¢Þ        ¡£
£¨2£©ËùÐèÒÇÆ÷Ϊ£ºÍÐÅÌÌìƽ¡¢»¹ÐèÒªÄÇЩʵÑéÒÇÆ÷²ÅÄÜÍê³É¸ÃʵÑ飬Çëд³ö£º                        ¡£
£¨3£©ÊÔ·ÖÎöÏÂÁвÙ×÷¶ÔËùÅäÈÜÒºµÄŨ¶ÈÓкÎÓ°Ïì¼°Ôì³É¸ÃÓ°ÏìµÄÔ­Òò¡£
¶¨Èݺ󣬼Ӹǵ¹×ªÒ¡ÔȺ󣬷¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÓֵμÓÕôÁóË®ÖÁ¿Ì¶È¡£¶ÔËùÅäÈÜҺŨ¶ÈµÄÓ°Ï죺             £¬Ô­ÒòÊÇ£º                       ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ʵÑéÐèÒª0.10mol/LNaOHÈÜÒº470mL£¬¸ù¾ÝÈÜÒºÅäÖÆÖÐÇé¿ö»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÊµÑéÖгýÁËÍÐÅÌÌìƽ¡¢ÉÕ±­¡¢²£Á§°ô¡¢Á¿Í²¡¢Ò©³×Í⻹ÐèÒªµÄÆäËüÒÇÆ÷ÓУº                                  ¡£
£¨2£©¸ù¾Ý¼ÆËãµÃÖª£¬ËùÐèNaOHµÄÖÊÁ¿Îª               g¡£
£¨3£©¶¨ÈÝʱ£¬´ýÈÝÁ¿Æ¿ÖÐÈÜÒºµÄ°¼ÒºÃæÕýºÃÓë¿Ì¶ÈÏßÏàÇУ¬¸ÇºÃÆ¿ÈûºóµÄÏÂÒ»²½²Ù×÷ÊÇ                                      ¡£
£¨4£©¶¨ÈÝʱ,Èô¼ÓÈëµÄË®³¬¹ý¿Ì¶ÈÏߣ¬±ØÐë²ÉÈ¡µÄ´ëÊ©ÊÇ£º               ¡£
£¨5£©ÏÂÁвÙ×÷¶ÔËùÅäŨ¶ÈÓкÎÓ°Ï죨Ìîд×Öĸ£©
Æ«µÍµÄÓР                    £»ÎÞÓ°ÏìµÄÓР                  ¡£

A£®³ÆÁ¿ÓÃÁËÉúÐâµÄíÀÂ룻
B£®½«NaOH·ÅÔÚÖ½ÕÅÉϳÆÁ¿£»
C£®NaOHÔÚÉÕ±­ÖÐÈܽâºó£¬Î´ÀäÈ´¾ÍÁ¢¼´×ªÒƵ½ÈÝÁ¿Æ¿ÖУ»
D£®¶¨ÈÝʱ¸©Êӿ̶ÈÏß
E¡¢ÈÝÁ¿Æ¿Î´¸ÉÔï¼´ÓÃÀ´ÅäÖÆÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ҫ׼ȷÕÆÎÕ»¯Ñ§ÓÃÓï¼°³£ÓüÆÁ¿·½·¨¡£°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©NA±íʾ°¢·ü¼ÓµÂÂÞ³£Êý¡£28gÒÒÏ©ºÍ»·¶¡Í飨C4H8£©µÄ»ìºÏÆøÌåÖк¬ÓÐ____NA¸ö̼ԭ×Ó£»·Ö×Ó×ÜÊýΪNA¸öµÄNO2ºÍCO2»ìºÏÆøÌ庬______ NA¸öÑõÔ­×ÓÊý£»1mol37ClÖУ¬ÖÐ×ÓÊý±ÈÖÊ×ÓÊý¶à_______ NA¸ö£»1L 1mol/LFe2(SO4)3ÈÜÒºÖк¬_____NA¸öSO42£­Àë×Ó¡£
£¨2£©Í¼ÖÐA¡¢B·Ö±ðÊÇij΢Á£µÄ½á¹¹Ê¾Òâͼ£¬»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙÈôA±íʾijԪËصÄÔ­×Ó£¬Ôòy£½              ¡£
¢ÚÈôB±íʾijϡÓÐÆøÌåÔªËصÄÔ­×Ó£¬Ôò¸ÃÔªËصĵ¥ÖʵĻ¯Ñ§Ê½Îª             £¬ÈôBÊÇÒõÀë×ӵĽṹʾÒâͼ£¬ÔòxµÄÈ¡Öµ·¶Î§ÊÇ________________¡£
£¨3£©RxO42£­ÖÐRµÄ»¯ºÏ¼ÛΪ___________(Óú¬x µÄʽ×Ó±íʾ)£¬µ±0.3 mol RxO42£­ÍêÈ«·´Ó¦£¬Éú³ÉRO2ʱ£¬×ªÒÆ0.6 molµç×Ó£¬Ôòx£½__________¡£
£¨4£©½«7.8 gþÂÁºÏ½ðÓë100mL Ï¡ÁòËáÇ¡ºÃÍêÈ«·´Ó¦£¬½«·´Ó¦ºóµÄÈÜÒº¼ÓÈÈÕô¸É£¬µÃµ½ÎÞË®ÁòËáÑÎ46.2 g£¬ÔòÔ­ÁòËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¼ÆËãÌâ

ÓÐÒ»ÁòËáºÍÏõËáµÄ»ìºÏÈÜÒº£¬È¡³ö20mL£¬¼ÓÈë×ãÁ¿BaCl2ÈÜÒº£¬¾­¹ýÂË¡¢Ï´µÓ¡¢ºæ¸ÉºóµÃ³Áµí9.32g£»ÂËÒºÓë4.0mol/LµÄÇâÑõ»¯ÄÆÈÜÒº35mLÇ¡ºÃÍêÈ«Öк͡£ÊÔÇó£º
£¨1£©Ô­»ìºÏÈÜÒºÖÐÁòËáºÍÏõËáµÄÎïÖʵÄÁ¿Å¨¶È¡£
£¨2£©ÁíÈ¡10mLÔ­ÈÜÒº£¬¼ÓÈë0.96gÍ­·Û¹²ÈÈ£¬Éú³ÉÒ»Ñõ»¯µªµÄÌå»ýΪ¶àÉÙ£¿
£¨3£©ÁíÈ¡10mLÔ­ÈÜÒº£¬¼ÓÈë1.92gÍ­·Û¹²ÈÈ,£¬ÖÁÉÙ»¹ÒªÏò·´Ó¦ºóµÄÈÜÒºÖмÓÈë¶àÉÙºÁÉý1mol/LÁòËá²ÅÄܽ«Í­·ÛÇ¡ºÃÍêÈ«Èܽ⣿

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸