ÒÑÖªA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÔªËØÖУ¬A¡¢B¡¢C¡¢DÊÇ×é³Éµ°°×ÖʵĻù±¾ÔªËØ£»AÓëBµÄÔ­×ÓÐòÊýÖ®ºÍµÈÓÚCÔ­×ÓºËÄÚµÄÖÊ×ÓÊý£»AÓëE¡¢DÓëF·Ö±ðλÓÚͬһÖ÷×壬EÓëFͬÖÜÆÚ£¬ÇÒFÔ­×ÓºËÄÚµÄÖÊ×ÓÊýÊÇDÔ­×ÓºËÍâµç×ÓÊýµÄ2±¶£®¾Ý´ËÇë»Ø´ð£º
£¨1£©FÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ
 
£®
£¨2£©ÓÉA¡¢C¡¢D¡¢F°´8£º2£º4£º1Ô­×Ó¸öÊý±È×é³ÉµÄ»¯ºÏÎï¼×Öк¬ÓеĻ¯Ñ§¼üÀàÐÍΪ
 
£»¼×ÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ
 
£¨ÓÃÀë×ÓŨ¶È·ûºÅ±íʾ£©£®
£¨3£©»¯ºÏÎïÒÒÓÉA¡¢C×é³ÉÇÒÏà¶Ô·Ö×ÓÖÊÁ¿Îª32£»»¯ºÏÎï±ûÓÉA¡¢D×é³ÉÇÒ·Ö×ÓÄÚµç×Ó×ÜÊýÓëÒÒ·Ö×ÓÄÚµç×Ó×ÜÊýÏàµÈ£»ÒÒÓë±ûµÄ·´Ó¦¿ÉÓÃÓÚ»ð¼ý·¢É䣨·´Ó¦²úÎï²»ÎÛȾ´óÆø£©£¬Ôò¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨4£©ÓÉA¡¢D¡¢E¡¢F×é³ÉµÄ»¯ºÏÎﶡÄÜÓëÁòËá·´Ó¦²¢·Å³ö´Ì¼¤ÐÔÆøζµÄÆøÌ壬Ôò¶¡µÄ»¯Ñ§Ê½Îª
 
£»ÊµÑé²âµÃ¶¡ÈÜÒºÏÔÈõËáÐÔ£¬ÓÉ´ËÄãÄܵóöµÄ½áÂÛÊÇ
 
£®
£¨5£©ÓÉB¡¢A°´1£º4Ô­×Ó¸öÊý±È×é³ÉµÄ»¯ºÏÎïÎìÓëDµÄ³£¼ûÆø̬µ¥Öʼ°NaOHÈÜÒº¹¹³ÉÔ­µç³Ø£¨Èçͼ£©£¬ÊÔ·ÖÎö£º
¢Ù±ÕºÏK£¬Ð´³ö×ó³ØXµç¼«µÄ·´Ó¦Ê½
 
£»
¢Ú±ÕºÏK£¬µ±Xµç¼«ÏûºÄ1.6g»¯ºÏÎïÎìʱ£¨¼ÙÉè¹ý³ÌÖÐÎÞÈκÎËðʧ£©£¬ÔòÓÒ³ØÁ½¼«¹²·Å³öÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ
 
Éý£®
¿¼µã£ºÎ»ÖýṹÐÔÖʵÄÏ໥¹ØϵӦÓÃ,Ô­µç³ØºÍµç½â³ØµÄ¹¤×÷Ô­Àí
רÌ⣺ԪËØÖÜÆÚÂÉÓëÔªËØÖÜÆÚ±íרÌâ
·ÖÎö£ºA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÔªËØÖУ¬A¡¢B¡¢C¡¢DÊÇ×é³Éµ°°×ÖʵĻù±¾ÔªËØ£¬·Ö±ðΪH¡¢C¡¢N¡¢OÖеÄÒ»ÖÖ£¬DÓëFλÓÚͬһÖ÷×壬ÇÒFÔ­×ÓºËÄÚµÄÖÊ×ÓÊýÊÇDÔ­×ÓºËÍâµç×ÓÊýµÄ2±¶£¬ÔòDΪÑõÔªËØ¡¢FΪÁòÔªËØ£»AÓëBµÄÔ­×ÓÐòÊýÖ®ºÍµÈÓÚCÔ­×ÓºËÄÚµÄÖÊ×ÓÊý£¬ÔòA¡¢B·Ö±ðΪÇ⡢̼ÖеÄÒ»ÖÖ£¬CΪµªÔªËØ£»ÓÉ£¨5£©ÖÐB¡¢AÄÜ°´1£º4Ô­×Ó¸öÊý±È×é³É»¯ºÏÎ¿ÉÍÆÖªAΪÇâ¡¢BΪ̼£»AÓëEͬÖ÷×壬EÓëFͬÖÜÆÚ£¬ÔòEΪNa£¬
£¨1£©ÖÜÆÚÊý=µç×Ó²ãÊý¡¢Ö÷×å×åÐòÊý=×îÍâ²ãµç×ÓÊý£»
£¨2£©ÓÉH¡¢N¡¢O¡¢S°´8£º2£º4£º1Ô­×Ó¸öÊý±È×é³ÉµÄ»¯ºÏÎï¼×Ϊ£¨NH4£©2SO4£¬º¬ÓÐÀë×Ó¼ü¡¢¹²¼Û¼ü£¬ÈÜÒºÖÐNH4+Àë×ÓË®½â£¬ÈÜÒº³ÊËáÐÔ£»
£¨3£©»¯ºÏÎïÒÒÓÉH¡¢N×é³ÉÇÒÏà¶Ô·Ö×ÓÖÊÁ¿Îª32£¬ÔòÒÒΪN2H4£¬»¯ºÏÎï±ûÓÉH¡¢O×é³ÉÇÒ·Ö×ÓÄÚµç×Ó×ÜÊýÓëÒÒ·Ö×ÓÄÚµç×Ó×ÜÊýÏàµÈ£¬Ôò±ûΪH2O2£»ÒÒÓë±ûµÄ·´Ó¦¿ÉÓÃÓÚ»ð¼ý·¢É䣬·´Ó¦²úÎï²»ÎÛȾ´óÆø£¬Ó¦Éú³ÉµªÆøÓëË®£»
£¨4£©ÓÉH¡¢O¡¢Na¡¢S×é³ÉµÄ»¯ºÏÎﶡÄÜÓëÁòËá·´Ó¦²¢·Å³ö´Ì¼¤ÐÔÆøζµÄÆøÌ壬Ôò¶¡ÎªNaHSO3£¬ÆäÈÜÒºÖÐHSO3-¼È·¢ÉúµçÀëÓÖ·¢ÉúË®½â£¬Á½¹ý³ÌµÄÏà¶ÔÇ¿Èõ¾ö¶¨ÈÜÒºËá¼îÐÔ£»
£¨5£©ÓÉC¡¢H°´1£º4Ô­×Ó¸öÊý±È×é³ÉµÄ»¯ºÏÎïÎìΪCH4£¬ÓëDµÄ³£¼ûÆø̬µ¥ÖÊO2¼°NaOHÈÜÒº¹¹³ÉÔ­µç³Ø£¬
¢Ù±ÕºÏK£¬×ó³ØXµç¼«Í¨Èë¼×Í飬·¢ÉúÑõ»¯·´Ó¦£¬¼îÐÔÌõ¼þÏÂÉú³É̼Ëá¸ùÓëË®£»
¢Ú×ó³ØÏûºÄ1.6g¼×Í飬ÆäÎïÖʵÄÁ¿=
1.6g
16g/mol
=0.1mol£¬×ªÒƵç×ÓÎïÖʵÄÁ¿=0.1mol¡Á8=0.8mol£¬ÓÒ³ØΪµç½â³Ø£¬Ñô¼«ÉÏÒõÀë×ÓÖ»ÓÐÇâÑõ¸ùÀë×ӷŵçÉú³ÉÑõÆø£¬¸ù¾Ýµç×ÓתÒÆÊغã¼ÆËãÉú³ÉµÄÑõÆøÎïÖʵÄÁ¿£¬ÈÜÒºÖÐAg+µÄÎïÖʵÄÁ¿=0.2L¡Á2mol/L=0.4mol£¬ÍêÈ«·ÅµçתÒÆ»ñµÃµç×ÓΪ0.4mol£¬¹ÊÒõ¼«ÉÏAg+¡¢H+·Åµç£¬¸ù¾Ýµç×ÓתÒÆÊغã¼ÆËãÉú³ÉÇâÆøµÄÎïÖʵÄÁ¿£¬½ø¶ø¼ÆËã»ñµÃÆøÌå×ÜÌå»ý£®
½â´ð£º ½â£ºA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÔªËØÖУ¬A¡¢B¡¢C¡¢DÊÇ×é³Éµ°°×ÖʵĻù±¾ÔªËØ£¬·Ö±ðΪH¡¢C¡¢N¡¢OÖеÄÒ»ÖÖ£¬DÓëFλÓÚͬһÖ÷×壬ÇÒFÔ­×ÓºËÄÚµÄÖÊ×ÓÊýÊÇDÔ­×ÓºËÍâµç×ÓÊýµÄ2±¶£¬ÔòDΪÑõÔªËØ¡¢FΪÁòÔªËØ£»AÓëBµÄÔ­×ÓÐòÊýÖ®ºÍµÈÓÚCÔ­×ÓºËÄÚµÄÖÊ×ÓÊý£¬ÔòA¡¢B·Ö±ðΪÇ⡢̼ÖеÄÒ»ÖÖ£¬CΪµªÔªËØ£»ÓÉ£¨5£©ÖÐB¡¢AÄÜ°´1£º4Ô­×Ó¸öÊý±È×é³É»¯ºÏÎ¿ÉÍÆÖªAΪÇâ¡¢BΪ̼£»AÓëEͬÖ÷×壬EÓëFͬÖÜÆÚ£¬ÔòEΪNa£¬
£¨1£©FΪÁòÔªËØ£¬Ò»ÖÖºËÍâÓÐ3¸öµç×Ӳ㡢×îÍâ²ãµç×ÓÊýΪ6£¬´¦ÓÚÖÜÆÚ±íÖеÚÈýÖÜÆÚµÚ¢öA×壬
¹Ê´ð°¸Îª£ºµÚÈýÖÜÆÚµÚ¢öA×壻
£¨2£©ÓÉH¡¢N¡¢O¡¢S°´8£º2£º4£º1Ô­×Ó¸öÊý±È×é³ÉµÄ»¯ºÏÎï¼×Ϊ£¨NH4£©2SO4£¬º¬ÓÐÀë×Ó¼ü¡¢¹²¼Û¼ü£¬ÈÜÒºÖÐNH4+Àë×ÓË®½â£¬ÈÜÒº³ÊËáÐÔ£¬c£¨H+£©£¾c£¨OH-£©£¬Ë®½â³Ì¶È²»´ó£¬¹Êc£¨NH4+£©£¾c£¨SO42-£©£¾c£¨H+£©£¬¹Ê´ð°¸Îª£ºÀë×Ó¼ü¡¢¹²¼Û¼ü£»c£¨NH4+£©£¾c£¨SO42-£©£¾c£¨H+£©£¾c£¨OH-£©£»
£¨3£©»¯ºÏÎïÒÒÓÉH¡¢N×é³ÉÇÒÏà¶Ô·Ö×ÓÖÊÁ¿Îª32£¬ÔòÒÒΪN2H4£¬»¯ºÏÎï±ûÓÉH¡¢O×é³ÉÇÒ·Ö×ÓÄÚµç×Ó×ÜÊýÓëÒÒ·Ö×ÓÄÚµç×Ó×ÜÊýÏàµÈ£¬Ôò±ûΪH2O2£»ÒÒÓë±ûµÄ·´Ó¦¿ÉÓÃÓÚ»ð¼ý·¢É䣬·´Ó¦²úÎï²»ÎÛȾ´óÆø£¬Ó¦Éú³ÉµªÆøÓëË®£¬Ôò¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºN2H4+2H2O2=N2+4H2O£¬
¹Ê´ð°¸Îª£ºN2H4+2H2O2=N2+4H2O£»
£¨4£©ÓÉH¡¢O¡¢Na¡¢S×é³ÉµÄ»¯ºÏÎﶡÄÜÓëÁòËá·´Ó¦²¢·Å³ö´Ì¼¤ÐÔÆøζµÄÆøÌ壬Ôò¶¡ÎªNaHSO3£¬ÆäÈÜÒºÖÐHSO3-¼È·¢ÉúµçÀëÓÖ·¢ÉúË®½â£¬ÊµÑé²âµÃ¶¡ÈÜÒºÏÔÈõËáÐÔ£¬ËµÃ÷HSO3-µÄµçÀëÄÜÁ¦±ÈË®½âÄÜÁ¦Ç¿£¬
¹Ê´ð°¸Îª£ºNaHSO3£»HSO3-µÄµçÀëÄÜÁ¦±ÈË®½âÄÜÁ¦Ç¿£»
£¨5£©ÓÉC¡¢H°´1£º4Ô­×Ó¸öÊý±È×é³ÉµÄ»¯ºÏÎïÎìΪCH4£¬ÓëDµÄ³£¼ûÆø̬µ¥ÖÊO2¼°NaOHÈÜÒº¹¹³ÉÔ­µç³Ø£¬
¢Ù±ÕºÏK£¬×ó³ØXµç¼«Í¨Èë¼×Í飬·¢ÉúÑõ»¯·´Ó¦£¬¼îÐÔÌõ¼þÏÂÉú³É̼Ëá¸ùÓëË®£¬µç¼«·´Ó¦Ê½Îª£ºCH4+10OH--8e-=CO32-+7H2O£¬
¹Ê´ð°¸Îª£ºCH4+10OH--8e-=CO32-+7H2O£»
¢Ú±ÕºÏK£¬µ±Xµç¼«ÏûºÄ1.6g¼×Í飬ÆäÎïÖʵÄÁ¿=
1.6g
16g/mol
=0.1mol£¬×ªÒƵç×ÓÎïÖʵÄÁ¿=0.1mol¡Á8=0.8mol£¬ÓÒ³ØΪµç½â³Ø£¬ÒõÀë×ÓÖ»ÓÐÇâÑõ¸ùÀë×ӷŵçÉú³ÉÑõÆø£¬¸ù¾Ýµç×ÓתÒÆÊغã¿ÉÖª£¬Éú³ÉÑõÆøµÄÎïÖʵÄÁ¿=
0.8mol
4
=0.2mol£¬ÈÜÒºÖÐAg+µÄÎïÖʵÄÁ¿=0.2L¡Á2mol/L=0.4mol£¬ÍêÈ«·ÅµçתÒÆ»ñµÃµç×ÓΪ0.4mol£¬¹Ê·´Ó¦µÄH+µÄÎïÖʵÄÁ¿=0.8mol-0.4mol=0.4mol£¬¹ÊÉú³ÉÇâÆøµÄÎïÖʵÄÁ¿=
0.4mol
2
=0.2mol£¬¹Ê¹²µÃµ½ÆøÌåÌå»ý=£¨0.2mol+0.2mol£©¡Á22.4L/mol=8.96L£¬
¹Ê´ð°¸Îª£º8.96£®
µãÆÀ£º±¾Ì⿼²é½á¹¹ÐÔÖÊλÖùØϵ¡¢ÔªËØ»¯ºÏÎïÍƶϡ¢ÑÎÀàË®½â¡¢Ô­µç³Ø¡¢µç½â³Ø¡¢»¯Ñ§¼ÆËãµÈ£¬ÌâÄ¿±È½Ï×ۺϣ¬ÍƶÏÔªËØÊǽâÌâµÄ¹Ø¼ü£¬×¢ÒâÀûÓõç×ÓתÒÆÊغã¼ÆËãÔ­µç³Ø¡¢µç½â³ØµÄÓйؼÆË㣬ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢Ca£¨ClO£©2ÈÜÒºÖУºc£¨Ca2+£©£¾c£¨ClO-£©£¾c£¨OH-£©£¾c£¨H+£©
B¡¢Í¬Å¨¶ÈµÄÏÂÁÐÈÜÒºÖУ¬¢ÙNH4HSO4 ¢ÚNH4Cl¢ÛNH3?H2O  c£¨NH4+£©ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ£º¢Ú£¾¢Ù£¾¢Û
C¡¢0.2 mol?L-1 CH3COOHÈÜÒººÍ0.2 mol?L-1 CH3COONaÈÜÒºµÈÌå»ý»ìºÏ£ºc£¨CH3COO-£©+c£¨OH-£©-c£¨H+£©=0.1 mol?L-1
D¡¢0.1mol/LNaHSÈÜÒº£ºc£¨OH-£©=c£¨H+£©+2c£¨H2S£©+c£¨HS-£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

°±µÄ´ß»¯Ñõ»¯ÊÇ°±ÖØÒªµÄ»¯Ñ§ÐÔÖÊ£®ÆäʵÑéÉè¼ÆÒ²ÊÇ´ó¼Ò̽¾¿µÄÖ÷ÒªÄÚÈÝ£®

ÑÝʾʵÑé¡°°±µÄ´ß»¯Ñõ»¯¡±×°Öü×Ò»°ãÈçͼ1£¬¸Ä½øʵÑé×°ÖÃÒÒÈçͼ2£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³ö°±´ß»¯Ñõ»¯·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£®
£¨2£©ÑÝʾʵÑéÖУ¬Ñ§Éú¿´µ½×¶ÐÎÆ¿ÖеÄÏÖÏóÊÇÆ¿¿Ú¸½½üÓкì×ØÉ«ÆøÌåÉú³É¡¢
 
£®¶øÔÚʵ¼ÊʵÑé¹ý³ÌÖУ¬Ñ§Éú»¹³£³£»á¿´µ½×¶ÐÎÆ¿ÖÐÓа×É«ÑÌÎí£¬³ýË®Í⣬ÐγɴËÑÌÎíµÄÎïÖÊ¿ÉÄÜÊÇ
 
¡¢
 
£®
£¨3£©¸Ä½øºóµÄʵÑéÓëÑÝʾʵÑé×°Öü×Ïà±È£¬³ýÁ˸ıäÁË´ß»¯¼ÁºÍ×°ÖÃÒÇÆ÷Í⣬»¹Ôö¼ÓÁËC¡¢EÁ½²¿·ÖʵÑé×°Öã¬ÆäÖÐC×°ÖÃÖиÉÔï¼ÁΪ
 
£¬EµÄ×÷ÓÃÊÇ
 
£®
£¨4£©¹¤ÒµÉÏÓð±ÆøÓë¿ÕÆø£¨ÑõÆøÕ¼¿ÕÆøµÄ20%£©»ìºÏ£¬Í¨¹ý¡°°±µÄ´ß»¯Ñõ»¯¡±µÈ·´Ó¦¹¤ÒµºÏ³ÉÏõËᣬÏõËṤҵÁ÷³ÌÈçͼËùʾ£¬¼×¡¢ÒÒ¡¢±ûÊǹ¤ÒµÉ豸£®ÆäÖз¢Éú°±´ß»¯Ñõ»¯µÄÉ豸ÊÇ£¨Ôڼס¢ÒÒ¡¢±ûÖÐÑ¡Ôñ£©
 
£¬¼××°ÖõÄ×÷ÓÃÊÇ
 
£®ÉÏ·½¹ÜµÀÖеÄÖ÷ÒªÆøÌå³É·ÖÊÇ
 
£®

£¨5£©ÎïÖÊÑ­»·ÀûÓÃÊǹ¤ÒµÉú²úÌá¸ß×ÛºÏЧÂʵÄÖØÒªÊֶΣ¬ÔÚÏõËṤҵÁ÷³ÌÖÐÑ­»·Ê¹ÓÃÎïÖÊËùÁ÷¾­µÄÉ豸ÊDZû¡ú
 
¡ú
 
£®
£¨6£©ÉÏÊö¹¤ÒµÉú²úÖÐÀíÂÛÉÏ1Ìå»ý°±ÆøÈ«²¿×ª»¯ÎªÏõËáÐèÒªÏàͬ״¿öϵĿÕÆø
 
Ìå»ý£¬Êµ¼Ê¹¤ÒµÉú²úÖг£ÒªÍ¨ÈëÊʵ±¹ýÁ¿µÄ¿ÕÆø£¬Æä×÷ÓóýÁËÌá¸ß»¯Ñ§·´Ó¦ËÙÂÊ¡¢Ìá¸ß°±´ß»¯Ñõ»¯·´Ó¦Öа±ÆøµÄת»¯ÂÊ£¬»¹¿ÉÒÔ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijѧУѧÉúÀûÓÃÏÂͼËùʾװÖÃÑéÖ¤ÂÈÆøÓë°±ÆøÖ®¼äµÄ·´Ó¦£¨²¿·Ö×°ÖÃÒÑÂÔÈ¥£©£®ÆäÖÐA¡¢B·Ö±ðΪÂÈÆøºÍ°±ÆøµÄ·¢Éú×°Öã¬CΪ´¿¾»¸ÉÔïµÄÂÈÆøÓë°±Æø·´Ó¦µÄ×°Öã®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©×°ÖÃAÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
 
£®
£¨2£©×°ÖÃBÖÐŨ°±Ë®ÓëNaOH¹ÌÌå»ìºÏ¿ÉÖÆÈ¡°±Æø£¬ÆäÔ­ÒòÊÇ
 
£®
£¨3£©ÒÔNH4ClÓëCa£¨OH£©2ΪԭÁÏÖÆNH3µÄ»¯Ñ§·½³Ìʽ
 
£®
£¨4£©×°ÖÃCÖÐÂÈÆøºÍ°±ÆøÏàÓö£¬ÓÐŨºñµÄ°×Ñ̲¢ÔÚÈÝÆ÷ÄÚ±ÚÄý½á£¬Í¬Ê±Éú³ÉÒ»ÖֿɲÎÓëÉú̬ѭ»·µÄÎÞ¶¾ÆøÌ壬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£¬Èô·´Ó¦ÖÐÏûºÄ±ê×¼×´¿öÏÂ6.72LCl2£¬Ôò±»Ñõ»¯µÄNH3µÄÖÊÁ¿Îª
 
g£¬´Ë·´Ó¦¿ÉÓÃÓÚ¼ì²éÊäËÍÂÈÆøµÄ¹ÜµÀÊÇ·ñ©Æø£®
£¨5£©ÈôÒª¼ìÑéC·´Ó¦Éú³ÉµÄ¹ÌÌåÖеÄÑôÀë×Ó£¬ÊµÑé²Ù×÷Ϊ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÌúÊÇÉú»îÖеij£¼û½ðÊô£¬Ò²ÊǸßÖÐËùѧµÄ¹ý¶ÉÔªËØ£¬ÓÉÓÚÆä½á¹¹Ìص㣬ÔÚÓëÆ仯ÎïÖÊ·´Ó¦Ê±ÍùÍù±íÏÖ³ö²»Í¬µÄ¼Û̬£®Ä³Ñо¿ÐÔѧϰС×é¶ÔÌúºÍÁò·´Ó¦²úÎïÖÐÌúµÄ»¯ºÏ¼Û½øÐÐÁË̽¾¿£¬²¢Éè¼ÆÁËÒÔÏÂʵÑ飬ÆäʵÑéµÄ²½ÖèΪ£º

²½ÖèÒ»£º½«Ìú·ÛºÍÁò·Û°´Ò»¶¨ÖÊÁ¿±È»ìºÏ£¬ÈçͼËùʾÔÚµªÆø»·¾³ÖУ¬¼ÓÈȺó·´Á¢¼´¿É·¢Éú²¢³ÖÐø½øÐУ¬ÀäÈ´ºóµÃµ½ºÚÉ«¹ÌÌ壮
²½Öè¶þ£ºÈ¡ºÚÉ«¹ÌÌåÉÙÐí£¬¼ÓÈëŨNaOHÈÜÒº²¢ÉÔÉÔ¼ÓÈÈ£¬ÀäÈ´ºó¹ýÂË£¬µÃµ½ºÚÉ«ÂËÔü£®
ÔÚÈ¡µÃºÚÉ«ÂËÔüºó£¬Í¬Ñ§ÃǾ­¹ýÌÖÂÛ·ÖÎö£¬¶ÔÆä³É·ÖÌá³öÁ˸÷ÖÖ²ÂÏ룬ÆäÖдú±íÐÔµÄΪ£º
²ÂÏëÒ»£ºÓÉÓÚÁòºÍÑõÊÇͬһÖ÷×åÔªËØ£¬ËùÒÔºÚÉ«¹ÌÌåµÄ¿ÉÄÜÊÇFe3S4
²ÂÏë¶þ£ºÁòºÍÑõËäÊÇͬһÖ÷×åÔªËØ£¬µ«ÁòµÄ·Ç½ðÊôÐÔÈõ£¬ÔòºÚÉ«¹ÌÌåΪFeS
ΪÁËÑéÖ¤ÒÔÉϲÂÏëÓÖ½øÐÐÁËÒÔϲÙ×÷£º
²½ÖèÈý£ºÔÚÂËÔüÖмÓÈëÒÑÖó·ÐµÄÏ¡ÁòËᣮ
²½ÖèËÄ£ºÈ¡²½ÖèÈýËùµÃÈÜÒºÉÙÐí£¬¡­£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²éÔÄ×ÊÁÏ£ºÊµÑéÊÒ¿ÉÒÔÓÃÂÈ»¯ï§±¥ºÍÈÜÒººÍÑÇÏõËáÄÆ£¨NaNO2£©±¥ºÍÈÜÒº»ìºÏ¼ÓÈÈ·´Ó¦µÃµªÆø£®Çëд³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º
 
£®
£¨2£©ÊµÑéÖÐͨÈ뵪ÆøµÄ×÷ÓÃ
 
£®
£¨3£©Èô²ÂÏëÒ»ÕýÈ·£¬Ôò²½ÖèÈýÓ¦¸Ã¹Û²ìµ½ÊÔ¹ÜÄÚÓÐ
 
³öÏÖ£¨ÌîʵÑéÏÖÏ󣩣®
£¨4£©Èô²ÂÏë¶þÕýÈ·£¬ÇëÍê³É²½ÖèËĵÄʵÑé²Ù×÷¡¢ÏÖÏóºÍ½áÂÛ
 
£®
£¨5£©²½ÖèÒ»ÖÐÌú·ÛºÍÁò·ÛµÄ»ìºÏ·ÛÄ©ÖÐÁòÒª¹ýÁ¿Ð©£¬ÆäÔ­ÒòÊÇ
 
£®
£¨6£©²½Öè¶þÖмÓÈëŨNaOHÈÜÒº²¢ÉÔÉÔ¼ÓÈȵÄÄ¿µÄÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏòÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖмÓÈë2molA¡¢0.6molCºÍÒ»¶¨Á¿µÄBÈýÖÖÆøÌ壮һ¶¨Ìõ¼þÏ·¢Éú·´Ó¦£¬¸÷ÎïÖʵÄÁ¿Å¨¶ÈËæʱ¼ä±ä»¯Èçͼ£¨¢ñ£©Ëùʾ£¬ÆäÖÐt0¡«t1½×¶Îc£¨B£©Î´»­³ö£®Í¼£¨¢ò£©Îªt2ʱ¿Ìºó¸Ä±ä·´Ó¦Ìõ¼þ£¬»¯Ñ§·´Ó¦ËÙÂÊËæʱ¼ä±ä»¯µÄÇé¿ö£¬Ëĸö½×¶Î¸Ä±äµÄÌõ¼þ¾ù²»Ïàͬ£¬Ã¿¸ö½×¶ÎÖ»¸Ä±äŨ¶È¡¢Ñ¹Ç¿¡¢Î¶ȡ¢´ß»¯¼ÁÖеÄÒ»¸öÌõ¼þ£¬ÆäÖÐt3¡«t4½×¶ÎΪʹÓô߻¯¼Á£®

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Èôt1=15min£¬Ôòt0¡«t1½×¶ÎÒÔCÎïÖʵÄŨ¶È±ä»¯±íʾµÄ·´Ó¦ËÙÂÊΪ
 
mol?L-1?min-1£®
£¨2£©t4¡«t5½×¶Î¸Ä±äµÄÌõ¼þΪ
 
£¬BµÄÆðʼÎïÖʵÄÁ¿Å¨¶ÈΪ
 
mol?L-1£®
£¨3£©t5¡«t6½×¶Î±£³ÖÈÝÆ÷ÄÚζȲ»±ä£¬ÈôAµÄÎïÖʵÄÁ¿¹²±ä»¯ÁË0.01mol£¬¶ø´Ë¹ý³ÌÖÐÈÝÆ÷ÓëÍâ½çµÄÈȽ»»»×ÜÁ¿Îªa kJ£¬Ð´³ö´ËζÈϸ÷´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¢ñ£®ÏÖ½«·´Ó¦2Fe3++2I-?2Fe2++I2Éè¼Æ³ÉÈçÏÂͼËùʾµÄÔ­µç³Ø

£¨1£©ÄÜ˵Ã÷·´Ó¦´ïµ½Æ½ºâµÄ±êÖ¾ÊÇ
 
£»£¨ÌîÐòºÅ£©
a£®µçÁ÷¼Æ¶ÁÊýΪÁã    b£®µçÁ÷¼ÆÖ¸Õë²»ÔÙƫתÇÒ²»ÎªÁã   c£®µçÁ÷¼ÆÖ¸Õëƫת½Ç¶È×î´ó
£¨2£©ÈôÑÎÇÅÖÐ×°ÓÐÇíÖ¬-±¥ºÍKClÈÜÒº£¬·´Ó¦¹ý³ÌÖеÄCl-ÒÆÏòÉÕ±­
 
£»£¨Ìî¡°¼×¡±»ò¡°ÒÒ¡±£©
£¨3£©·´Ó¦´ïµ½Æ½ºâºó£¬Ïò¼×ÖмÓÈëÊÊÁ¿FeCl2¹ÌÌ壬´Ëʱ
 
£¨Ìî¡°¼×¡±»ò¡°ÒÒ¡±£©ÖÐʯīµç¼«Îª¸º¼«£¬¶ÔÓ¦µÄµç¼«·´Ó¦·½³ÌʽΪ
 
£»
¢ò£®ÈçÏÂͼËùʾµÄ×°Öã¬C¡¢D¡¢E¡¢F¡¢X¡¢Y¶¼ÊǶèÐԵ缫£®½«µçÔ´½Óͨºó£¬ÏòÒÒÖеÎÈë·Ó̪ÊÔÒº£¬ÔÚF¼«¸½½üÏÔºìÉ«£®

£¨4£©×°ÖÃÒÒÖе缫FµÄµç¼«·´Ó¦Ê½
 
£»
£¨5£©ÏàͬÌõ¼þÏ£¬×°Öüס¢ÒÒµÄC¡¢Eµç¼«Éú³ÉÎïÖʵÄÌå»ýÖ®±ÈΪ
 
£»
£¨6£©ÓûÓÃ×°Öñû½øÐдÖÍ­¾«Á¶£¬µç¼«GÓ¦¸ÃÊÇ
 
£»
£¨7£©×°Öö¡Öе缫
 
¸½½üºìºÖÉ«±äÉ˵Ã÷ÇâÑõ»¯Ìú½ºÁ£´øÕýµçºÉ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖªÔÚ1¡Á105Pa£¬298KÌõ¼þÏ£¬1molÇâÆøȼÉÕÉú³ÉË®ÕôÆø·Å³ö242kJÈÈÁ¿£¬ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢H2£¨g£©+
1
2
O2£¨g£©=H2O£¨g£©£»¡÷H=+242kJ?mol-1
B¡¢H2O£¨g£©=H2£¨g£©+
1
2
O2£¨g£©£»¡÷H=+242kJ?mol-1
C¡¢2H2£¨g£©+O2£¨g£©=2H2O£¨l£©£»¡÷H=-484kJ?mol-1
D¡¢2H2£¨g£©+O2£¨g£©=2H2O£¨g£©£»¡÷H=+484kJ?mol-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ò»ÖÖÓɼ״¼ºÍÑõÆøÒÔ¼°Ç¿¼î×öµç½âÖÊÈÜÒºµÄÐÂÐÍÊÖ»úµç³Ø£¬2CH3OH+3O2+4OH-
·Åµç
³äµç
2CO32-+6H2O£¬ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢¹¹³É¸Ãµç³ØµÄÕý¼«ºÍ¸º¼«±ØÐëÊÇÁ½ÖÖ»îÐÔ²»Í¬µÄ½ðÊô
B¡¢³äµçʱÓÐCH3OHÉú³ÉµÄµç¼«ÎªÑô¼«
C¡¢·ÅµçʱÈÜÒºÖеÄOH-ÒÆÏòÕý¼«
D¡¢·Åµçʱ¸º¼«µÄµç¼«·´Ó¦Îª£ºCH3OH-6e-+8OH-¨TCO32-+6H2O

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸