ÌâÄ¿ÄÚÈÝ

ȡһ¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒº100mL£¬È»ºóÏòÆäͨÈëÒ»¶¨Á¿µÄCO2ÆøÌ壬µÃµ½ÈÜÒºA£¬ÏòAÖÐÖðµÎ»ºÂý¼ÓÈë0.1mol/LµÄHClÈÜÒº£¬²úÉúµÄCO2ÆøÌåÌå»ý£¨±ê×¼×´¿ö£©ÓëËù¼Ó
HClÈÜÒºµÄÌå»ýÖ®¼ä¹ØÏµÈçͼËùʾ£¬ÏÂÁÐÓйØËµ·¨²»ÕýÈ·µÄÊÇ

A£®AÈÜÒºÖÐÈÜÖÊNa2CO3ºÍNaHCO3ÎïÖʵÄÁ¿±ÈΪ1£º1
B£®AÈÜÒºÖÐÈÜÖÊNaOHºÍNa2CO3ÎïÖʵÄÁ¿±ÈΪ1£º1
C£®Ô­NaOHÈÜÒºÎïÖʵÄÁ¿Å¨¶ÈΪ0.075mol/L
D£®Í¨ÈëCO2ÆøÌåÔÚ±ê¿öϵÄÌå»ýΪ56mL

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¼×¡¢ÒÒÁ½Í¬Ñ§ÄâÓÃʵÑéÈ·¶¨Ä³ËáHAÊÇÈõµç½âÖÊ£®ËûÃǵķ½°¸·Ö±ðÊÇ£º
¼×£º¢Ù³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄHAÅäÖÆ0.1mol/LµÄÈÜÒº100mL£»¢ÚÓÃpHÊÔÖ½²â³ö¸ÃÈÜÒºµÄpH£¬¼´¿ÉÖ¤Ã÷HAÊÇÈõµç½âÖÊ£®
ÒÒ£º¢ÙÓÃÒÑÖªÎïÖʵÄÁ¿Å¨¶ÈµÄHAÈÜÒº¡¢ÑÎËᣬ·Ö±ðÅäÖÆpH=1µÄÁ½ÖÖËáÈÜÒº¸÷100mL£»¢Ú·Ö±ðÈ¡ÕâÁ½ÖÖÈÜÒº¸÷10mL£¬¼ÓˮϡÊÍΪ100mL£»
¢Û¸÷È¡ÏàͬÌå»ýµÄÁ½ÖÖÏ¡ÊÍҺװÈëÁ½¸öÊԹܣ¬Í¬Ê±¼ÓÈë´¿¶ÈÏàͬµÄпÁ££¬¹Û²ìÏÖÏ󣬼´¿ÉÖ¤Ã÷HAÊÇÈõµç½âÖÊ£®
£¨1£©ÔÚÁ½¸ö·½°¸µÄµÚ¢Ù²½ÖУ¬¶¼ÒªÓõ½µÄ¶¨Á¿ÒÇÆ÷ÊÇ
100mLÈÝÁ¿Æ¿
100mLÈÝÁ¿Æ¿
£®
£¨2£©¼×·½°¸ÖУ¬ËµÃ÷HAÊÇÈõµçÖʵÄÀíÓÉÊDzâµÃÈÜÒºµÄpH
£¾
£¾
1£¨Ñ¡Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®¼òҪ˵Ã÷pHÊÔÖ½µÄʹÓ÷½·¨£º
ȡһС¶ÎpHÊÔÖ½·ÅÔÚ²£Á§Æ¬ÉÏ£¬Óò£Á§°ôպȡ´ý²âÒºµãÔÚpHÊÔÖ½ÖÐÑ룬¶ÔÕÕ±ÈÉ«¿¨£¬¶Á³öÈÜÒºµÄpH
ȡһС¶ÎpHÊÔÖ½·ÅÔÚ²£Á§Æ¬ÉÏ£¬Óò£Á§°ôպȡ´ý²âÒºµãÔÚpHÊÔÖ½ÖÐÑ룬¶ÔÕÕ±ÈÉ«¿¨£¬¶Á³öÈÜÒºµÄpH
£®
£¨3£©ÒÒ·½°¸ÖУ¬ËµÃ÷HAÊÇÈõµç½âÖʵÄÏÖÏóÊÇ
×°HAÈÜÒºµÄÊÔ¹ÜÖзųöH2µÄËÙÂÊ¿ì
×°HAÈÜÒºµÄÊÔ¹ÜÖзųöH2µÄËÙÂÊ¿ì
£®
£¨4£©ÇëÄãÔÙÌá³öÒ»¸öºÏÀí¶ø±È½ÏÈÝÒ×½øÐеķ½°¸£¨Ò©Æ·¿ÉÈÎÈ¡£©£¬²¢×÷¼òÃ÷¶óÒª±íÊö£º
ÓÃÕôÁóË®ÅäÖÆÉÙÁ¿NaAÈÜÒº£¬ÔÙ²âÆäpH£¬ÈôpH£¾7£¬ËµÃ÷HAÊÇÈõµç½âÖÊ
ÓÃÕôÁóË®ÅäÖÆÉÙÁ¿NaAÈÜÒº£¬ÔÙ²âÆäpH£¬ÈôpH£¾7£¬ËµÃ÷HAÊÇÈõµç½âÖÊ
£®

¢ñ£¨1£©²£Á§°ôÊÇÖÐѧ»¯Ñ§ÊµÑéÖг£ÓõÄÒÇÆ÷£®ÏÂÁйý³ÌÖУ¬Ò»°ã²»ÐèÒª²£Á§°ôµÄÊÇ______£¨ Ìîд±àºÅ¡¡£©
¢ÙÓÃPHÊÔÖ½²â¶¨Na2CO3ÈÜÒºµÄPH
¢ÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÂÈ»¯ÄÆÈÜÒº
¢Û½«ÊÊÁ¿ÂÈ»¯Ìú±¥ºÍÈÜÒºµÎÈë·ÐË®ÖÐÖÆ±¸ÇâÑõ»¯Ìú½ºÌå
¢Ü̽¾¿Ba£¨OH£©2?8H2O¾§ÌåºÍNH4Cl¾§Ìå·´Ó¦¹ý³ÌÖеÄÄÜÁ¿±ä»¯£®
¢ÝʵÑéÊÒÓÃÐÂÖÆ±¸µÄFeSO4ÈÜÒººÍÔ¤´¦Àí¹ýµÄNaOHÈÜÒºÖÆ±¸Fe£¨OH£©2°×É«³Áµí
£¨2£©Óá°´óÓÚ¡±¡°µÈÓÚ¡±¡°Ð¡ÓÚ¡±Ìî¿Õ
¢Ù¶ÁÁ¿Í²ÖÐÒºÌåµÄÌå»ýʱ£¬ÊÓÏ߯«¸ß£¬¶ÁÈ¡µÄÌå»ýÊý______ʵ¼ÊÌå»ýÊý
¢ÚÓÃÍÐÅÌÌìÆ½³ÆÈ¡10.4gʳÑΣ¬½«íÀÂëºÍʳÑεÄλÖõߵ¹£¬Ëù³ÆÊ³ÑεÄÖÊÁ¿______10.4g
¢ÛÅäÖÆ500ml 0.1mol/lNaOHÈÜÒº£¬¶¨ÈÝʱ¸©Êӿ̶ÈÏߣ¬ËùµÃÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È______0.1mol/l
¢ÜÓÃÖк͵ζ¨·¨²âijNaOHÈÜÒºµÄŨ¶È£¬Á¿È¡´ý²âҺδÓøÃÈÜÒºÈóÏ´µÎ¶¨¹Ü£¬²âµÃµÄÈÜÒºµÄŨ¶È______ʵ¼ÊŨ¶È
¢ò¡¢Ñõ»¯»¹Ô­·´Ó¦ÖÐʵ¼ÊÉϰüº¬Ñõ»¯ºÍ»¹Ô­Á½¸ö¹ý³Ì£®ÏÂÃæÊÇÒ»¸ö»¹Ô­¹ý³ÌµÄ·´Ó¦Ê½£º
2Êýѧ¹«Ê½+10H++8e-¡úN2O+5H2O
Fe£¨SO4£©3¡¢KMnO4¡¢Na2CO3¡¢FeSO4ËÄÖÖÎïÖÊÖеÄÒ»ÖÖÎïÖÊÄÜʹÉÏÊö»¹Ô­¹ý³Ì·¢Éú£®
£¨1£©Ð´³ö¸ÃÑõ»¯»¹Ô­·´Ó¦µÄ·½³Ìʽ£¨²»Å䯽£©£º______£®
£¨2£©¸Ã·´Ó¦ÖÐÑõ»¯¼ÁºÍ»¹Ô­¼ÁÎïÖʵÄÁ¿Ö®±ÈΪ______£®
£¨3£©·´Ó¦ÖÐÏõËáÌåÏÖÁË______¡¢______ÐÔÖÊ£®
£¨4£©·´Ó¦ÖÐÈô²úÉú0.1molÆøÌ壬Ôò×ªÒÆµç×ÓµÄÎïÖʵÄÁ¿ÊÇ______£®

¢ñ£¨1£©²£Á§°ôÊÇÖÐѧ»¯Ñ§ÊµÑéÖг£ÓõÄÒÇÆ÷£®ÏÂÁйý³ÌÖУ¬Ò»°ã²»ÐèÒª²£Á§°ôµÄÊÇ______£¨ Ìîд±àºÅ  £©
¢ÙÓÃPHÊÔÖ½²â¶¨Na2CO3ÈÜÒºµÄPH
¢ÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÂÈ»¯ÄÆÈÜÒº
¢Û½«ÊÊÁ¿ÂÈ»¯Ìú±¥ºÍÈÜÒºµÎÈë·ÐË®ÖÐÖÆ±¸ÇâÑõ»¯Ìú½ºÌå
¢Ü̽¾¿Ba£¨OH£©2?8H2O¾§ÌåºÍNH4Cl¾§Ìå·´Ó¦¹ý³ÌÖеÄÄÜÁ¿±ä»¯£®
¢ÝʵÑéÊÒÓÃÐÂÖÆ±¸µÄFeSO4ÈÜÒººÍÔ¤´¦Àí¹ýµÄNaOHÈÜÒºÖÆ±¸Fe£¨OH£©2°×É«³Áµí
£¨2£©Óá°´óÓÚ¡±¡°µÈÓÚ¡±¡°Ð¡ÓÚ¡±Ìî¿Õ
¢Ù¶ÁÁ¿Í²ÖÐÒºÌåµÄÌå»ýʱ£¬ÊÓÏ߯«¸ß£¬¶ÁÈ¡µÄÌå»ýÊý______ʵ¼ÊÌå»ýÊý
¢ÚÓÃÍÐÅÌÌìÆ½³ÆÈ¡10.4gʳÑΣ¬½«íÀÂëºÍʳÑεÄλÖõߵ¹£¬Ëù³ÆÊ³ÑεÄÖÊÁ¿______10.4g
¢ÛÅäÖÆ500ml 0.1mol/lNaOHÈÜÒº£¬¶¨ÈÝʱ¸©Êӿ̶ÈÏߣ¬ËùµÃÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È______0.1mol/l
¢ÜÓÃÖк͵ζ¨·¨²âijNaOHÈÜÒºµÄŨ¶È£¬Á¿È¡´ý²âҺδÓøÃÈÜÒºÈóÏ´µÎ¶¨¹Ü£¬²âµÃµÄÈÜÒºµÄŨ¶È______ʵ¼ÊŨ¶È
¢ò¡¢Ñõ»¯»¹Ô­·´Ó¦ÖÐʵ¼ÊÉϰüº¬Ñõ»¯ºÍ»¹Ô­Á½¸ö¹ý³Ì£®ÏÂÃæÊÇÒ»¸ö»¹Ô­¹ý³ÌµÄ·´Ó¦Ê½£º
2
NO-3
+10H++8e-¡úN2O+5H2O
Fe£¨SO4£©3¡¢KMnO4¡¢Na2CO3¡¢FeSO4ËÄÖÖÎïÖÊÖеÄÒ»ÖÖÎïÖÊÄÜʹÉÏÊö»¹Ô­¹ý³Ì·¢Éú£®
£¨1£©Ð´³ö¸ÃÑõ»¯»¹Ô­·´Ó¦µÄ·½³Ìʽ£¨²»Å䯽£©£º______£®
£¨2£©¸Ã·´Ó¦ÖÐÑõ»¯¼ÁºÍ»¹Ô­¼ÁÎïÖʵÄÁ¿Ö®±ÈΪ______£®
£¨3£©·´Ó¦ÖÐÏõËáÌåÏÖÁË______¡¢______ÐÔÖÊ£®
£¨4£©·´Ó¦ÖÐÈô²úÉú0.1molÆøÌ壬Ôò×ªÒÆµç×ÓµÄÎïÖʵÄÁ¿ÊÇ______£®
¢ñ£¨1£©²£Á§°ôÊÇÖÐѧ»¯Ñ§ÊµÑéÖг£ÓõÄÒÇÆ÷£®ÏÂÁйý³ÌÖУ¬Ò»°ã²»ÐèÒª²£Á§°ôµÄÊÇ______£¨ Ìîд±àºÅ  £©
¢ÙÓÃPHÊÔÖ½²â¶¨Na2CO3ÈÜÒºµÄPH
¢ÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÂÈ»¯ÄÆÈÜÒº
¢Û½«ÊÊÁ¿ÂÈ»¯Ìú±¥ºÍÈÜÒºµÎÈë·ÐË®ÖÐÖÆ±¸ÇâÑõ»¯Ìú½ºÌå
¢Ü̽¾¿Ba£¨OH£©2?8H2O¾§ÌåºÍNH4Cl¾§Ìå·´Ó¦¹ý³ÌÖеÄÄÜÁ¿±ä»¯£®
¢ÝʵÑéÊÒÓÃÐÂÖÆ±¸µÄFeSO4ÈÜÒººÍÔ¤´¦Àí¹ýµÄNaOHÈÜÒºÖÆ±¸Fe£¨OH£©2°×É«³Áµí
£¨2£©Óá°´óÓÚ¡±¡°µÈÓÚ¡±¡°Ð¡ÓÚ¡±Ìî¿Õ
¢Ù¶ÁÁ¿Í²ÖÐÒºÌåµÄÌå»ýʱ£¬ÊÓÏ߯«¸ß£¬¶ÁÈ¡µÄÌå»ýÊý______ʵ¼ÊÌå»ýÊý
¢ÚÓÃÍÐÅÌÌìÆ½³ÆÈ¡10.4gʳÑΣ¬½«íÀÂëºÍʳÑεÄλÖõߵ¹£¬Ëù³ÆÊ³ÑεÄÖÊÁ¿______10.4g
¢ÛÅäÖÆ500ml 0.1mol/lNaOHÈÜÒº£¬¶¨ÈÝʱ¸©Êӿ̶ÈÏߣ¬ËùµÃÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È______0.1mol/l
¢ÜÓÃÖк͵ζ¨·¨²âijNaOHÈÜÒºµÄŨ¶È£¬Á¿È¡´ý²âҺδÓøÃÈÜÒºÈóÏ´µÎ¶¨¹Ü£¬²âµÃµÄÈÜÒºµÄŨ¶È______ʵ¼ÊŨ¶È
¢ò¡¢Ñõ»¯»¹Ô­·´Ó¦ÖÐʵ¼ÊÉϰüº¬Ñõ»¯ºÍ»¹Ô­Á½¸ö¹ý³Ì£®ÏÂÃæÊÇÒ»¸ö»¹Ô­¹ý³ÌµÄ·´Ó¦Ê½£º
2+10H++8e-¡úN2O+5H2O
Fe£¨SO4£©3¡¢KMnO4¡¢Na2CO3¡¢FeSO4ËÄÖÖÎïÖÊÖеÄÒ»ÖÖÎïÖÊÄÜʹÉÏÊö»¹Ô­¹ý³Ì·¢Éú£®
£¨1£©Ð´³ö¸ÃÑõ»¯»¹Ô­·´Ó¦µÄ·½³Ìʽ£¨²»Å䯽£©£º______£®
£¨2£©¸Ã·´Ó¦ÖÐÑõ»¯¼ÁºÍ»¹Ô­¼ÁÎïÖʵÄÁ¿Ö®±ÈΪ______£®
£¨3£©·´Ó¦ÖÐÏõËáÌåÏÖÁË______¡¢______ÐÔÖÊ£®
£¨4£©·´Ó¦ÖÐÈô²úÉú0.1molÆøÌ壬Ôò×ªÒÆµç×ÓµÄÎïÖʵÄÁ¿ÊÇ______£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø