18£®³£ÎÂÏÂijÈÜÒºÖÐÖ»º¬ÓÐNa+¡¢H+¡¢OH-¡¢A-ËÄÖÖÀë×Ó£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÈôÈÜÖÊΪNaA£¬ÈÜÒºÖпÉÄÜ´æÔÚ£ºc £¨OH-£©=c£¨H+£©+c£¨HA£©
B£®Èôc£¨OH-£©£¾c£¨H+£©£¬ÈÜÒºÖв»¿ÉÄÜ´æÔÚc£¨Na+£©£¾c£¨OH-£©£¾c£¨A-£©£¾c£¨H+£©
C£®ÈÜÒºÖпÉÄÜ´æÔÚc£¨Na+£©£¾c£¨A-£©£¾c£¨H+£©£¾c£¨OH-£©
D£®ÈôÈÜÖÊΪNaAºÍHA£¬ÔòÒ»¶¨´æÔÚc£¨A-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©

·ÖÎö ÈÜÒºÖÐÖ»º¬ÓÐNa+¡¢H+¡¢OH-¡¢A-ËÄÖÖÀë×Ó£¬ÔòÈÜÒºÖеÄÈÜÖÊ¿ÉÄÜΪNaA»òNaA¡¢HA»òNaA¡¢NaOH£¬ÀûÓõçÀëºÍË®½âÀ´·ÖÎöÀë×ÓŨ¶ÈµÄ¹Øϵ£¬
A¡¢¸ù¾ÝÈÜÒºÖÐÖÊ×ÓÊغã½øÐÐÅжϣ»
B¡¢Èôc£¨OH-£©£¾c£¨H+£©£¬ÀûÓÃÈÜÖÊ¿ÉÄÜΪNaA¡¢NaOHÀ´·ÖÎö
C¡¢ÑôÀë×Ó´óÓÚÒõÀë×ÓŨ¶È£¬ÈÜÒºµçºÉ²»Êغ㣻
D¡¢ÈÜÖÊΪNaA¡¢HA£¬¸ù¾ÝËáµÄµçÀë³Ì¶ÈºÍÑεÄË®½â³Ì¶È½øÐзÖÎö£®

½â´ð ½â£ºA¡¢Ë®µçÀëµÄÇâÀë×ÓÓëÇâÑõ¸ùÀë×ÓŨ¶ÈÏàµÈ£¬µ±HAΪÈõËáʱ£¬¸ù¾ÝÖÊ×ÓÊغã¿ÉÖª£ºc £¨OH-£©=c£¨H+£©+c£¨HA£©£¬¹ÊA´íÎó£»
B¡¢Èôc£¨OH-£©£¾c£¨H+£©£¬ÈÜÖÊ¿ÉÄÜΪNaA¡¢NaOH£¬ÇÒNaOHµÄÎïÖʵÄÁ¿½Ï´óʱ¿ÉÒÔ´æÔÚ£ºc£¨Na+£©£¾c£¨OH-£©£¾c£¨A-£©£¾c£¨H+£©£¬¹ÊB´íÎó£»
C¡¢ÑôÀë×Ó´óÓÚÒõÀë×ÓŨ¶È£¬ÈÜÒºµçºÉ²»Êغ㣬Èç¹ûc£¨Na+£©£¾c£¨A-£©£¬ÔòÓ¦¸ÃÂú×㣺c£¨OH-£©£¾c£¨H+£©£¬¹ÊC´íÎó£»
D¡¢ÈôÈÜÖÊΪNaA¡¢HA£¬¶þÕßµÈÎïÖʵÄÁ¿Ê±ÇÒËáµçÀë´óÓÚÑεÄË®½â´æÔÚc£¨A-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©£¬Èô¶þÕßµÈÎïÖʵÄÁ¿ÇÒÑεÄË®½â´óÓÚËáµÄµçÀ룬Ôò´æÔÚc£¨Na+£©£¾c£¨A-£©£¾c£¨OH-£©£¾c£¨H+£©£¬¹ÊD´íÎó£»
¹ÊÑ¡A£®

µãÆÀ ±¾Ì⿼²éÀë×ÓŨ¶È´óСµÄ±È½Ï£¬ÌâÄ¿ÄѶÈÖеȣ¬×¢ÒâÕÆÎÕÅжÏÈÜÒºÖÐÀë×ÓŨ¶È´óСµÄ·½·¨£¬Ã÷È·ÈÜÒºÖеÄÈÜÖʼ°µçÀ롢ˮ½âµÄ³Ì¶ÈÊǽâ´ð±¾ÌâµÄ¹Ø¼ü£¬²¢Ñ§»áÀûÓõçºÉÊغãÀ´·ÖÎö½â´ð£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

8£®µ±¹âÊøͨ¹ýÏÂÁзÖɢϵʱ£¬ÄܲúÉú¶¡´ï¶ûЧӦµÄÊÇ£¨¡¡¡¡£©
A£®NaClÈÜÒºB£®Fe£¨OH£©3½ºÌåC£®Å¨ÁòËáD£®Äཬ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

9£®ÓÃÏÂÁз½·¨£º¢ÙKMnO4ÊÜÈȷֽ⣻¢ÚKClO3ÊÜÈȷֽ⣨ÓÃMnO2×÷´ß»¯¼Á£©£»
¢Û2H2O2$\frac{\underline{\;´ß»¯¼Á\;}}{\;}$2H2O+O2¡ü  ¢Ü2Na2O2+2H2O¨T4NaOH+O2¡ü£¬ÈôÒªÖƵÃÏàͬÖÊÁ¿µÄÑõÆø£¬ÉÏÊöËÄÖÖ·½·¨ÖÐËùתÒƵĵç×ÓÊýÄ¿Ö®±ÈÊÇ£¨¡¡¡¡£©
A£®3£º2£º1£º4B£®1£º1£º1£º1C£®1£º2£º1£º2D£®2£º2£º1£º1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

6£®¿ÆÑÐÈËÔ±ÉèÏëÓÃÈçͼËùʾװÖÃÉú²úÁòËᣬÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
 
A£®aΪÕý¼«£¬bΪ¸º¼«
B£®Éú²ú¹ý³ÌÖÐH+Ïòaµç¼«ÇøÓòÔ˶¯
C£®µç×Ó´Ób¼«Ïòa¼«Òƶ¯
D£®¸º¼«·´Ó¦Ê½Îª£º2H2+SO2-2e-=SO42-+4H+

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

13£®ÏÂÁÐÈÜÒºÖеμӷÓ̪ÊÔÒººó³öÏÖºìÉ«ÇÒ²Ù×÷ºó¿ÉʹÑÕÉ«ÍÊÈ¥µÄÊÇ£¨¡¡¡¡£©
A£®Ã÷·¯ÈÜÒº¼ÓÈÈB£®CH3COONaÈÜÒº¼ÓÈÈ
C£®°±Ë®ÖмÓÈëÉÙÁ¿NH4ClD£®Na2CO3ÈÜÒºÖмÓÈëBaCl2¹ÌÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

3£®½â´ðÏÂÁÐÎÊÌâ
£¨1£©³ýÈ¥Fe2O3 ÖлìÓÐAl2O3µÄÊÔ¼ÁÊÇNaOH£¬Àë×Ó·½³ÌʽΪAl2O3+2OH-¨T2AlO2-+H2O
£¨2£©³ýÈ¥Na2CO3·ÛÄ©ÖлìÈëµÄNaHCO3ÔÓÖÊÓüÓÈÈ·½·¨£¬»¯Ñ§·½³ÌʽΪ2NaHCO3$\frac{\underline{\;\;¡÷\;\;}}{\;}$Na2CO3+H2O+CO2¡ü
£¨3£©Ó¡Ë¢µç·°åÊÇÓɸ߷Ö×Ó²ÄÁϺÍÍ­²­¸´ºÏ¶ø³É£¬¿ÌÖÆÓ¡Ë¢µç·°åÊÇ£¬ÒªÓÃFeCl3ÈÜÒº×÷Ϊ¡°¸¯Ê´Òº¡±£®Íê³ÉFeCl3 ÈÜÒºµÄÓйØÎÊÌ⣮
¢Ùд³öFeCl3 ÈÜÒºÓëÍ­·´Ó¦µÄÀë×Ó·½³Ìʽ£º2Fe3++Cu=2Fe2++Cu2+
¢Ú½«1¡«2mL FeCl3 ±¥ºÍÈÜÒºÖ𽥵μӵ½·ÐË®ÖУ¬¼ÌÐøÖó·ÐÖÁÈÜÒº³ÊºìºÖÉ«£¬¿ÉÒÔÖƵýºÌ壮

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

10£®£¨1£©ÈçͼÊÇÅäÖÆ50mL NaOH±ê×¼ÈÜÒºµÄ¹ý³ÌʾÒâͼ£®ÇëÄã¹Û²ìͼʾÅжÏÆäÖв»ÕýÈ·µÄ²Ù×÷ÓТ٢ڢۢݣ¨ÌîÐòºÅ£©£¬ÆäÖÐÈ·¶¨50mLÈÜÒºÌå»ýµÄÈÝÆ÷ÊÇ50mLÈÝÁ¿Æ¿£¬Èç¹û°´ÕÕͼʾµÄ²Ù×÷ËùÅäÖƵÄÈÜÒº½øÐÐʵÑ飬ÔÚÆäËû²Ù×÷¾ùÕýÈ·µÄÇé¿öÏ£¬Ëù²âµÃµÄʵÑé½á¹û½«Æ«Ð¡£¨Ìî¡°Æ«´ó¡±»ò¡°Æ«Ð¡¡±£©£®

£¨2£©Ê¹ÓÃÈÝÁ¿Æ¿ÅäÖÆÈÜҺʱ£¬ÓÉÓÚ²Ù×÷²»µ±£¬»áÒýÆðÎó²î£¬ÏÂÁÐÇé¿ö»áʹËùÅäÈÜҺŨ¶ÈÆ«µÍµÄÊÇA
¢ÙÓÃÌìƽ£¨Ê¹ÓÃÓÎÂ룩³ÆÁ¿Ê±£¬±»³ÆÁ¿ÎïÓëíÀÂëµÄλÖ÷ŵߵ¹ÁË
¢ÚŨÁòËáÅäÖÆÏ¡ÁòËᣬÓÃÁ¿Í²Á¿È¡Å¨ÁòËáʱÑöÊӿ̶ÈÏß
¢ÛÈÜҺתÒƵ½ÈÝÁ¿Æ¿ºó£¬ÉÕ±­¼°²£Á§°ôδÓÃÕôÁóˮϴµÓ
¢ÜתÒÆÈÜҺǰÈÝÁ¿Æ¿ÄÚÓÐÉÙÁ¿ÕôÁóË®
¢Ý¶¨ÈÝʱ£¬ÑöÊÓÈÝÁ¿Æ¿µÄ¿Ì¶ÈÏß
¢Þ¶¨ÈݺóÒ¡ÔÈ£¬·¢ÏÖÒºÃæ½µµÍ£¬ÓÖ²¹¼ÓÉÙÁ¿Ë®£¬ÖØдﵽ¿Ì¶ÈÏß
A£®¢Ù¢Û¢Ý¢ÞB£®¢Ù¢Ú¢Ý¢ÞC£®¢Ú¢Û¢Ü¢ÞD£®¢Û¢Ü¢Ý¢Þ
£¨3£©ÔÚʹÓÃÈÝÁ¿Æ¿Ö®Ç°±ØÐë½øÐеIJÙ×÷ÊÇ£»ÔÚÈÝÁ¿Æ¿É϶¼±êÓп̶ÈÏßÒÔ¼°ÈÝ»ýºÍζÈÈý¸öÌØÕ÷±êÖ¾£®
£¨4£©ÔÚʵÑé¹ý³ÌÖгöÏÖÈçÏÂÇé¿öÓ¦ÈçºÎ´¦Àí£¿¼ÓÕôÁóˮʱ²»É÷³¬¹ýÁ˿̶ȣ¬Ó¦£»¼ÓÕôÁóË®µ½¿Ì¶ÈÔ¼1cm¡«2cm´¦Ó¦¸ÄÓýºÍ·µÎ¹ÜÖðµÎµÎ¼Óµ½°¼ÒºÃæ×îµÍ´¦Óë¿Ì¶ÈÏßÏàÇУ»¸©Êӹ۲쵽ҺÃæ¸ÕºÃµ½´ï¿Ì¶ÈÏߣ¬Ó¦£»ÒÆҺʱ²»É÷½«ÉÙÁ¿ÒºµÎµôÔÚÈÝÁ¿Æ¿ÍâÃ棬Ӧ£®NH4++OH¨TNH3¡ü+H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÍƶÏÌâ

7£®´ÓúºÍʯÓÍÖпÉÒÔÌáÁ¶³ö»¯¹¤Ô­ÁÏAºÍB£¬AÊÇÒ»ÖÖ¹ûʵ´ßÊì¼Á£¬ËüµÄ²úÁ¿ÓÃÀ´ºâÁ¿Ò»¸ö¹ú¼ÒµÄʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½£®BÊÇÒ»ÖÖ±ÈË®ÇáµÄÓÍ×´ÒºÌ壬B½öÓÉ̼ÇâÁ½ÖÖÔªËØ×é³É£¬Ì¼ÔªËØÓëÇâÔªËصÄÖÊÁ¿±ÈΪ12£º1£¬BµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª78£®»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AµÄµç×Óʽ£¬BµÄ½á¹¹¼òʽ»ò£®
£¨2£©±ÈA¶àÒ»¸ö̼ԭ×ÓµÄAµÄͬϵÎïÄÜʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«µÄ»¯Ñ§·´Ó¦·½³ÌʽCH2=CHCH3+Br2¡úCH2BrCHBrCH3£®
£¨3£©µÈÖÊÁ¿µÄA¡¢BÍêȫȼÉÕʱÏûºÄO2µÄÎïÖʵÄÁ¿A£¾B£¨Ìî¡°A£¾B¡±¡¢¡°A£¼B¡±»ò¡°A=B¡±£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

8£®ÔÚ»¯Ñ§·´Ó¦X+2Y=R+2MÖУ¬ÒÑÖªRºÍMµÄĦ¶ûÖÊÁ¿Ö®±ÈΪ22£º9£¬µ±1.6¿ËXÓëYÇ¡ºÃÍêÈ«·´Ó¦ºó£¬Éú³É4.4¿ËR£¬Í¬Ê±Éú³ÉMµÄÖÊÁ¿Îª3.6g¿Ë£¬YÓëMµÄÖÊÁ¿Ö®±ÈΪ16£º9£¬½â´ð´ËÌâʱʹÓõ½µÄ»¯Ñ§¶¨ÂÉÊÇÖÊÁ¿Êغ㶨ÂÉ£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸