15£®Ä³½á¾§Ë®ºÏÎﺬÓÐÁ½ÖÖÑôÀë×ÓºÍÒ»ÖÖÒõÀë×Ó£®³ÆÈ¡Á½·ÝÖÊÁ¿¾ùΪ10.00gµÄ¸Ã½á¾§Ë®ºÏÎ·Ö±ðÖƳÉÈÜÒº£®
Ò»·Ý¼ÓÈë×ãÁ¿NaOHÈÜÒº£¬Éú³ÉÀ¶É«³Áµí£¬½«´ËÐü×ÇÒº¼ÓÈÈ£¬ÒݳöµÄÆøÌåÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬´ËÆøÌåÇ¡ºÃ¿É±»50.00mL1.00mol/LÑÎËáÍêÈ«ÎüÊÕ£®Í¬Ê±À¶É«³Áµí±äΪºÚÉ«³Áµí£»
ÁíÒ»·Ý¼ÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬Éú³É²»ÈÜÓÚÏ¡ÏõËá°×É«³Áµí£¬¾­¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï³ÆÆäÖÊÁ¿Îª11.65g£®
£¨1£©¸Ã½á¾§Ë®ºÏÎïÖк¬ÓеÄÁ½ÖÖÑôÀë×ÓÊÇCu2+ºÍNH4+£¬ÒõÀë×ÓÊÇSO42-£®
£¨2£©ÊÔͨ¹ý¼ÆËãÈ·¶¨¸Ã½á¾§Ë®ºÏÎïµÄ»¯Ñ§Ê½£¨NH4£©2Cu£¨SO4£©2•6H2O»ò£¨NH4£©2SO4•CuSO4•6H2O£¨Ð´³ö¼ÆËã¹ý³Ì£¬×¢Òâ½âÌâ¹æ·¶£©£®

·ÖÎö Ò»·Ý¼ÓÈë×ãÁ¿NaOHÈÜÒº£¬Éú³ÉÀ¶É«³Áµí£¬½«´ËÐü×ÇÒº¼ÓÈÈ£¬ÒݳöµÄÆøÌåÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬ËµÃ÷Éú³É°±Æø£¬ÑôÀë×Óº¬ÓÐNH4+£¬ÇÒn£¨NH4+£©=n£¨NH3£©=n£¨HCl£©=0.05L¡Á1mol/L=0.05mol£¬À¶É«³ÁµíΪÇâÑõ»¯Í­£¬ËµÃ÷º¬ÓÐCu2+£¬ÁíÒ»·Ý¼ÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬Éú³É²»ÈÜÓÚÏ¡ÏõËá°×É«³Áµí£¬¾­¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï³ÆÆäÖÊÁ¿Îª11.65g£¬Ó¦ÎªBaSO4£¬¿ÉÖªn£¨SO42-£©=n£¨BaSO4£©=$\frac{11.65g}{233g/mol}$=0.05mol£¬ÒÔ´Ë¿ÉÈ·¶¨n£¨£¨NH4£©2SO4£©£º£¨CuSO4£©£¬½áºÏÖÊÁ¿È·¶¨Ë®µÄÁ¿£¬¿ÉÈ·¶¨»¯Ñ§Ê½£®

½â´ð ½â£º£¨1£©Ò»·Ý¼ÓÈë×ãÁ¿NaOHÈÜÒº£¬Éú³ÉÀ¶É«³Áµí£¬½«´ËÐü×ÇÒº¼ÓÈÈ£¬ÒݳöµÄÆøÌåÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬ËµÃ÷Éú³É°±Æø£¬ÑôÀë×Óº¬ÓÐNH4+£¬À¶É«³ÁµíΪÇâÑõ»¯Í­£¬ËµÃ÷º¬ÓÐCu2+£¬ÁíÒ»·Ý¼ÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬Éú³É²»ÈÜÓÚÏ¡ÏõËá°×É«³Áµí£¬Ó¦ÎªBaSO4£¬ËµÃ÷º¬ÓÐSO42-£¬
¹Ê´ð°¸Îª£ºCu2+¡¢NH4+£»SO42-£»
£¨2£©°±Æø±»50.00mL1.00mol/LÑÎËáÍêÈ«ÎüÊÕ£¬Ôòn£¨NH4+£©=n£¨NH3£©=n£¨HCl£©=0.05L¡Á1mol/L=0.05mol£¬ÓɳÁµíÖÊÁ¿¿ÉÖªn£¨SO42-£©=n£¨BaSO4£©=$\frac{11.65g}{233g/mol}$=0.05mol£¬
ÓɵçºÉÊغã¿ÉÖª2n£¨Cu2+£©+n£¨NH4+£©=2n£¨SO42-£©£¬n£¨Cu2+£©=0.025mol£¬
Ôòn£¨H2O£©=$\frac{10g-0.05mol¡Á18g/mol-0.025mol¡Á64g/mol-0.05mol¡Á96g/mol}{18g/mol}$=0.15mol£¬Ôòn£¨£¨NH4£©2SO4£©£ºn£¨CuSO4£©£ºn£¨H2O£©=0.025£º0.0525£º0.15=1£º1£º6£¬Ôò»¯Ñ§Ê½Îª£¨NH4£©2SO4•CuSO4•6H2O»ò£¨NH4£©2Cu£¨SO4£©2•6H2O£®
¹Ê´ð°¸Îª£º£¨NH4£©2Cu£¨SO4£©2•6H2O»ò£¨NH4£©2SO4•CuSO4•6H2O£®

µãÆÀ ±¾Ì⿼²éÎÞ»úÎïµÄÍƶϣ¬Îª¸ßƵ¿¼µã£¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦¡¢ÊµÑéÄÜÁ¦ºÍ¼ÆËãÄÜÁ¦µÄ¿¼²é£¬½â´ðʱעÒâ´ÓµçºÉÊغãºÍÖÊÁ¿ÊغãµÄ½Ç¶È·ÖÎö£¬ÄѵÀÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

5£®ÔÚºãÈݾøÈÈ£¨²»ÓëÍâ½ç½»»»ÄÜÁ¿£©Ìõ¼þϽøÐÐ2A£¨g£©+B£¨g£©?3C£¨g£©+2D£¨s£©·´Ó¦£¬°´Ï±íÊý¾ÝͶÁÏ£¬·´Ó¦´ïµ½Æ½ºâ״̬£¬²âµÃÌåϵѹǿÉý¸ß£®
ÎïÖÊABCD
ÆðʼͶÁÏ/mol2230
¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪK=$\frac{{c}^{3}£¨C£©}{{c}^{2}£¨A£©¡Ác£¨B£©}$£®Éý¸ßζȣ¬Æ½ºâ³£Êý½«¼õС£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

6£®ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜҺʱ£¬Ôì³ÉËùÅäÈÜҺŨ¶ÈÆ«µÍµÄÊÇ£¨¡¡¡¡£©
A£®NaOHÈܽâºóδ¾­ÀäÈ´£¬Ñ¸ËÙתÒƵ½ÈÝÁ¿Æ¿ÖÐ
B£®ÈÝÁ¿Æ¿Î´¸ÉÔï
C£®¶¨ÈÝʱ¸©ÊÓÒºÃæ
D£®Ï´µÓÉÕ±­ºÍ²£Á§°ôµÄÈÜҺδתÒƵ½ÈÝÁ¿Æ¿ÖÐ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

3£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¼õС·´Ó¦ÎïŨ¶È£¬¿É¼õСµ¥Î»Ìå»ýÄڻ·Ö×ӵİٷÖÊý£¬´Ó¶øʹÓÐЧÅöײ´ÎÊý¼õÉÙ
B£®¶ÔÓÚÓÐÆøÌå²Î¼ÓµÄ»¯Ñ§·´Ó¦£¬Èô¼õСѹǿ£¨¼´À©´ó·´Ó¦ÈÝÆ÷µÄÌå»ý£©£¬¿É¼õС»î»¯·Ö×ӵİٷÖÊý£¬´Ó¶øʹ·´Ó¦ËÙÂʼõС
C£®¸Ä±äÌõ¼þÄÜʹ»¯Ñ§·´Ó¦ËÙÂÊÔö´óµÄÖ÷ÒªÔ­ÒòÊÇÔö´óÁË·´Ó¦Îï·Ö×ÓÖл·Ö×ÓµÄÓÐЧÅöײ¼¸ÂÊ
D£®ÄÜÁ¿¸ßµÄ·Ö×ÓÒ»¶¨ÄÜ·¢ÉúÓÐЧÅöײ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

10£®ÏÂÁл¯Ñ§ÓÃÓïʹÓÃÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®µí·ÛµÄ×î¼òʽ£ºCH2OB£®ÁÚôÇ»ù±½¼×ËáµÄ½á¹¹¼òʽ£º
C£®2-ÒÒ»ù-1£¬3-¶¡¶þÏ©µÄ¼üÏßʽ£ºD£®±½·Ö×ÓÇò¹÷Ä£ÐÍ£º

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

3£®Ä³Ñ§ÉúΪ²â¶¨Î´ÖªÅ¨¶ÈµÄÏ¡ÁòËáÈÜÒº£¬ÊµÑéÈçÏ£ºÒÔ0.14mol•L-1µÄNaOHÈÜÒºµÎ¶¨ÉÏÊöÏ¡H2SO425.00mL£¬µÎ¶¨ÖÕֹʱÏûºÄNaOHÈÜÒº15.00mL£¬¸ÃѧÉúÓñê×¼0.14mol•L-1NaOHÈÜÒºµÎ¶¨ÁòËáµÄʵÑé²Ù×÷ÈçÏ£ºa¼ì²éµÎ¶¨¹ÜÊÇ·ñ©ˮbÈ¡Ò»¶¨Ìå»ýµÄ´ý²âÒºÓÚ׶ÐÎÆ¿ÖÐcÓñê×¼ÈÜÒºÈóÏ´Ê¢±ê×¼ÈÜÒºµÄµÎ¶¨¹Ü£¬Óôý²âÒºÈóÏ´Ê¢´ý²âÒºµÄµÎ¶¨¹Üd×°±ê×¼ÈÜÒººÍ´ý²âÒº²¢µ÷ÕûÒºÃ棨¼Ç¼³õ¶ÁÊý£©eÓÃÕôÁóˮϴµÓ²£Á§ÒÇÆ÷f°Ñ׶ÐÎÆ¿·ÅÔڵζ¨¹ÜÏÂÃ棬ƿϵæÒ»ÕÅ°×Ö½£¬±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿Ö±ÖÁµÎ¶¨Öյ㣬¼ÇÏµζ¨¹ÜÒºÃæËùÔڿ̶È
¢ÙµÎ¶¨²Ù×÷µÄÕýȷ˳ÐòÊÇaecdbf£¨ÓÃÐòºÅÌîд£©£»
¢Ú¼îʽµÎ¶¨¹ÜÓÃÕôÁóË®ÈóÏ´ºó£¬Î´Óñê×¼ÒºÈóÏ´µ¼Öµζ¨½á¹ûÆ«´ó£¨ÌƫС¡±¡¢¡°Æ«´ó¡±»ò¡°ÎÞÓ°Ï족£©£»¹Û²ì¼îʽµÎ¶¨¹Ü¶ÁÊýʱ£¬ÈôµÎ¶¨Ç°ÑöÊÓ£¬µÎ¶¨ºó¸©ÊÓ£¬Ôò½á¹û»áµ¼Ö²âµÃµÄÏ¡H2SO4ÈÜҺŨ¶È²â¶¨ÖµÆ«Ð¡£¨Ñ¡Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

10£®ºìÆÏÌѾÆÃÜ·â´¢´æʱ¼äÔ½³¤£¬ÖÊÁ¿Ô½ºÃ£¬Ô­ÒòÖ®Ò»ÊÇ´¢´æ¹ý³ÌÖÐÉú³ÉÁËÓÐÏãζµÄõ¥£®ÔÚʵÑéÊÒÒ²¿ÉÒÔÓÃÈçͼËùʾµÄ×°ÖÃÖÆÈ¡ÒÒËáÒÒõ¥£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣮
£¨1£©ÒÒ´¼·Ö×ÓÖйÙÄÜÍŵÄÃû³ÆÊÇôÇ»ù£®
£¨2£©ÊÔ¹ÜaÖмÓÈ뼸¿éËé´ÉƬµÄÄ¿µÄÊÇ·ÀÖ¹ÒºÌ屩·Ð£®
£¨3£©ÊÔ¹ÜaÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCH3COOH+CH3CH2OH $?_{¡÷}^{ŨÁòËá}$CH3COOC2H5+H2O£¬·´Ó¦ÀàÐÍÊÇÈ¡´ú·´Ó¦£®
£¨4£©·´Ó¦¿ªÊ¼Ç°£¬ÊÔ¹ÜbÖÐÊ¢·ÅµÄÈÜÒºÊDZ¥ºÍNa2CO3ÈÜÒº£®
£¨5£©¿ÉÓ÷ÖÒºµÄ·½·¨°ÑÖƵõÄÒÒËáÒÒõ¥·ÖÀë³öÀ´£®
£¨6£©½«1molÒÒ´¼£¨ÆäÖеÄÑõÓà18O ±ê¼Ç£©ÔÚŨÁòËá´æÔÚÌõ¼þÏÂÓë×ãÁ¿ÒÒËá³ä·Ö·´Ó¦£®ÏÂÁÐÐðÊö²»ÕýÈ·µÄÊÇB
A£®Éú³ÉµÄÒÒËáÒÒõ¥Öк¬ÓÐ18O       B£®Éú³ÉµÄË®·Ö×ÓÖк¬ÓÐ18O
C£®¿ÉÄÜÉú³É44gÒÒËáÒÒõ¥         D£®²»¿ÉÄÜÉú³É90gÒÒËáÒÒõ¥£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

7£®ÒÒËáµÄõ¥»¯·´Ó¦
ʵÑ飺£¨1£©¡¢ÏòÊÔ¹ÜÖÐÏȼÓÈëÉÙÁ¿Ëé´ÉƬ£¬ÔÙ¼ÓÈë3mlÒÒ´¼¡¢2mlÒÒËᣬÔÙ±ßÕñµ´ÊԹܱßÂýÂý¼ÓÈë2mlŨÁòËᣬÁ¬½ÓºÃ×°Ö㬻ºÂý¼ÓÈÈ£®
£¨2£©µ¼Æø¹Ü²»Òª²åÈëÒºÃæÏ£®
£¨3£©×¢Òâ¹Û²ì±¥ºÍ̼ËáÄÆÈÜÒºµÄÒºÃæÉϵõ½µÄÎïÖʵÄÑÕÉ«ºÍ״̬£¬²¢Îŵ½Ïãζ£®
¢Ù·´Ó¦·½³Ìʽ£ºCH3COOH+CH3CH2OH$?_{¡÷}^{ŨH_{2}SO_{4}}$CH3COOCH2CH3+H2O
¢ÚŨÁòËáÔÚõ¥»¯·´Ó¦ÖÐÆð´ß»¯¼Á¡¢ÎüË®¼Á×÷Óã¿
¢Û±¥ºÍ̼ËáÄÆÈÜÒºµÄ×÷Óã¿¢ÙÖкÍÒÒËá¢ÚÈܽâÒÒ´¼¢Û½µµÍÒÒËáÒÒõ¥µÄÈܽâ¶È
¢Üµ¼Æø¹ÜΪʲô²»ÄܲåÈëÒºÃæÏ£¿·ÀÖ¹µ¹Îü
¢ÝʵÑéÖмÓÈÈÊԹܵÄÄ¿µÄÊÇʲô£¿¼Ó¿ì»¯Ñ§·´Ó¦ËÙÂÊ
¢Þõ¥»¯·´Ó¦ÊôÓÚÄÄÒ»ÀàÐÍÓлú·´Ó¦£¿È¡´ú·´Ó¦£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

8£®µªÆøÓëÇâÆø·´Ó¦Éú³É°±ÆøµÄƽºâ³£Êý¼ûÏÂ±í£º
N2+3H2?2NH3
ζÈ25¡æ200¡æ400¡æ600¡æ
ƽºâ³£ÊýK5¡Á1086500.5070.01
£¨1£©¹¤ÒµÉϺϳɰ±µÄζÈÒ»°ã¿ØÖÆÔÚ500¡æ£¬Ô­ÒòÊÇ´ß»¯¼ÁµÄ»îÐÔζÈÇÒ·´Ó¦ËÙÂʽϿ죮
£¨2£©ÔÚ2LÃܱÕÈÝÆ÷ÖмÓÈë1molµªÆøºÍ3molÇâÆø£¬½øÐй¤ÒµºÏ³É°±µÄÄ£ÄâʵÑ飬Èô2·ÖÖÓºó£¬ÈÝÆ÷ÄÚѹǿΪԭÀ´µÄ0.8±¶£¬Ôò0µ½2·ÖÖÓ£¬°±ÆøµÄ·´Ó¦ËÙÂÊΪ0.2mol/£¨L•min£©£®
£¨3£©ÏÂÁÐ˵·¨ÄܱíÃ÷¸Ã·´Ó¦´ïµ½Æ½ºâµÄÊÇAB
A£®ÆøÌåµÄƽ¾ù·Ö×ÓÁ¿²»Ôٱ仯             B£®ÃܱÕÈÝÆ÷ÄÚµÄѹǿ²»Ôٱ仯
C£®v £¨N2£©=2v £¨NH3£©                      D£®ÆøÌåµÄÃܶȲ»Ôٱ仯
£¨4£©ÏÂÁдëÊ©£¬¼ÈÄܼӿì¸Ã·´Ó¦µÄ·´Ó¦ËÙÂÊ£¬ÓÖÄÜÔö´óת»¯ÂʵÄÊÇB
A£®Ê¹Óô߻¯¼Á     B£®ËõСÈÝÆ÷Ìå»ý      C£®Ìá¸ß·´Ó¦Î¶Ƞ    D£®ÒÆ×ßNH3
£¨5£©³£ÎÂÏ£¬ÔÚ°±Ë®ÖмÓÈëÒ»¶¨Á¿µÄÂÈ»¯ï§¾§Ì壬ÏÂÁÐ˵·¨´íÎóµÄÊÇAD£®
A£®ÈÜÒºµÄpHÔö´ó    B£®°±Ë®µÄµçÀë¶È¼õС    C£®c£¨OH-£©¼õС    D£®c£¨NH4+£©¼õС
£¨6£©½«°±Ë®ÓëÑÎËáµÈŨ¶ÈµÈÌå»ý»ìºÏ£¬ÏÂÁÐ×ö·¨ÄÜʹc£¨NH4+£©Óëc£¨Cl-£©±ÈÖµ±ä´óµÄÊÇAC
A£® ¼ÓÈë¹ÌÌåÂÈ»¯ï§                B£®Í¨ÈëÉÙÁ¿ÂÈ»¯Çâ
C£® ½µµÍÈÜҺζȠ                 D£®¼ÓÈëÉÙÁ¿¹ÌÌåÇâÑõ»¯ÄÆ£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸