A£® | °×É«¹ÌÌåÒ»¶¨Ö»º¬ÓÐÁòËáï§ | |
B£® | ÈôÇâÑõ»¯ÄÆÈÜÒº×ãÁ¿£¬ÔòÉú³É°±ÆøµÄÌå»ýӦΪ6.72L£¨±ê¿ö£© | |
C£® | ´ÓÉÏÊöÊý¾ÝÄÜÇóËã³ö°×É«¹ÌÌåÖУ¨NH4£©2SO4¡¢NH4HSO4µÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º2 | |
D£® | ½«Ä³°×É«¹ÌÌå¼ÓÈȷֽ⣬Èô²úÉúµÄÆøÌå²»ÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬Ôò¸Ã¹ÌÌåÖÐÒ»¶¨²»º¬ï§ÑÎ |
·ÖÎö A£®Éè²ÎÓë·´Ó¦µÄ£¨NH4£©2SO4ºÍNH4HSO4£¬·Ö±ðΪxmol£¬ymol£¬
£¨NH4£©2SO4+2NaOH=2NH3¡ü+Na2SO4+2H2O
1 2 2
x 2x 2x
NH4HSO4+2NaOH=NH3¡ü+Na2SO4+2H2O
1 2 1
y 2y y
n£¨NH3£©=$\frac{2.24L}{22.4L/mol}$=0.1mol£¬Ôò2x+y=0.1¢Ú£¬n£¨NaOH£©=2x+2y=0.05L¡Á4mol/L¢Ú£¬ÁªÁ¢¢Ù¢Ú£¬¾Ý´Ë½øÐзÖÎö¹ÌÌåµÄ³É·Ö£»
B£®ÓÉAµÃn£¨NH4HSO4£©=0.1mol£¬m£¨£¨NH4£©2SO4£©=24.7g-0.1mol¡Á115g/mol=13.2g£¬¹Ên£¨£¨NH4£©2SO4£©=$\frac{13.2g}{132g/mol}$=0.1mol£¬²¢¾ÝµªÔªËØÊغãn£¨NH3£©=2n£¨£¨NH4£©2SO4£©+n£¨NH4HSO4£©=2¡Á0.1mol+0.1mol=0.3mol£¬²¢¾ÝV=nVm½øÐмÆË㣻
C£®½áºÏABÖÐÊý¾Ý·ÖÎö£»
D£®¾Ýï§ÑμÓÈÈÌõ¼þÏ¿ÉÄÜÉú³ÉµªÆø·ÖÎö£®
½â´ð ½â£ºA£®Éè²ÎÓë·´Ó¦µÄ£¨NH4£©2SO4ºÍNH4HSO4£¬·Ö±ðΪxmol£¬ymol£¬
£¨NH4£©2SO4+2NaOH=2NH3¡ü+Na2SO4+2H2O
1 2 2
x 2x 2x
NH4HSO4+2NaOH=NH3¡ü+Na2SO4+2H2O
1 2 1
y 2y y
n£¨NH3£©=$\frac{2.24L}{22.4L/mol}$=0.1mol£¬Ôò2x+y=0.1¢Ú£¬n£¨NaOH£©=2x+2y=0.05L¡Á4mol/L¢Ú£¬ÁªÁ¢¢Ù¢Ú£¬½âµÃx=0£¬y=0.1£¬¹Ê¹ÌÌåÖп϶¨ÓÐNH4HSO4£¬¹Êm£¨NH4HSO4£©=0.1mol¡Á115g/mol=11.5g£¬11.5g£¼24.70g£¬¹Êº¬ £¨NH4£©2SO4ºÍNH4HSO4£¬¹ÊA´íÎó£»
B£®ÓÉAµÃn£¨NH4HSO4£©=0.1mol£¬m£¨£¨NH4£©2SO4£©=24.7g-0.1mol¡Á115g/mol=13.2g£¬¹Ên£¨£¨NH4£©2SO4£©=$\frac{13.2g}{132g/mol}$=0.1mol£¬²¢¾ÝµªÔªËØÊغãn£¨NH3£©=2n£¨£¨NH4£©2SO4£©+n£¨NH4HSO4£©=2¡Á0.1mol+0.1mol=0.3mol£¬²¢¾ÝV=nVm=0.3mol¡Á22.4L/mol=6.72L£¬¹ÊBÕýÈ·£»
C£®ÓÉA£¬BµÃn£¨NH4HSO4£©=0.1mol£¬n£¨£¨NH4£©2SO4£©=$\frac{13.2g}{132g/mol}$=0.1mol£¬¹Ê£¨NH4£©2SO4¡¢NH4HSO4µÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º1£¬¹ÊC´íÎó£»
D£®Èôï§ÑμÓÈÈÌõ¼þÏÂÉú³ÉµªÆø£¬ÔòµªÆø²»ÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬¹ÊD´íÎó£»
¹ÊÑ¡B£®
µãÆÀ ±¾Ì⿼²éÁË»ìºÏÎï×é³ÉµÄÍƶϣ¬Íê³É´ËÌ⣬¿ÉÒÔ½áºÏÌâ¸ÉÌṩµÄÐÅÏ¢½øÐзÖÎö£¬²¢½áºÏÔ×ÓÊغã½øÐмÆË㣬ÄѶÈÖеȣ®
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | Ò»¸ö¼×È©·Ö×ÓÓÉ1¸ö̼Ô×Ó¡¢2¸öÇâÔ×Ó¡¢1¸öÑõÔ×Ó¹¹³ÉµÄ | |
B£® | ¼×È©ÖÐ̼¡¢Çâ¡¢ÑõÈýÖÖÔªËصÄÖÊÁ¿±ÈΪ1£º2£º1 | |
C£® | ¼×È©ÓÉ̼¡¢Çâ¡¢ÑõÈýÖÖÔªËØ×é³ÉµÄ | |
D£® | ¼×È©ÖÐÑõÔªËصÄÖÊÁ¿·ÖÊýΪ53.3% |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
Ñ¡Ïî | ʵÑé²Ù×÷ºÍÏÖÏó | ½áÂÛ |
A | È¡ÉÙÁ¿´ý¼ìÑéÈÜÒº£¬ÏòÆäÖмÓÈëÉÙÁ¿ÐÂÖÆÂÈË®£¬ÔڵμÓKSCNÈÜÒº£¬ÈÜÒº³ÊѪºìÉ« | ´ý²âÒºÖк¬ÓÐFe3+ |
B | ÊÒÎÂÏ£¬ÏòŨ¶È¾ùΪ0.1mol•L-1µÄBaCl2ºÍCaCl2»ìºÏÈÜÒºÖеμÓNa2SO4ÈÜÒº£¬³öÏÖ°×É«³Áµí£® | Ksp£¨Ba2SO4£©£¼Ksp£¨Ca2SO4£© |
C | ÊÒÎÂÏ£¬ÏòFeCl3ÈÜÒºÖеμÓÉÙÁ¿KIÈÜÒº£¬ÔٵμӼ¸µÎµí·ÛÈÜÒº£¬ÈÜÒº±äÀ¶É« | Fe3+µÄÑõ»¯ÐÔ±ÈI2µÄÇ¿ |
D | ÊÒÎÂÏ£¬ÓÃPHÊÔÖ½²âµÃ£º0.1mol•L-1 Na2SO3ÈÜÒºµÄPHԼΪ10£»0.1mol•L-1 NaHSO3ÈÜÒºµÄPHԼΪ5 | HSO3-½áºÏH+µÄÄÜÁ¦±ÈSO32-µÄÇ¿ |
A£® | A | B£® | B | C£® | C | D£® | D |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | µâË®ÖмÓÈëÉÙÁ¿ÆûÓÍÕñµ´£¬¾²ÖúóÉϲãÑÕÉ«±ädz£¬Ï²ãÑÕÉ«±äΪ×ϺìÉ« | |
B£® | ºìÈȵÄÍË¿ÔÚÂÈÆøÖÐȼÉÕ£¬²úÉúÁË×Ø»ÆÉ«µÄÑÌÎí | |
C£® | ±Ëص¥ÖÊ£¨X2£©ÓëË®·´Ó¦¾ù¿ÉÓÃX2+H2O=HX+HXO±íʾ | |
D£® | ä廯ÄÆÈÜÒºÖмÓÈëÉÙÁ¿ÐÂÖƵÄÂÈË®Õñµ´£¬ÔÙ¼ÓÈëÉÙÁ¿ËÄÂÈ»¯Ì¼Õñµ´£¬¾²ÖúóÉϲãÑÕÉ«±ädz£¬Ï²ãÑÕÉ«±äΪ³ÈºìÉ« |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
»¯Ñ§¼ü | S=O£¨SO2£© | S=O£¨SO3£© | O=O£¨O2£© |
ÄÜÁ¿/kJ | 535 | 472 | 496 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ³ýÈ¥»ìÔÚCu·ÛÖеÄÉÙÁ¿Mg·ÛºÍAl·Û£¬¼ÓÏ¡ÑÎËáºó¹ýÂË | |
B£® | ·ÖÀëÆûÓͺÍúÓÍ£¬¿ÉÓÃÝÍÈ¡µÄ·½·¨ | |
C£® | ·ÖÀëÏõËá¼ØºÍÂÈ»¯ÄƹÌÌåµÄ»ìºÏÎ¿ÉÓÃÈܽ⡢¹ýÂ˵ķ½·¨ | |
D£® | ½«ÑõÆøºÍÇâÆøµÄ»ìºÏÆøÌåͨ¹ý×ÆÈȵÄÑõ»¯Í£¬ÒÔ³ýÈ¥ÆäÖеÄÇâÆø |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¼ÆËãÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com