·ÖÎö £¨1£©¸ù¾Ý±ûÏ©È©µÄ»¯Ñ§Ê½ºÍÔªËصÄÏà¶ÔÔ×ÓÖÊÁ¿£¬¿ÉÒÔ¼ÆËã³ö±ûÏ©È©µÄÏà¶Ô·Ö×ÓÖÊÁ¿£»
£¨2£©¸ù¾Ý±ûÏ©È©µÄ»¯Ñ§Ê½ºÍÔªËصÄÖÊÁ¿·ÖÊý¹«Ê½¿ÉÒÔÇó³ö±ûÏ©È©ÖÐÑõÔªËصÄÖÊÁ¿·ÖÊý£»
£¨3£©¸ù¾Ý±ûÏ©È©µÄ»¯Ñ§Ê½ºÍÔªËصÄÖÊÁ¿·ÖÊý¹«Ê½ÏȼÆËã³ö±ûÏ©È©ÖÐ̼ԪËصÄÖÊÁ¿·ÖÊý£¬ÔÙ¸ù¾Ý̼ԪËصÄÖÊÁ¿=224g¡Á±ûÏ©È©ÖÐ̼ԪËصÄÖÊÁ¿·ÖÊý¼ÆËã¼´¿É£®
½â´ð ½â£º
£¨1£©¸ù¾Ý±ûÏ©È©µÄ»¯Ñ§Ê½£¨C2H3CHO£©¿ÉÖª£¬ËüµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª£º12¡Á3+1¡Á4+16=56£®
£¨2£©±ûÏ©È©ÖÐ̼¡¢Çâ¡¢ÑõÈýÖÖÔªËصÄÖÊÁ¿±ÈΪ£º£¨12¡Á3£©£º£¨1¡Á4£©£º16=36£º4£º16=9£º1£º4£¬¹Ê´ð°¸Îª£º9£º1£º4£®
£¨3£©±ûÏ©È©ÖÐ̼ԪËصÄÖÊÁ¿·ÖÊýΪ£º$\frac{12¡Á3}{56}¡Á100%$£¬Ôò112g±ûÏ©È©ÖÐ̼ԪËصÄÖÊÁ¿Îª£º112g¡Á$\frac{12¡Á3}{56}¡Á100%$=72g£®
¹Ê´ð°¸Îª£º
£¨1£©56£®£¨2£©9£º1£º4£®£¨3£©72g£®
µãÆÀ ±¾ÌâÖ÷Òª¿¼²éѧÉúÔËÓû¯Ñ§Ê½ºÍÏà¶ÔÔ×ÓÖÊÁ¿ÒÔ¼°ÔªËصÄÖÊÁ¿·ÖÊý¹«Ê½½øÐмÆËãµÄÄÜÁ¦£®
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¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
·´ Ó¦ Îï | »¯ ѧ ·½ ³Ì ʽ | |
1 | ÇâÑõ»¯¸ÆÓëÑÎËá·´Ó¦ | Ca£¨OH£©2+2HCl=CaCl2+2H2O |
2 | ÇâÑõ»¯ÄÆÓëÏ¡ÁòËá·´Ó¦ | 2NaOH+H2SO4=Na2SO4+2H2O |
3 | ÓÃθÊæƽ£¨ÇâÑõ»¯ÂÁ£©Öк͹ý¶àµÄθËá | Al£¨OH£©3+3HCl=AlCl3+3H2O |
4 | ÓÃÊìʯ»ÒÖкͺ¬ÓÐÁòËáµÄÎÛË® | Ca£¨OH£©2+H2SO4=CaSO4+2H2O |
5 | ̼ËáÇâÄÆÓëÏ¡ÑÎËá·´Ó¦ | NaHCO3+HCl=NaCl+H2O+CO2¡ü |
6 | ÓÃÏ¡ÑÎËá³ýÌúÐâ | 6HCl+Fe2O3=2FeCl3+3H2O |
7 | ÓÃÏ¡ÁòËá³ýÌúÐâ | 3H2SO4+Fe2O3=Fe2£¨SO4£©3+3H2O |
8 | ÓÃ̼ËáÄÆÖÆÈ¡ÇâÑõ»¯ÄÆ | Ca£¨OH£©2+Na2CO3=CaCO3¡ý+2NaOH |
9 | ÂÈ»¯±µÓëÏ¡ÁòËá·´Ó¦ | BaCl2+H2SO4=BaSO4¡ý+2HCl |
10 | ÇâÑõ»¯ÄÆÓëÈýÑõ»¯Áò·´Ó¦ | 2NaOH+SO3=Na2SO4+H2O |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | Ëü²»ÊÇÔ×ÓµÄÖÊÁ¿ | B£® | ¿ÉÒÔÓÃǧ¿Ë×÷µ¥Î» | ||
C£® | ËüÊǸö±ÈÖµ | D£® | ÊýÖµÉÏÔ¼µÈÓÚÖÊ×ÓÊýÓëÖÐ×ÓÊýÖ®ºÍ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ×ó±ßÉÕ±ÖÐÈÜÒºµÄÖÊÁ¿¼õÇá | B£® | ÓÒ±ßÉÕ±ÖÐÈÜÒºµÄÖÊÁ¿¼õÇá | ||
C£® | Á½±ßÉÕ±ÖÐÈÜÒºµÄÖÊÁ¿¶¼¼õÇá | D£® | Ƭ¿ÌºóÁ½±ßÈÔȻƽºâ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | H2-2¸öÇâÔ×Ó | B£® | 2O-2¸öÑõÔªËØ | C£® | P2O5-ÎåÑõ»¯¶þÁ× | D£® | $\stackrel{+2}{Ca}$-¸ÆÀë×Ó |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ¸É±ù·Ö×ÓÖÊÁ¿±ä´ó | B£® | ¸É±ù·Ö×Ó¼äµÄ¼ä¸ô±ä´ó | ||
C£® | ¸É±ù·Ö×ÓÌå»ý±äС | D£® | ¸É±ù·Ö×ÓÔ˶¯ËÙÂʲ»±ä |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ
A£® | ×ÏɫʯÈïÊÔÒº¡°¾¯²ì¡± | B£® | Ï¡ÑÎËá¡°¾¯²ì¡± | ||
C£® | ÎÞÉ«·Ó̪ÊÔÒº¡°¾¯²ì¡± | D£® | Ë®¡°¾¯²ì¡± |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ÇâÆøºÍÒºÇⶼ¿É×öȼÁÏ£¨ÏàͬÎïÖʵķÖ×Ó£¬Æ仯ѧÐÔÖÊÏàͬ£© | |
B£® | Óþ¯È®ËѾȵØÕðÖб»ÂñÈËÔ±£¨·Ö×ÓÔÚ²»¶ÏÔ˶¯£© | |
C£® | ÓÃË®ÒøζȼƲâÁ¿ÌåΣ¨Î¶ÈÉý¸ß£¬Ô×Ó¼ä¼ä¸ô±ä´ó£© | |
D£® | Ë®ÉÕ¿ªºóÒװѺø¸Ç³åÆð£¨Î¶ÈÉý¸ß£¬·Ö×Ó±ä´ó£© |
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