ÎÒ¹úÖƼҵµÄÏÈÇý¡ª¡ªºîµÂ°ñ̽Ë÷·¢Ã÷ÁË¡°ºîÊÏÖƼ¡±£¬ÆäÉú²ú¹ý³ÌÉæ¼°µÄÖ÷Òª»¯Ñ§·´Ó¦ÈçÏ£º
¢ÙNH2£«CO2£«X£½NH4HCO3[
¢ÚNH4HCO3£«NaCl£½NH4Cl£«NaHCO3¡ý
¢Û2NaHCO3Na2CO3£«H2O£«CO2¡ü
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©·´Ó¦¢ÙÖÐXµÄ»¯Ñ§Ê½Îª¡£
£¨2£©³ýÈ¥»ìÔÚNa2CO3·ÛÄ©ÖÐÉÙÁ¿µÄNaHCO3µÄ·½·¨ÊÇ ¡£
£¨3£©¹¤Òµ´¿¼îÖк¬ÓÐÂÈ»¯ÄÆ£¬È¡55g¹¤Òµ´¿¼î£¬ÏòÆäÖмÓÈë269.5gÏ¡ÑÎËᣬǡºÃÍêÈ«·´Ó¦£¬Éú³É22g¶þÑõ»¯Ì¼£¬Çó£º
¢Ù¹¤Òµ´¿¼îÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊý¡££¨¼ÆËã½á¹û±£Áôµ½0.1%£©
¢Ú·´Ó¦ºóÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý¡£
H2O½«»ìºÏÎï¼ÓÈÈ
½â£ºÉè¢Ù55g¹¤Òµ´¿¼îÖÐ̼ËáÄƵÄÖÊÁ¿Îªx£¬¢Ú·´Ó¦ºóÉú³ÉÂÈ»¯ÄƵÄÖÊÁ¿Îªy¡£
Na2CO3£«2HCl=2NaCl£«H2O£«CO2¡ü
106 117 44
x y 22g
£½ x£½53g
¹¤Òµ´¿¼îÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊýΪ¡Á100%£½96.4%
£½ y£½58.5g
Ôò·´Ó¦ºóÈÜÒºÖÐÈÜÖÊÂÈ»¯ÄƵÄÖÊÁ¿·ÖÊýΪ¡Á100%£½20%
´ð£º¢Ù¹¤Òµ´¿¼îÖÐ̼ËáµÄÖÊÁ¿·ÖÊýΪ96.4%¡£¢Ú·´Ó¦ºóÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ20%¡£
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£º
ΪʵÏÖÏÂÁÐʵÑéÄ¿µÄ£¬ËùÑ¡µÄÊÔ¼Á»ò·½·¨ÕýÈ·µÄÊÇ£¨£©
Ñ¡Ïî ʵÑéÄ¿µÄ ËùÓÃÊÔ¼Á»ò·½·¨
A¡¢ Çø·ÖNH4NO3ÈÜÒººÍK2SO4ÈÜÒº ¼ÓBa£¨NO3£©2¡¡ÈÜÒº
B¡¢ ³ýÈ¥Éúʯ»ÒÖк¬ÓеÄÔÓÖÊʯ»Òʯ ¼ÓË®»òÏ¡ÑÎËá
C¡¢ ³ýÈ¥CO2ÖеÄCOÆøÌå ÔÚÑõÆøÖеãȼ
D¡¢ ³ýÈ¥Ñõ»¯Í·ÛÄ©ÖеÄÍ·Û¼ÓÈë×ãÁ¿Ï¡ÑÎËᣬ ¹ýÂË
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁйý³ÌÖÐÉæ¼°»¯Ñ§±ä»¯µÄÊÇ£¨¡¡¡¡£©
¡¡ | A£® | ÓþÛÒÒÏ©ËÜÁÏÖƵÃʳƷ´ü | B£® | ÓÃʳÑÎÖƵô¿¼î |
¡¡ | C£® | ½«ÒºÌ¬¿ÕÆø·ÖÀëÖƵÃÑõÆø | D£® | ½«Ê¯ÓÍ·ÖÀëÖƵÃÆûÓÍ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
È¡10g̼Ëá¸ÆÓëÈÜÖÊÖÊÁ¿·ÖÊýΪ10%µÄÏ¡ÑÎËá·´Ó¦£¬Ç¡ºÃÍêÈ«·´Ó¦Ê±¼äµÃµ½ÈÜÒºµÄÖÊÁ¿Îªa g£»ÁíÈ¡10g̼Ëá¸Æ¸ßÎÂÍêÈ«·Ö½âºó£¬È¡Éú³ÉµÄCaOÓëÉÏÊöͬŨ¶ÈµÄÏ¡ÑÎËá·´Ó¦£¬Ç¡ºÃÍêÈ«·´Ó¦Ê±µÃµ½ÈÜÒºµÄÖÊÁ¿Îªb g£¬ÔòaÓëbµÄ´óС¹ØϵΪ£¨£©
¡¡A£® a=b B£® a£¾b C£® a£¼b D£® ²»ÄÜÈ·¶¨
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
СÃ÷ÔÚʵÑéÊÒ·¢ÏÖһƿδ֪Ũ¶ÈµÄÇâÑõ»¯ÄÆÈÜÒº£¬Îª·½±ãÒÔºóʹÓã¬Ëû¶ÔÆäŨ¶È½øÐÐÁ˲ⶨ£®È¡20.0g´ËÇâÑõ»¯ÄÆÈÜÒºÓÚÉÕ±ÖУ¬ÖðµÎµÎ¼ÓÈÜÖÊÖÊÁ¿·ÖÊýΪ7.3%µÄÏ¡ÑÎËᣬ²¢Ëæʱ¶Ô·´Ó¦ºóµÄÈÜÒºÓÃpH¼Æ£¨Ò»ÖֲⶨÈÜÒºpHµÄÒÇÆ÷£©²â¶¨ÈÜÒºµÄpH£¬ËùµÃÊý¾ÝÈç±í£º
¼ÓÈëÏ¡ÑÎËá µÄÖÊÁ¿/g | 9.6 | 9.8 | 9.9 | 10.0 | 10.1 |
ÈÜÒºµÄpH | 12.4 | 12.1 | 11.8 | 7.0 | 2.2 |
ÊԻشð£º
£¨1£©µ±µÎ¼ÓÏ¡ÑÎËáµÄÖÊÁ¿Îª9.8gʱ£¬ÈÜÒºÖеÄÈÜÖÊÊÇ¡¡¡¡¡¢¡¡£»
£¨2£©¼ÆËãËù²âÇâÑõ»¯ÄÆÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁйØÓÚ»¯Ñ§·´Ó¦2X£«Y===2ZµÄÐðÊö£¬²»ÕýÈ·µÄÊÇ(¡¡¡¡)¡£
A£®ZÒ»¶¨ÊÇ»¯ºÏÎ²¢ÇÒ¿ÉÄÜÊÇÑõ»¯Îï
B£®ÔÚ·´Ó¦ÖÐX¡¢Y¡¢ZÈýÖÖÎïÖʵÄÁ£×ÓÊýÄ¿±ÈΪ2¡Ã1¡Ã2
C£®ÈôXºÍYµÄÏà¶Ô·Ö×ÓÖÊÁ¿·Ö±ðΪMºÍN£¬ÔòZµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª(M£«N)
D£®·´Ó¦ÊôÓÚ»¯ºÏ·´Ó¦ÀàÐÍ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
·¢ÉäͨÐÅÎÀÐǵĻð¼ýÓÃÁª°±(N2H4)×÷ȼÁÏ£¬ÓÃËÄÑõ»¯¶þµª(N2O4)Öúȼ£¬Éú³ÉÎï²»»á¶Ô´óÆøÔì³ÉÎÛȾ¡£
(1)·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2N2H4£«N2O4===3________£«4H2O£¬ÇëÔÚºáÏßÉÏÌîд»¯Ñ§Ê½ÒÔÍê³É¸Ã»¯Ñ§·½³Ìʽ¡£
(2)Çë¼ÆËã9.6 kg N2H4ÍêȫȼÉÕÐèÒªÖúȼÎïN2O4µÄÖÊÁ¿¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÅäÖƲ¢Ï¡ÊÍÒ»¶¨ÖÊÁ¿·ÖÊýµÄNa2SO4ÈÜÒº£®
£¨1£©ÅäÖÆ50gÖÊÁ¿·ÖÊýΪ6%µÄNa2SO4ÈÜÒº£®
¢Ù¼ÆË㣺ÐèÒªNa2SO4 3.0g£¬Ë® 47.0g
¢Ú³ÆÁ¿£ºÓÃÍÐÅÌÌìƽ³ÆÁ¿3.0gµÄNa2SO4£®Ììƽµ÷Áãºó£¬·Ö±ðÔÚÌìƽ×óÓÒÍÐÅÌ·ÅÉÏÖÊÁ¿ÏàͬµÄֽƬ£¬ÏÈ¡¡¡¡£¬È»ºó¡¡¡¡£¬ÖÁÍÐÅÌÌìƽǡºÃƽºâ£®
¢ÛÁ¿È¡£ºÓÃÁ¿Í²Á¿È¡47.0mLË®£®ÇëÔÚÈçͼÖл³ö47.0mLË®µÄÒºÃæλÖã®
¢ÜÈܽ⣮
£¨2£©Ï¡ÊÍÈÜÒº£®£¨ÓÉÓÚÕû¸öÅäÖƹý³ÌÖÐÈÜÒººÜÏ¡£¬ÆäÃܶȿɽüËÆ¿´×ö1g/mL£©
¢ÙÈ¡1mL 6%µÄNa2SO4ÈÜÒº¼ÓˮϡÊÍÖÁ100mL£¬µÃµ½ÈÜÒºa£»
¢ÚÈôÓÃ3.0gNa2SO4ÅäÖÆÓëÈÜÒºaŨ¶ÈÏàͬµÄÈÜÒº£¬ÆäÌå»ýÊÇ¡¡mL£®
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
´ò»ð»ú¸øÈËÃÇ´øÀ´ÁË·½±ã£¬Í¼Ê¾ÊÇÆÕͨµç×Ó´ò»ð»úµÄʵÎïÕÕƬ¡£ÏÂÃæ¿ÉÒÔ×öµ½·ûºÏȼÉÕÌõ¼þÄÜ´ò»ðµÄ×éºÏÊÇ(¡¡)
A£®¢Ù¢Û¢Ý B£®¢Ù¢Û¢Ü C£®¢Ù¢Ú¢Û D£®¢Ù¢Ú¢Ü
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