ÒÑÖª£ºBaSO4(s) + 4C(s)4CO(g) + BaS(s)¹¤ÒµÉÏÒÔÖؾ§Ê¯¿ó£¨Ö÷Òª³É·ÖBaSO4£¬ÔÓÖÊΪFe2O3¡¢SiO2£©ÎªÔ­ÁÏ£¬Í¨¹ýÏÂÁÐÁ÷³ÌÉú²úÂÈ»¯±µ¾§Ì壨BaCl2¡¤nH2O£©¡£

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²»ÈÜÎïAµÄ»¯Ñ§Ê½ÊÇ_________£»ÈôÔÚʵÑéÊÒ½øÐбºÉÕʱ£¬Ëù²úÉúµÄÆøÌåµÄ´¦Àí·½·¨ÊÇ       
a.ÓÃNaOHÈÜÒºÎüÊÕ       b.ÓÃŨÁòËáÎüÊÕ      c.µãȼ
£¨2£©Óõ¥Î»Ìå»ýÈÜÒºÖÐËùº¬µÄÈÜÖÊÖÊÁ¿ÊýÀ´±íʾµÄŨ¶È½ÐÖÊÁ¿-Ìå»ýŨ¶È£¬¿ÉÒÔÓÃg/L±íʾ,ÏÖÓÃ38%µÄŨÑÎËáÅäÖƺ¬ÈÜÖÊ109.5g/LµÄÏ¡ÑÎËá500mL,ËùÐèÒªµÄ²£Á§ÒÇÆ÷³ýÁ˲£Á§°ô»¹ÓР                          ¡£
£¨3£©³Áµí·´Ó¦ÖÐËù¼ÓµÄÊÔ¼ÁR¿ÉÒÔÊÇÏÂÁÐÊÔ¼ÁÖеĠ               
a.NaOHÈÜÒº    b.BaO¹ÌÌå      c.°±Ë®       d.Éúʯ»Ò
Ö¤Ã÷³ÁµíÒѾ­ÍêÈ«µÄ·½·¨ÊÇ________________________________________________________¡£
£¨4£©Éè¼ÆÒ»¸öʵÑéÈ·¶¨²úÆ·ÂÈ»¯±µ¾§Ì壨BaCl2¡¤nH2O£©ÖеÄnÖµ£¬ÍêÉÆÏÂÁÐʵÑé²½Ö裺
¢Ù³ÆÁ¿ÑùÆ·¢Ú_______ ¢ÛÖÃÓÚ_________£¨ÌîÒÇÆ÷Ãû³Æ£©ÖÐÀäÈ´ ¢Ü³ÆÁ¿ ¢ÝºãÖزÙ×÷¡£
ºãÖزÙ×÷ÊÇÖ¸____________________________________________                   _£»
µÚ¢Û²½ÎïÆ·Ö®ËùÒÔ·ÅÔÚ¸ÃÒÇÆ÷ÖнøÐÐʵÑéµÄÔ­ÒòÊÇ                                  ¡£
£¨5£©½«Öؾ§Ê¯¿óÓë̼ÒÔ¼°ÂÈ»¯¸Æ¹²Í¬±ºÉÕ£¬¿ÉÒÔÖ±½ÓµÃµ½ÂÈ»¯±µ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌΪ
BaSO4+ 4C+CaCl24CO + CaS+ BaCl2¡£ÇëÄãÍêÉÆÏÂÁдӱºÉÕºóµÄ¹ÌÌåÖзÖÀëµÃµ½ÂÈ»¯±µ¾§ÌåµÄʵÑéÁ÷³ÌµÄÉè¼Æ£¨ÒÑÖªÁò»¯¸Æ²»ÈÜÓÚË®£¬Ò×ÈÜÓÚÑÎËᣩ¡£

£¨·½¿òÄÚÌîд²Ù×÷Ãû³Æ£©

£¨1£©SiO2  c
£¨2£©ÉÕ±­  500mLÈÝÁ¿Æ¿ ½ºÍ·µÎ¹Ü
£¨3£©b£»È¡ÉϲãÇåÒºÓÚСÊÔ¹ÜÖУ¬µÎÈëÇâÑõ»¯ÄÆÈÜÒº£¬ÈôÎÞ³Áµí²úÉú£¬ËµÃ÷³ÁµíÍêÈ«¡£
£¨4£©¼ÓÈÈ¡¢¸ÉÔïÆ÷£»ÔÙ½øÐмÓÈÈ¡¢ÀäÈ´¡¢³ÆÁ¿£¬Ö±µ½Á¬ÐøÁ½´Î³ÆÁ¿µÄ½á¹ûÏà²î²»³¬¹ý0.001gΪֹ£»·ÀÖ¹ÀäÈ´¹ý³ÌÖÐÎüÊÕ¿ÕÆøÖеÄË®ÕôÆø£¬Ôì³ÉʵÑéÎó²î¡£
£¨5£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¹¤ÒµÉÏÒÔÖؾ§Ê¯¿óµÄÖ÷Òª³É·ÖBaSO4£¬ÔÓÖÊΪFe2O3¡¢SiO2£¬¼ÓËáºó²»ÈݵÄÊÇSiO2£¬¸ù¾ÝÒÑÖªÌõ¼þ¸øµÄ»¯Ñ§·½³Ìʽ¿ÉÖªËù²úÉúµÄÆøÌåΪCO£¬´¦ÀíCOµÄ·½·¨Êǽ«COµãȼ¡£
£¨2£©ÅäÖÆÈÜÒºËùÐè²£Á§ÒÇÆ÷ÓУºÉÕ±­¡¢500mLÈÝÁ¿Æ¿¡¢½ºÍ·µÎ¹Ü¡¢²£Á§°ô¡£
£¨3£©BaOÓëË®·´Ó¦Éú³ÉÇâÑõ»¯±µ£¬Äܹ»½«ÌúÀë×Ó³Áµí¡£ÇÒ²»ÒýÈëеÄÔÓÖÊ£¬¹ÊÑ¡ÓõÄÊÔ¼ÁΪBaO¹ÌÌ壻֤Ã÷³ÁµíÒѾ­ÍêÈ«µÄ·½·¨ÊÇÈ¡ÉϲãÇåÒºÓÚСÊÔ¹ÜÖУ¬µÎÈëÇâÑõ»¯ÄÆÈÜÒº£¬ÈôÎÞ³Áµí²úÉú£¬ËµÃ÷³ÁµíÍêÈ«¡£
£¨4£©´ËʵÑéµÄÔ­ÀíÊÇÀûÓüÓÈȺóÓë¼ÓÈÈÇ°µÄÖÊÁ¿²î»»Ëã³öÂÈ»¯±µ¾§ÌåÖеÄË®µÄÎïÖʵÄÁ¿¡£ÊµÑé¹ý³ÌÈçÏ¢ٳÆÁ¿ÑùÆ·¢Ú¼ÓÈÈ¢Û¸ÉÔïÆ÷ÖÐÀäÈ´ ¢Ü³ÆÁ¿ ¢ÝºãÖزÙ×÷¡£ºãÖزÙ×÷ÊÇÖ¸Ö±µ½Á¬ÐøÁ½´Î³ÆÁ¿µÄ½á¹ûÏà²î²»³¬¹ý0.001gΪֹ¡£µÚ¢Û²½ÎïÆ·Ö®ËùÒÔ·ÅÔÚ¸ÃÒÇÆ÷ÖнøÐÐʵÑéµÄÔ­ÒòÊÇΪÁË·ÀÖ¹ÀäÈ´¹ý³ÌÖÐÎüÊÕ¿ÕÆøÖеÄË®ÕôÆø£¬Ôì³ÉʵÑéÎó²î¡£
£¨5£©¸ù¾ÝBaSO4+ 4C+CaCl24CO + CaS+ BaCl2¿ÉÖª·´Ó¦ºóÒ»Ñõ»¯Ì¼Òݳö£¬ÒªÏë´Ó±ºÉÕºóµÄ¹ÌÌåÖзÖÀëµÃµ½ÂÈ»¯±µ¾§ÌåÐèÒª´Ó»ìºÏÈÜÒºÖзÖÀëCaS£¬ÓÖCaS΢ÈÜÓÚË®£¬¹ÊÖ±½Ó¹ýÂ˼´¿ÉµÃµ½BaCl2ÈÜÒº£¬ÊµÑé¹ý³ÌÈçÏ£º
¿¼µã£ºÎÞ¼«Íƶϡ¢ÊµÑéÒÇÆ÷µÄʹÓá¢ÊµÑéÊÔ¼ÁµÄÑ¡Ôñ¡¢ÊµÑéÉè¼Æ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

CoCl2¡¤6H2OÊÇÒ»ÖÖËÇÁÏÓªÑøÇ¿»¯¼Á¡£Ò»ÖÖÀûÓÃË®îÜ¿ó(Ö÷Òª³É·ÖΪCo2O3¡¢Co(OH)3£¬»¹º¬ÉÙÁ¿Fe2O3¡¢Al2O3¡¢MnOµÈ)ÖÆÈ¡CoCl2¡¤6H2OµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£º¢Ù½þ³öÒºº¬ÓеÄÑôÀë×ÓÖ÷ÒªÓÐH+¡¢Co2+¡¢Fe2+¡¢Mn2+¡¢Al3+µÈ£»
¢Ú²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ³ÁµíʱÈÜÒºµÄpH¼ûÏÂ±í£º(½ðÊôÀë×ÓŨ¶ÈΪ£º0.01mol/L)

³ÁµíÎï
Fe(OH)3
Fe(OH)2
Co(OH)2
Al(OH)3
Mn(OH)2
¿ªÊ¼³Áµí
2.7
7.6
7.6
4.0
7.7
ÍêÈ«³Áµí
3.7
9.6
9.2
5.2
9.8
 
¢ÛCoCl2¡¤6H2OÈÛµãΪ86¡æ£¬¼ÓÈÈÖÁ110~120¡æʱ£¬Ê§È¥½á¾§Ë®Éú³ÉÎÞË®ÂÈ»¯îÜ¡£
£¨1£©Ð´³ö½þ³ö¹ý³ÌÖÐCo2O3·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ________________________¡£
£¨2£©Ð´³öNaClO3·¢Éú·´Ó¦µÄÖ÷ÒªÀë×Ó·½³Ìʽ_____________________________£»Èô²»É÷Ïò¡°½þ³öÒº¡±ÖмӹýÁ¿NaClO3ʱ£¬¿ÉÄÜ»áÉú³ÉÓж¾ÆøÌ壬д³öÉú³É¸ÃÓж¾ÆøÌåµÄÀë×Ó·½³Ìʽ_______________¡£
£¨3£©¡°¼ÓNa2CO3µ÷pHÖÁa¡±,¹ýÂËËùµÃµ½µÄ³Áµí³É·ÖΪ                   ¡£
£¨4£©¡°²Ù×÷1¡±Öаüº¬3¸ö»ù±¾ÊµÑé²Ù×÷£¬ËüÃÇÒÀ´ÎÊÇ_________¡¢__________ºÍ¹ýÂË¡£ÖƵõÄCoCl2¡¤6H2OÔÚºæ¸ÉʱÐè¼õѹºæ¸ÉµÄÔ­ÒòÊÇ__________________¡£
£¨5£©ÝÍÈ¡¼Á¶Ô½ðÊôÀë×ÓµÄÝÍÈ¡ÂÊÓëpHµÄ¹ØϵÈçͼ¡£Ïò¡°ÂËÒº¡±ÖмÓÈëÝÍÈ¡¼ÁµÄÄ¿µÄÊÇ_________£»ÆäʹÓõÄ×î¼ÑpH·¶Î§ÊÇ________________¡£

A£®2.0~2.5     B£®3.0~3.5         C£®4.0~4.5     D£®5.0~5.5
£¨6£©Îª²â¶¨´Ö²úÆ·ÖÐCoCl2¡¤6H2Oº¬Á¿£¬³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄ´Ö²úÆ·ÈÜÓÚË®£¬¼ÓÈë×ãÁ¿AgNO3ÈÜÒº£¬¹ýÂË¡¢Ï´µÓ£¬½«³Áµíºæ¸Éºó³ÆÆäÖÊÁ¿¡£Í¨¹ý¼ÆËã·¢ÏÖ´Ö²úÆ·ÖÐCoCl2¡¤6H2OµÄÖÊÁ¿·ÖÊý´óÓÚ100£¥£¬ÆäÔ­Òò¿ÉÄÜÊÇ_____________________¡££¨´ðÒ»Ìõ¼´¿É£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

îÜËáï®(LiCoO2)ï®Àë×Óµç³ØÊÇÒ»ÖÖÓ¦Óù㷺µÄÐÂÐ͵çÔ´£¬ÊµÑéÊÒ³¢ÊÔÀûÓ÷ϾÉîÜËáï®ï®Àë×Óµç³Ø»ØÊÕÂÁ¡¢Ìú¡¢Í­¡¢îÜ¡¢ï®ÔªËØ£¬ÊµÑé¹ý³ÌÈçÏ£º

(1)¼î½þÅݹý³ÌÖУ¬ÂÁ±»ÈܽâµÄÀë×Ó·½³ÌʽΪ__________________________
(2)ÂËÒºAÖмÓÈë²ÝËáï§ÈÜÒº£¬Ê¹CoÔªËØÒÔCoC2O4¡¤2H2O³ÁµíÐÎʽÎö³ö¡£²ÝËáîÜÊÇÖƱ¸Ñõ»¯îܼ°îÜ·ÛµÄÖØÒªÔ­ÁÏ¡£ÔÚ¿ÕÆøÖÐCoC2O4¡¤2H2OµÄÈÈ·Ö½âʧÖØÊý¾Ý¼ûÏÂ±í£¬Çë²¹³äÍêÕû±íÖеÄÈȷֽⷽ³Ìʽ¡£

ÐòºÅ
ζȷ¶Î§/¡æ
Èȷֽⷽ³Ìʽ
¹ÌÌåʧÖØÂÊ
¢Ù
120¡«220
 
19.67%
¢Ú
280¡«310
 
56.10%
 
(3)¹ýÂËLi2CO3ʱ·¢ÏÖÂËÒºÖÐÓÐÉÙÁ¿»ë×Ç£¬´ÓʵÑé²Ù×÷µÄ½Ç¶È¸ø³öÁ½ÖÖ¿ÉÄܵÄÔ­Òò£º_____________________________________________________________
(4)×îÖÕËùµÃµÄFeCl3ÈÜÒº¿É×÷¾»Ë®¼Á£¬ÊÔ½áºÏÀë×Ó·½³Ìʽ½âÊÍÆ侻ˮԭÀí________________________________________________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

º¬±½·ÓµÄ¹¤Òµ·ÏË®´¦ÀíµÄÁ÷³ÌͼÈçͼËùʾ£º

£¨1£©ÉÏÊöÁ÷³ÌÀÉ豸¢ñÖнøÐеÄÊÇ_________²Ù×÷£¨Ìîд²Ù×÷Ãû³Æ£©¡£ÊµÑéÊÒÀïÕâÒ»²½
²Ù×÷¿ÉÒÔÓÃ____________£¨ÌîдÒÇÆ÷Ãû³Æ£©½øÐС£
£¨2£©ÓÉÉ豸¢ò½øÈëÉ豸¢óµÄÎïÖÊAÊÇ_____    ____¡£ÓÉÉ豸¢ó½øÈëÉ豸¢ôµÄÎïÖÊBÊÇ______________¡££¨¾ùÌîдÎïÖʵĻ¯Ñ§Ê½£©
£¨3£©ÔÚÉ豸¢óÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º_______                    ________¡£
£¨4£©ÔÚÉ豸¢ôÖУ¬ÎïÖÊBµÄË®ÈÜÒººÍCaO·´Ó¦ºó£¬²úÎïÊÇNaOH¡¢H2OºÍ________¡£Í¨¹ý________²Ù×÷£¨Ìîд²Ù×÷Ãû³Æ£©£¬¿ÉÒÔʹ²úÎïÏ໥·ÖÀë¡£
£¨5£©Í¼ÖУ¬ÄÜÑ­»·Ê¹ÓõÄÎïÖÊÊÇC6H6¡¢CaO¡¢________¡¢__________¡££¨¾ùÌîдÎïÖʵĻ¯Ñ§Ê½£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÑÇÂÈËáÄÆ£¨NaClO2£©Ö÷ÒªÓÃÓÚÃÞ·Ä¡¢ÔìÖ½ÒµµÄƯ°×¼Á£¬Ò²ÓÃÓÚʳƷÏû¶¾¡¢Ë®´¦ÀíµÈ£¬ÑÇÂÈËáÄÆÊÜÈÈÒ׷ֽ⡣ÒÔÂÈËáÄƵÈΪԭÁÏÖƱ¸ÑÇÂÈËáÄƵŤÒÕÁ÷³ÌÈçÏ£º

£¨1£©Ìá¸ß¡°·´Ó¦l¡±·´Ó¦ËÙÂʵĴëÊ©ÓÐ________________________£¨Ð´³öÒ»Ìõ¼´¿É£©¡£
£¨2£©¡°·´Ó¦2¡±µÄÑõ»¯¼ÁÊÇ_____________£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________¡£
£¨3£©²ÉÈ¡¡°¼õѹÕô·¢¡±¶ø²»Óá°³£Ñ¹Õô·¢¡±£¬Ô­ÒòÊÇ__________________¡£
£¨4£©´Ó¡°Ä¸Òº¡±ÖпɻØÊÕµÄÖ÷ÒªÎïÖÊÊÇ__________________________¡£
£¨5£©¡°ÀäÈ´½á¾§¡±ºó¾­______________£¨Ìî²Ù×÷Ãû³Æ£©¼´¿É»ñµÃ´Ö²úÆ·¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÓÐÒ»¹ÌÌå»ìºÏÎ¿ÉÄÜÓÉNaI¡¢KCl¡¢Na2CO3¡¢Na2SO4¡¢CaCl2¡¢Cu(NO3)2ÖеÄÒ»ÖÖ»ò¼¸ÖÖ×é³É£¬ÎªÁ˼ìÑéËùº¬µÄÎïÖÊ£¬×öÁËÒÔÏÂʵÑ飺
¢ÙÈ¡ÉÙÐí¹ÌÌåÈÜÓÚË®£¬µÃµ½ÎÞɫ͸Ã÷ÈÜÒº£»
¢ÚÍù´ËÈÜÒºÖеμÓÂÈ»¯±µÈÜÒº£¬Óа×É«³ÁµíÉú³É£»
¢Û¹ýÂË£¬Íù³ÁµíÖмÓÈë×ãÁ¿µÄÏ¡ÏõËᣬ·¢ÏÖ³ÁµíûÓÐÈ«²¿ÈܽâÇÒÓÐÎÞÉ«ÎÞζµÄÆøÌåÉú³É¡£
¢ÜÍùÂËÒºÖмÓÈë×ãÁ¿µÄÐÂÖƵÄÂÈË®£¬ÔÙ¼ÓÈëÉÙÐíÆûÓÍ£¬Õñµ´£¬¾²Öã¬ÉϲãÒºÌå³Ê×ϺìÉ«¡£
£¨1£©ÊÔÅжϣº¹ÌÌå»ìºÏÎïÖп϶¨º¬ÓР                                £¬
Ò»¶¨Ã»ÓР                                        £¬
¿ÉÄܺ¬ÓР                                        ¡£
£¨2£©¶Ô¿ÉÄܺ¬ÓеÄÎïÖÊ£¬ÈçºÎ½øÐÐʵÑéÒÔ½øÒ»²½¼ìÑé¡£
                                                                      ¡£
£¨3£©ÊµÑé¢ÜÖз¢ÉúµÄ»¯Ñ§·´Ó¦ÊôÓÚ                  ·´Ó¦£¨Ìî·´Ó¦ÀàÐÍ£©£¬Ö÷ҪʵÑé²Ù×÷Ãû³Æ½Ð               ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

̼Ëá﮹㷺ӦÓÃÓÚÌմɺÍÒ½Ò©µÈÁìÓò,ÒԦ¡ªï®»Ôʯ(Ö÷Òª³É·ÖΪLi2O¡¤Al2O3¡¤4SiO2)ΪԭÁÏÖƱ¸Li2CO3µÄ¹¤ÒÕÁ÷³ÌÈçÏÂ:

ÒÑÖª:Fe3+¡¢Al3+¡¢Fe2+ºÍMg2+ÒÔÇâÑõ»¯ÎïÐÎʽÍêÈ«³Áµíʱ,ÈÜÒºµÄpH·Ö±ðΪ3.2¡¢5.2¡¢9.7ºÍ12.4;Li2SO4¡¢LiOHºÍLi2CO3ÔÚ303 KϵÄÈܽâ¶È·Ö±ðΪ34.2 g¡¢12.7 gºÍ1.3 g¡£
£¨1£©²½Öè¢ñÇ°,¦Â¡ªï®»ÔʯҪ·ÛËé³Éϸ¿ÅÁ£µÄÄ¿µÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
£¨2£©²½Öè¢ñÖÐ,Ëá½þºóµÃµ½µÄËáÐÔÈÜÒºÖк¬ÓÐLi+¡¢SO42-£»Áíº¬ÓÐAl3+¡¢Fe3+¡¢Fe2+¡¢Mg2+¡¢Ca2+¡¢Na+µÈÔÓÖÊ,ÐèÔÚ½Á°èϼÓÈë¡¡¡¡¡¡(Ìʯ»Òʯ¡±¡¢¡°ÂÈ»¯¸Æ¡±»ò¡°Ï¡ÁòËᡱ)ÒÔµ÷½ÚÈÜÒºµÄpHµ½6.0~6.5,³Áµí²¿·ÖÔÓÖÊÀë×Ó,È»ºó·ÖÀëµÃµ½½þ³öÒº¡£
£¨3£©²½Öè¢òÖÐ,½«ÊÊÁ¿µÄH2O2ÈÜÒº¡¢Ê¯»ÒÈéºÍNa2CO3ÈÜÒºÒÀ´Î¼ÓÈë½þ³öÒºÖÐ,¿É³ýÈ¥µÄÔÓÖʽðÊôÀë×ÓÓС¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
£¨4£©²½Öè¢óÖÐ,Éú³É³ÁµíµÄÀë×Ó·½³ÌʽΪ¡¡                                      ¡£
£¨5£©´ÓĸҺÖпɻØÊÕµÄÖ÷ÒªÎïÖÊÊÇ¡¡¡¡¡¡                                  ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¶à¾§¹è£¨¹èµ¥ÖʵÄÒ»ÖÖ£©±»³ÆΪ¡°Î¢µç×Ó´óÏõĻùʯ¡±£¬ÖƱ¸Öи±²úÎïÒÔSiCl4ΪÖ÷£¬Ëü»·¾³ÎÛȾºÜ´ó£¬ÄÜÓöˮǿÁÒË®½â£¬·Å³ö´óÁ¿µÄÈÈ¡£Ñо¿ÈËÔ±ÀûÓÃSiCl4Ë®½âÉú³ÉµÄÑÎËáºÍ±µ¿ó·Û£¨Ö÷Òª³É·ÖΪBaCO3£¬ÇÒº¬ÓÐÌú¡¢Ã¾µÈÀë×Ó£©£¬ÖƱ¸BaCl2 ? 2H2O£¬¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£º ¢Ù ³£ÎÂÏÂFe3+¡¢Mg2+ ÍêÈ«³ÁµíµÄpH·Ö±ðÊÇ3.4¡¢12.4£»
¢Ú BaCO3µÄÏà¶Ô·Ö×ÓÖÊÁ¿ÊÇ197£» BaCl2 ? 2H2OµÄÏà¶Ô·Ö×ÓÖÊÁ¿ÊÇ244£»
£¨1£©SiCl4·¢ÉúË®½â·´Ó¦µÄ»¯Ñ§·½³Ìʽ__________________________________
£¨2£©¸ßÎÂÏ£¬SiCl4 (g) ÓÃH2»¹Ô­¿ÉÖÆÈ¡´¿¶ÈºÜ¸ßµÄ¹è£¬µ±·´Ó¦ÖÐÓÐ1molµç×ÓתÒÆʱÎüÊÕ
59 kJÈÈÁ¿£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ__________________________________
£¨3£©¼Ó±µ¿ó·Ûµ÷½ÚpH=7µÄ×÷ÓÃÊÇ£º
¢ÙʹBaCO3ת»¯ÎªBaCl2     ¢Ú_______________________________
£¨4£©Éú³ÉÂËÔüAµÄÀë×Ó·½³Ìʽ________________________________________
£¨5£©BaCl2ÂËÒº¾­__________¡¢_________¡¢¹ýÂË¡¢Ï´µÓ£¬ÔÙ¾­Õæ¿Õ¸ÉÔïºóµÃµ½BaCl2 ? 2H2O
£¨6£©10¶Öº¬78.8% BaCO3µÄ±µ¿ó·ÛÀíÂÛÉÏ×î¶àÄÜÉú³ÉBaCl2 ? 2H2O___________¶Ö¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

³ýÈ¥À¨ºÅÄÚµÄÔÓÖÊ£¬Ð´³öËù¼ÓÊÔ¼ÁµÄ»¯Ñ§Ê½
£¨1£©Cl£­(SO42£­)£º      £¨2£©SO42£­(CO32£­)£º     £¨3£©Fe2£«(Cu2£«)£º     

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸