¡¾ÌâÄ¿¡¿Îª³ýÈ¥´ÖÑÎÖеÄCa2+. Mg2+. SO42-ÒÔ¼°ÄàɳµÈÔÓÖÊ£¬Ä³Í¬Ñ§Éè¼ÆÁËÒ»ÖÖÖƱ¸¾«ÑεÄʵÑé·½°¸£¬²½ÖèÈçÏ£¨ÓÃÓÚ³ÁµíµÄÊÔ¼ÁÉÔ¹ýÁ¿£©
£¨1£©µÚ¢Ü²½ÖУ¬Ð´³öÏàÓ¦µÄÀë×Ó·½³Ìʽ£¨Éè´ÖÑÎÈÜÒºÖÐCa2+µÄÖ÷Òª´æÔÚÐÎʽΪCaCl2£©_______________ £»____________¡£
£¨2£©ÊµÑé·½°¸µÄ£¨1£©ÖÐӦʹÓóýÔÓÊÔ¼ÁµÄ»¯Ñ§Ê½__________£¬·¢ÉúµÄÀë×Ó·½³ÌʽÊÇ__________£¬ÔÚʵÑé·½°¸µÄ£¨2£©ÖеIJÙ×÷Ãû³ÆÊÇ_______¡£
£¨3£©´ÓʵÑéÉè¼Æ·½°¸ÓÅ»¯µÄ½Ç¶È·ÖÎö²½Öè¢ÚºÍ¢Ü¿É·ñµßµ¹____________£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£¬Èç¹û¡°·ñ¡±£¬Çë˵Ã÷ÀíÓÉ¡£___________________________________________£»
£¨4£©ÅжÏBaCl2ÒѹýÁ¿µÄ·½·¨ÊÇ_________________________________________________¡£
¡¾´ð°¸¡¿Ca2£« + CO32¡ª = CaCO3¡ý Ba2£«+ CO32£ = BaCO3¡ý NaOH 2OH-+Mg2+=Mg(OH)2¡ý ¹ýÂË ·ñ ¹ýÁ¿µÄBaCl2±ØÐëÒªÓÃNa2CO3³ýÈ¥ ¾²ÖÃÈ¡ÉϲãÇåÒº¼ÌÐøµÎ¼ÓBaCl2ÈÜÒº£¬ÈôÎÞ³ÁµíÉú³É£¬ÔòBaCl2¹ýÁ¿£¨ÆäËû´ð°¸ºÏÀí¾ù¿É£©
¡¾½âÎö¡¿
£¨1£©Ca2+ºÍCO32-·´Ó¦Éú³ÉCaCO3³Áµí£¬Ba2ºÍCO32-·´Ó¦Éú³ÉBaCO3³Áµí£»
£¨2£©ÇâÑõ»¯ÄÆ¿ÉÒÔ½«ÈÜÒºÖеÄþÀë×Ó³ýµô£»
£¨3£©Ì¼ËáÄƱØÐë·ÅÔÚÂÈ»¯±µµÄºóÃæ¼ÓÈë,²»Äܵߵ¹£¬ÇâÑõ»¯ÄƺÍÂÈ»¯±µµÄ¼ÓÈëÎÞ˳ÐòÒªÇó£»
£¨4£©ÂÈ»¯±µ¹ýÁ¿Ê±,ÈÜÒºÖв»»áº¬ÓÐÁòËá¸ùÀë×Ó,¿ÉÒÔ¼ìÑéÊÇ·ñº¬ÓÐÁòËá¸ùÀë×ÓÀ´È·¶¨ÂÈ»¯±µÊÇ·ñ¹ýÁ¿£»
£¨1£©Ì¼ËáÄƵÄ×÷ÓÃÊǽ«ÈÜÒºÖеĸÆÀë×Ӻ͹ýÁ¿µÄ±µÀë×Ó³ÁµíÏÂÀ´,·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ·Ö±ðΪ£ºCa2£« + CO32¡ª = CaCO3¡ý£¬Ba2£«+ CO32£ = BaCO3¡ý£»
±¾Ìâ´ð°¸Îª£ºCa2£« + CO32¡ª = CaCO3¡ý£»Ba2£«+ CO32£ = BaCO3¡ý¡£
£¨2£©ÇâÑõ»¯ÄÆ¿ÉÓëÈÜÒºÖеÄþÀë×ÓÐγÉMg(OH)2³Áµí£¬È»ºó¹ýÂËÒÔ³ýÈ¥Mg2+£»Àë×Ó·½³ÌʽΪ£º2OH-+Mg2+=Mg(OH)2¡ý£»
±¾Ìâ´ð°¸Îª£ºNaOH£»2OH-+Mg2+=Mg(OH)2¡ý£»¹ýÂË¡£
£¨3£©Ì¼ËáÄƱØÐë·ÅÔÚÂÈ»¯±µµÄºóÃæ¼ÓÈë,ÕâÑù̼ËáÄƼȿÉÒÔ½«ÔÓÖÊÀë×Ó¸ÆÀë×Ó³ýÈ¥,ÓÖ¿ÉÒÔ½«¹ýÁ¿µÄ±µÀë×Ó³ýÈ¥,·ñÔòBa2+ÎÞ·¨³ýµô£¬Ó°Ï쾫ÑεĴ¿¶È£¬¹Ê²»Äܵߵ¹£¬±¾Ìâ´ð°¸Îª£º·ñ£»¹ýÁ¿µÄBaCl2±ØÐëÒªÓÃNa2CO3³ýÈ¥£»
£¨4£©ÂÈ»¯±µ¹ýÁ¿Ê±£¬ÈÜÒºÖв»»áº¬ÓÐÁòËá¸ùÀë×Ó£¬¿ÉÒÔ¼ìÑéÇåÒºÖÐÊÇ·ñº¬ÓÐÁòËá¸ùÀë×ÓÀ´È·¶¨ÂÈ»¯±µÊÇ·ñ¹ýÁ¿£¬¾ßÌå×ö·¨ÊÇ:È¡¾²ÖúóµÄÉϲãÇåÒºµÎÓÚµãµÎ°åÉÏ(»òÈ¡ÉÙÁ¿ÉϲãÇåÒºÓÚÊÔ¹ÜÖÐ)£¬ÔÙµÎÈë1~2µÎBaCl2ÈÜÒº,ÈôÈÜҺδ±ä»ë×Ç,Ôò±íÃ÷BaCl2ÒѹýÁ¿£¬±¾ÌâÕýÈ·´ð°¸ÊÇ: ¾²ÖÃÈ¡ÉϲãÇåÒº¼ÌÐøµÎ¼ÓBaCl2ÈÜÒº£¬ÈôÎÞ³ÁµíÉú³É£¬ÔòBaCl2¹ýÁ¿¡£
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£º
¡¾ÌâÄ¿¡¿ÏÖÓÃÂÈË®À´ÖÆÈ¡º¬ÓдÎÂÈËáµÄÈÜÒº£¬¼ÈÒªÌá¸ßÈÜÒºÖÐHClOÎïÖʵÄÁ¿Å¨¶È£¬ÓÖÒª½µµÍÈÜÒºÖÐHClŨ¶È£¬ÏÂÁдëÊ©¿ÉÒÔ²ÉÓõÄÊÇ(¡¡¡¡)
A. ¼ÓÈȻӷ¢HCl B. ¼ÓCaSO3
C. ¼ÓNaOHÖкÍHCl D. ¼ÓCaCO3ÖкÍHCl
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ( )
A. ÔÚ³£Î¡¢³£Ñ¹Ï£¬11.2 L N2º¬ÓеķÖ×ÓÊýΪ0.5NA
B. ±ê×¼×´¿öÏ£¬22.4 L H2ºÍO2µÄ»ìºÏÆøÌåËùº¬·Ö×ÓÊýΪNA
C. 18 g H2OµÄÎïÖʵÄÁ¿ÊÇ1mol
D. ±ê¿öÏ£¬1 mol SO2µÄÌå»ýÊÇ22.4 L
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ò©ÎïF¾ßÓп¹Ö×Áö¡¢½µÑªÌÇ¡¢½µÑªÑ¹µÈ¶àÖÖÉúÎï»îÐÔ£¬ÆäºÏ³É·ÏßÈçÏ£º
ÒÑÖª£ºMµÄ½á¹¹¼òʽΪ£º¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)AµÄ»¯Ñ§Ãû³ÆÊÇ______________________¡£
(2)CÖйÙÄÜÍŵÄÃû³ÆÊÇ_________________________¡£
(3)д³öFµÄ½á¹¹¼òʽ____________________________¡£
(4)ÒÑÖªAÔÚÒ»¶¨Ìõ¼þÏÂÄÜÉú³É¿É½µ½âµÄ¾Ûõ¥£¬Çëд³ö¸Ã·´Ó¦»¯Ñ§·½³Ìʽ£º_________________¡£
(5)Âú×ãÏÂÁÐÌõ¼þµÄMµÄͬ·ÖÒì¹¹ÌåÓÐ_____ÖÖ(²»º¬Á¢ÌåÒì¹¹)¡£
¢ÙÄܹ»·¢ÉúÒø¾µ·´Ó¦¡£
¢Úº¬ÓÐÏõ»ù(-NO2)£¬ÇÒÏõ»ùÖ±½ÓÁ¬ÔÚ±½»·ÉÏ¡£
¢Ûº¬Óб½»·ÇÒ±½»·ÉÏÖ»ÓÐÁ½¸öÈ¡´ú»ù¡£
ÆäÖк˴Ź²ÕñÇâÆ×ΪËÄ×é·åÇÒ·åÃæ»ýÖ®±ÈΪ6£º2£º2£º1µÄ½á¹¹¼òʽΪ______________(д³öÒ»ÖÖ¼´¿É)¡£
(6)д³öÓÃÒÒȩΪÔÁÏÖƱ¸¸ß·Ö×Ó»¯ºÏÎï¾Û±ûÏ©ëæµÄºÏ³É·Ïß(ÎÞ»úÊÔ¼ÁÈÎÑ¡)£º_____________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÉèijԪËصÄÔ×ÓºËÄÚµÄÖÊ×ÓÊýΪm£¬ÖÐ×ÓÊýΪn£¬ÔòÏÂÊöÂÛ¶ÏÕýÈ·µÄÊÇ£¨ £©
A. ²»ÄÜÓÉ´ËÈ·¶¨¸ÃÔªËصÄÏà¶ÔÔ×ÓÖÊÁ¿
B. ÕâÖÖÔªËصÄÏà¶ÔÔ×ÓÖÊÁ¿Îª(m+n)g
C. ̼Ô×ÓÖÊÁ¿Îªwg£¬´ËÔ×ÓµÄÖÊÁ¿Îª(m+n)wg
D. ºËÄÚÖÐ×ÓµÄ×ÜÖÊÁ¿Ð¡ÓÚÖÊ×ÓµÄ×ÜÖÊÁ¿
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¹¤ÒµÉÏÀûÓ÷ÏÄø´ß»¯¼Á(Ö÷Òª³É·ÖΪNi£¬»¹º¬ÓÐÒ»¶¨Á¿µÄZn¡¢Fe¡¢SiO2¡¢CaOµÈ)ÖƱ¸²ÝËáÄø¾§ÌåµÄÁ÷³ÌÈçÏ£º
(1)Çëд³öÒ»ÖÖÄÜÌá¸ß¡°Ëá½þ¡±ËÙÂʵĴëÊ©£º________________________£»ÂËÔüIµÄ³É·ÖÊÇ____________(Ìѧʽ)¡£
(2)³ýÌúʱ£¬¿ØÖƲ»Í¬µÄÌõ¼þ¿ÉÒԵõ½²»Í¬µÄÂËÔüII¡£ÒÑÖªÂËÔüIIµÄ³É·ÖÓëζȡ¢pHµÄ¹ØϵÈçͼËùʾ£º
¢ÙÈô¿ØÖÆζÈ40¡æ¡¢pH=8£¬ÔòÂËÔüIIµÄÖ÷Òª³É·ÖΪ_________________________(Ìѧʽ)¡£
¢ÚÈô¿ØÖÆζÈ80¡æ¡¢pH=2£¬¿ÉµÃµ½»ÆÌú·¯ÄÆ[Na2Fe6(SO4)4(OH)12](ͼÖÐÒõÓ°²¿·Ö)£¬Ð´³öÉú³É»ÆÌú·¯ÄƵÄÀë×Ó·½³Ìʽ£º___________________________________________¡£
(3)ÒÑÖª³ýÌúºóËùµÃ100 mLÈÜÒºÖÐc(Ca2+)=0.01mol¡¤L-1£¬¼ÓÈë100 mL NH4FÈÜÒº£¬Ê¹Ca2+Ç¡ºÃ³ÁµíÍêÈ«¼´ÈÜÒºÖÐc(Ca2+)=1¡Á10-5 mol¡¤L-1£¬ÔòËù¼Óc(NH4F)=_________mol¡¤L-1¡£[ÒÑÖªKsp(CaF2)=5.29¡Á10-9]
(4)¼ÓÈëÓлúÝÍÈ¡¼ÁµÄ×÷ÓÃÊÇ________________________¡£
(5)ij»¯Ñ§¶ÆÄøÊÔ¼ÁµÄ»¯Ñ§Ê½ÎªMxNi(SO4)y(MΪ+1¼ÛÑôÀë×Ó£¬NiΪ+2¼Û£¬x¡¢y¾ùΪÕýÕûÊý)¡£Îª²â¶¨¸Ã¶ÆÄøÊÔ¼ÁµÄ×é³É£¬½øÐÐÈçÏÂʵÑ飺
I£®³ÆÁ¿28.7g¶ÆÄøÊÔ¼Á£¬ÅäÖÆ100 mLÈÜÒºA£»
¢ò£®×¼È·Á¿È¡10.00 mLÈÜÒºA£¬ÓÃ0.40 mol¡¤L-1µÄEDTA(Na2H2Y)±ê×¼ÈÜÒºµÎ¶¨ÆäÖеÄNi2+(Àë×Ó·½³ÌʽΪNi2++H2Y2-=NiY2-+2H+)£¬ÏûºÄEDTA±ê×¼ÈÜÒº25.00mL£»
¢ó£®ÁíÈ¡10.00 mLÈÜÒºA£¬¼ÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬µÃµ½°×É«³Áµí4.66g¡£
¢ÙÅäÖÆ100 mL¶ÆÄøÊÔ¼Áʱ£¬ÐèÒªµÄÒÇÆ÷³ýÒ©³×¡¢ÍÐÅÌÌìƽ¡¢²£Á§°ô¡¢ÉÕ±¡¢Á¿Í²¡¢½ºÍ·µÎ¹ÜÍ⣬»¹ÐèÒª________________________¡£
¢Ú¸Ã¶ÆÄøÊÔ¼ÁµÄ»¯Ñ§Ê½Îª________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÔC2H2ΪÔÁϺϳɻ·¶¡Ï©À໯ºÏÎïLµÄ·ÏßÈçÏ£º
(1)AÓÉ̼¡¢Çâ¡¢ÑõÈýÖÖÔªËØ×é³É£¬Ïà¶Ô·Ö×ÓÖÊÁ¿ÊÇ30£¬ÔòAµÄÃû³ÆΪ____________¡£
(2)B·Ö×ÓÖеÄ̼Ô×Ó¶¼ÔÚͬһֱÏßÉÏ£¬BµÄ½á¹¹¼òʽΪ________________¡£
(3)C¡úD¡¢J¡úKµÄ·´Ó¦ÀàÐÍ·Ö±ðΪ________________¡¢________________¡£
(4)Éè¼ÆC¡úD¡¢E¡úFÁ½²½·´Ó¦µÄÄ¿µÄÊÇ________________________¡£
(5)GºÍHÉú³ÉIµÄ»¯Ñ§·½³ÌʽΪ______________________¡£
(6)»¯ºÏÎïXÊÇIµÄͬ·ÖÒì¹¹Ì壬¿ÉÓëFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£¬ÇÒÄÜÓëNaHCO3ÈÜÒº·´Ó¦·Å³öCO2£¬X¹²ÓÐ__ÖÖ(²»¿¼ÂÇÁ¢ÌåÒì¹¹)£»ÆäÖк˴Ź²ÕñÇâÆ×ΪÎå×é·å£¬·åÃæ»ý±ÈΪ2£º2£º2£º1£º1µÄ½á¹¹¼òʽΪ____________________¡£
(7)д³öÓúÍÒÒȲΪÔÁÏÖƱ¸µÄºÏ³É·Ïß(ÆäËûÊÔ¼ÁÈÎÑ¡)_____¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÎïÖÊ·ÖÀàÖУ¬Ç°Õß°üÀ¨ºóÕßµÄÊÇ£¨¡¡¡¡£©
A.µç½âÖÊ Ê¯Ä«
B.½ºÌå ·Öɢϵ
C.»ìºÏÎï Ư°×·Û
D.¼îÐÔÑõ»¯Îï CO2
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ï´µÓº¬SO2µÄÑÌÆø£¬ÒÔÏÂÎïÖÊ¿É×÷ΪϴµÓ¼ÁµÄÊÇ
A. Ca(OH)2B. CaCl2C. NaHSO3D. HCl
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