¡¾ÌâÄ¿¡¿µ¨·¯£¨CuSO4¡¤5H2O£©Óй㷺µÄÓÃ;¡£Ä³Ñо¿ÐÔѧϰС×éÀûÓÃij´ÎʵÑéºóµÄÏ¡ÁòËᡢϡÏõËá»ìºÏÒºÖƱ¸µ¨·¯¡£ÊµÑéÁ÷³ÌÈçÏ£º
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²Ù×÷XΪ___£¬___¡£
£¨2£©NOÐèÒª»ØÊÕÀûÓã¬Ð´³öNOÓë¿ÕÆø¡¢H2O·´Ó¦Éú³ÉÏõËáµÄ»¯Ñ§·½³Ìʽ___¡£
£¨3£©ÏÖÓÐ48gº¬CuOÖÊÁ¿·ÖÊýΪ20%µÄÍ·Û£¬ÓëÒ»¶¨Á¿µÄÏ¡ÁòËᡢϡÏõËá»ìºÏҺǡºÃÍêÈ«·´Ó¦Éú³ÉCuSO4¡£ÊÔÇó£º
¢ÙÀíÂÛÉÏÉú³Éµ¨·¯µÄÖÊÁ¿Îª___g¡£
¢ÚÔ»ìºÏÒºÖÐÁòËáºÍÏõËáµÄÎïÖʵÄÁ¿Ö®±È¡£___£¨Ð´³ö¼ÆËã¹ý³Ì£©
¡¾´ð°¸¡¿Õô·¢Å¨Ëõ ÀäÈ´½á¾§ 4NO+3O2+2H2O=4HNO3 180g ÍÓëÏõËá·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ3Cu+8H++2NO3-=3Cu2++2NO¡ü+4H2O
ÔòÏûºÄµÄÏõËán£¨HNO3£©=0.4mol£¬
¸ù¾ÝÁòÔªËØÊغãn£¨H2SO4£©=n£¨CuSO4¡¤5H2O£©=0.72mol£¬
Ôòn£¨H2SO4£©£ºn£¨HNO3£©=0.72mol£º0.4mol=9£º5
¡¾½âÎö¡¿
£¨1£©Í·ÛÓëÏ¡ÁòËᡢϡÏõËá·´Ó¦Éú³ÉÁòËáÍÈÜÒº£¬½«ÈÜÒº½øÐÐÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§²Ù×÷£¬µÃµ½¾§Ì壻
£¨2£©NOÓë¿ÕÆø¡¢H2O·´Ó¦Éú³ÉÏõËᣬ½áºÏµÃʧµç×ÓÊغ㡢ÖÊÁ¿ÊغãÅäƽ·½³Ìʽ£»
£¨3£©¼ÆËãCuO¡¢CuµÄÖÊÁ¿£¬½áºÏÍÔªËØÖÊÁ¿Êغã¿É¼ÆËãCuSO4¡¤5H2OµÄÖÊÁ¿£»½áºÏ·´Ó¦µÄÀë×Ó·½³Ìʽ¼ÆËãÔ»ìºÏÒºÖÐÁòËáºÍÏõËáµÄÎïÖʵÄÁ¿Ö®±È¡£
£¨1£©Í·ÛÓëÏ¡ÁòËᡢϡÏõËá·´Ó¦Éú³ÉÁòËáÍÈÜÒº£¬ÈçµÃµ½µ¨·¯£¬Ó¦½«ÈÜÒº½øÐÐÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§²Ù×÷£¬È»ºó¾¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¿ÉµÃµ½¾§Ì壻
£¨2£©NOÓë¿ÕÆø¡¢H2O·´Ó¦Éú³ÉÏõËᣬ·´Ó¦µÄ·½³ÌʽΪ4NO+3O2+2H2O=4HNO3£»
£¨3£©¢Ùn£¨CuO£©==0.12mol£¬ n£¨Cu£©==0.6mol£¬
n£¨CuSO4¡¤5H2O£©=0.12mol+0.6mol=0.72mol£»
m£¨CuSO4¡¤5H2O£©=0.72mol¡Á250g/mol=180g£¬ÀíÂÛÉÏÉú³Éµ¨·¯µÄÖÊÁ¿Îª180g£»
¢ÚÍÓëÏõËá·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ3Cu+8H++2NO3-=3Cu2++2NO¡ü+4H2O
ÔòÏûºÄµÄÏõËán£¨HNO3£©=0.4mol£¬
¸ù¾ÝÁòÔªËØÊغãn£¨H2SO4£©=n£¨CuSO4¡¤5H2O£©=0.72mol£¬
Ôòn£¨H2SO4£©£ºn£¨HNO3£©=0.72mol£º0.4mol=9£º5¡£
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¿¡¿ÏÂÁÐÓйØÑõ»¯»¹Ô·´Ó¦µÄÐðÊöÕýÈ·µÄÊÇ( )
A.Ñõ»¯»¹Ô·´Ó¦ÖÐÓÐÒ»ÖÖÔªËر»Ñõ»¯Ê±£¬Ò»¶¨ÓÐÁíÒ»ÖÖÔªËر»»¹Ô
B.ij½ðÊôÔªËØMÓÉ»¯ºÏ̬±äΪÓÎÀë̬£¬MÒ»¶¨±»»¹Ô
C.ÓÃË«ÏßÇűíʾÏÂÁз´Ó¦µÄµç×ÓתÒÆ
D.·Ç½ðÊôµ¥ÖÊÔÚ·´Ó¦ÖÐÖ»ÄÜ×÷Ñõ»¯¼Á
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ËÄÂÈ»¯Îý¿ÉÓÃ×÷ýȾ¼Á¡£ÀûÓÃÈçͼËùʾװÖÿÉÒÔÖƱ¸ËÄÂÈ»¯Îý£¨²¿·Ö¼Ð³Ö×°ÖÃÒÑÂÔÈ¥£©
ÓйØÐÅÏ¢ÈçÏÂ±í£º
»¯Ñ§Ê½ | SnCl2 | SnCl4 |
ÈÛµã/¡æ | 246 | 33 |
·Ðµã/¡æ | 652 | 144 |
ÆäËûÐÔÖÊ | ÎÞÉ«¾§Ì壬Ò×Ñõ»¯ | ÎÞÉ«ÒºÌ壬Ò×Ë®½â |
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¼××°ÖÃÖÐÒÇÆ÷AµÄÃû³ÆΪ___________¡£
£¨2£©Óü××°ÖÃÖÆÂÈÆø£¬MnO4±»»¹ÔΪMn2+£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ________________¡£
£¨3£©½«×°ÖÃÈçͼÁ¬½ÓºÃ£¬¼ì²éÆøÃÜÐÔ£¬ÂýÂýµÎÈëŨÑÎËᣬ´ý¹Û²ìµ½__________£¨ÌîÏÖÏ󣩺󣬿ªÊ¼¼ÓÈȶ¡×°Öã¬ÎýÈÛ»¯ºóÊʵ±Ôö´óÂÈÆøÁ÷Á¿£¬¼ÌÐø¼ÓÈȶ¡×°Ö㬴Ëʱ¼ÌÐø¼ÓÈȶ¡×°ÖõÄÄ¿µÄÊÇ£º¢Ù´Ù½øÂÈÆøÓëÎý·´Ó¦£»¢Ú_______________________¡£
£¨4£©ÒÒ×°ÖõÄ×÷ÓÃ____________£¬Èç¹ûȱÉÙÒÒ×°Ö㬿ÉÄÜ·¢ÉúµÄ¸±·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________£»¼º×°ÖõÄ×÷ÓÃÊÇ_____£¨ÌîÐòºÅ£©¡£
A£®·ÀÖ¹¿ÕÆøÖÐCO2ÆøÌå½øÈëÎì×°ÖÃ
B£®³ýȥδ·´Ó¦µÄÂÈÆø£¬·ÀÖ¹ÎÛȾ¿ÕÆø
C£®·ÀֹˮÕôÆø½øÈëÎì×°ÖõÄÊÔ¹ÜÖÐʹ²úÎïË®½â
D£®·ÀÖ¹¿ÕÆøÖÐO2½øÈëÎì×°ÖõÄÊÔ¹ÜÖÐʹ²úÎïÑõ»¯
£¨5£©Ä³Í¬Ñ§ÈÏΪ¶¡×°ÖÃÖеķ´Ó¦¿ÉÄܲúÉúSnCl2ÔÓÖÊ£¬ÒÔÏÂÊÔ¼ÁÖв»¿ÉÓÃÓÚ¼ì²âÊÇ·ñ²úÉúSnCl2 µÄÓÐ_______£¨ÌîÐòºÅ£©¡£
A£®H2O2ÈÜÒº B£®ËáÐÔ¸ßÃÌËá¼ØÈÜÒº C£®AgNO3ÈÜÒº D£®äåË®
£¨6£©·´Ó¦ÖÐÓÃÈ¥ÎýÁ£1.19 g£¬·´Ó¦ºóÔÚÎì×°ÖõÄÊÔ¹ÜÖÐÊÕ¼¯µ½2.04 g SnCl4£¬ÔòSnCl4µÄ²úÂÊΪ_______£¨±£Áô2λÓÐЧÊý×Ö£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÀë×Ó·½³Ìʽ»òµçÀë·½³ÌʽÕýÈ·µÄÊÇ
A. NaHSO3ÈÜÒº³ÊËáÐÔ£º
B. ÏòNa2SiO3ÈÜÒºÖÐͨÈëÉÙÁ¿CO2£º
C. ¹¤ÒµÖÆƯ°×·ÛµÄ·´Ó¦£º
D. ÔÚNa2S2O3ÈÜÒºÖеμÓÏ¡ÏõË᣺
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A. 22.4LCl2ÈÜÓÚ×ãÁ¿Ë®£¬ËùµÃÈÜÒºÖÐCl2¡¢Cl-¡¢HClOºÍClO-ËÄÖÖ΢Á£×ÜÊýΪNA
B. ±ê×¼×´¿öÏ£¬38g3H2O2Öк¬ÓÐ3NA¹²¼Û¼ü
C. ³£ÎÂÏ£¬½«5.6gÌú¿éͶÈë×ãÁ¿Å¨ÏõËáÖУ¬×ªÒÆ0.3NAµç×Ó
D. 0.1molL-1MgCl2ÈÜÒºÖк¬ÓеÄMg2+ÊýÄ¿Ò»¶¨Ð¡ÓÚ0.1NA
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÔÚ·´Ó¦ 3Cl2£«6KOHKClO3£«5KCl£«3H2O ÖУ¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£º( )
¢ÙCl2ÊÇÑõ»¯¼Á£¬KOHÊÇ»¹Ô¼Á ¢ÚKClÊÇ»¹Ô²úÎKClO3ÊÇÑõ»¯²úÎï ¢Û·´Ó¦ÖÐÿÏûºÄ3 mol Cl2·Ö×ÓÖÐÓÐ5mol µç×Ó·¢ÉúתÒÆ ¢Ü±»Ñõ»¯Óë±»»¹ÔµÄÂÈÔ×ÓÎïÖʵÄÁ¿Ö®±ÈΪ5¡Ã1
A.¢Ù¢ÜB.¢Ú¢ÛC.¢Ú¢Û¢ÜD.¢Ú¢Ü
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ClO2ÊÇÒ»ÖÖÓÅÁ¼µÄÏû¶¾¼Á£¬³£½«ÆäÖƳÉNaClO2¹ÌÌ壬ÒÔ±ãÔËÊäºÍÖü´æ£¬¹ýÑõ»¯Çâ·¨±¸NaClO2¹ÌÌåµÄʵÑé×°ÖÃÈçͼËùʾ¡£
ÒÑÖª£º¢Ù2NaC1O3+H2O2+H2SO4=2C1O2¡ü+O2¡ü+Na2SO4+2H2O
2ClO2+H2O2+2NaOH=2NaClO2+O2¡ü+2H2O
¢ÚClO2ÈÛµã-59¡æ¡¢·Ðµã11¡æ£¬Å¨¶È¹ý¸ßʱÒ×·¢Éú·Ö½â£»
¢ÛH2O2·Ðµã150¡æ
£¨1£©ÒÇÆ÷CµÄÃû³ÆÊÇ__________________£¬ÒÇÆ÷AµÄ×÷ÓÃÊÇ_________________£¬±ùˮԡÀäÈ´µÄÄ¿µÄÊÇ_____________________ºÍ___________________________¡£
£¨2£©¸Ã×°Öò»ÍêÉƵķ½ÃæÊÇ________________________¡£
£¨3£©¿ÕÆøÁ÷ËÙ¹ý¿ì»ò¹ýÂý£¬¾ù½µµÍNaClO2²úÂÊ£¬ÊÔ½âÊÍÆäÔÒò£¬¿ÕÆøÁ÷ËÙ¹ýÂýʱ£¬___________£»¿ÕÆøÁ÷ËÙ¹ý¿ìʱ£¬________________¡£
£¨4£©Cl-´æÔÚʱ»á´ß»¯ClO2µÄÉú³É¡£·´Ó¦¿ªÊ¼Ê±ÔÚCÖмÓÈëÉÙÁ¿ÑÎËᣬClO2µÄÉú³ÉËÙÂÊ´ó´óÌá¸ß£¬²¢²úÉú΢Á¿ÂÈÆø¡£¸Ã¹ý³Ì¿ÉÄܾÁ½²½Íê³É£¬Ç뽫Æä²¹³äÍêÕû£º
¢Ù_____________________________£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©
¢ÚH2O2+Cl2=2Cl-+O2+2H+
£¨5£©NaClO2´¿¶È²â¶¨£º
¢Ù׼ȷ³ÆÈ¡ËùµÃNaClO2ÑùÆ·10.0gÓÚÉÕ±ÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®ºÍ¹ýÁ¿µÄµâ»¯¼Ø¾§Ì壬ÔÙµÎÈëÊÊÁ¿µÄÏ¡ÁòËᣬ³ä·Ö·´Ó¦£¨C1O2-µÄ²úÎïΪCl-£©£¬½«ËùµÃ»ìºÏÒºÅä³É250mL´ý²âÈÜÒº£»
¢ÚÈ¡25.00mL´ý²âÒº£¬ÓÃ2.0mol¡¤L-1Na2S2O3±ê×¼ÒºµÎ¶¨(I2+2S2O32-=2I-+S4O62-)£¬ÒÔµí·ÛÈÜÒº×öָʾ¼Á£¬´ïµ½µÎ¶¨ÖÕµãʱµÄÏÖÏóΪ__________________________£¬Öظ´µÎ¶¨3´Î£¬²âµÃNa2S2O3±ê׼Һƽ¾ùÓÃÁ¿Îª20.00mL£¬Ôò¸ÃÑùÆ·ÖÐNaClO2µÄÖÊÁ¿·ÖÊýΪ_________________¡££¨M(NaClO2)=90.5g/mol£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿²ÝËáÊÇÒ»ÖÖ¶þÔªÈõËᣬ¿ÉÓÃ×÷»¹Ô¼Á¡¢³Áµí¼ÁµÈ¡£Ä³Ð£¿ÎÍâС×éµÄͬѧÉè¼ÆÀûÓÃC2H2ÆøÌåÖÆÈ¡H2C2O4¡¤2H2O¡£»Ø´ðÏÂÁÐÎÊÌ⣺
(1)¼××éµÄͬѧÒÔµçʯ(Ö÷Òª³É·ÖCaC2£¬ÉÙÁ¿CaS¼°Ca3P2ÔÓÖʵÈ)ΪÔÁÏ£¬²¢ÓÃÏÂͼ1×°ÖÃÖÆÈ¡C2H2¡£
¢ÙµçʯÓëË®·´Ó¦ºÜ¿ì£¬ÎªÁ˼õ»º·´Ó¦ËÙÂÊ£¬×°ÖÃAÖгýÓñ¥ºÍʳÑÎË®´úÌæˮ֮Í⣬»¹¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊÇ__________£¨Ð´Ò»ÖÖ¼´¿É£©¡£
¢Ú×°ÖÃBÖУ¬NaClO½«H2S¡¢PH3 Ñõ»¯ÎªÁòËá¼°Á×Ëᣬ±¾Éí±»»¹ÔΪNaCl£¬ÆäÖÐPH3±»Ñõ»¯µÄÀë×Ó·½³ÌʽΪ______¡£¸Ã¹ý³ÌÖУ¬¿ÉÄܲúÉúеÄÔÓÖÊÆøÌåCl2£¬ÆäÔÒòÊÇ£º _____________£¨ÓÃÀë×Ó·½³Ìʽ»Ø´ð£©¡£
(2)ÒÒ×éµÄͬѧ¸ù¾ÝÎÄÏ××ÊÁÏ£¬ÓÃHg(NO3)2×÷´ß»¯¼Á£¬Å¨ÏõËáÑõ»¯C2H2ÖÆÈ¡H2C2O4¡¤2H2O¡£ÖƱ¸×°ÖÃÈçÉÏͼ2Ëùʾ£º
¢Ù×°ÖÃDÖжà¿×ÇòÅݵÄ×÷ÓÃÊÇ______________________¡£
¢Ú×°ÖÃDÖÐÉú³ÉH2C2O4µÄ»¯Ñ§·½³ÌʽΪ____________________________¡£
¢Û´Ó×°ÖÃDÖеõ½²úÆ·£¬»¹Ðè¾¹ý_____________£¨Ìî²Ù×÷Ãû³Æ£©¡¢¹ýÂË¡¢Ï´µÓ¼°¸ÉÔï¡£
(3)±û×éÉè¼ÆÁ˲ⶨÒÒ×é²úÆ·ÖÐH2C2O4¡¤2H2OµÄÖÊÁ¿·ÖÊýʵÑé¡£ËûÃǵÄʵÑé²½ÖèÈçÏ£º×¼È·³ÆÈ¡m g²úÆ·ÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈëÊÊÁ¿µÄÕôÁóË®Èܽ⣬ÔÙ¼ÓÈëÉÙÁ¿Ï¡ÁòËᣬȻºóÓÃc mol¡¤L£1ËáÐÔKMnO4±ê×¼ÈÜÒº½øÐеζ¨ÖÁÖյ㣬¹²ÏûºÄ±ê×¼ÈÜÒºV mL¡£
¢ÙµÎ¶¨ÖÕµãµÄÏÖÏóÊÇ______________________¡£
¢ÚµÎ¶¨¹ý³ÌÖз¢ÏÖÍÊÉ«ËÙÂÊ¿ªÊ¼ºÜÂýºóÖ𽥼ӿ죬·ÖÎö¿ÉÄܵÄÔÒòÊÇ_______________¡£
¢Û²úÆ·ÖÐH2C2O4¡¤2H2OµÄÖÊÁ¿·ÖÊýΪ_______________£¨Áгöº¬ m¡¢c¡¢V µÄ±í´ïʽ£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÁòµÄº¬Á¿Ó°Ïì¸ÖÌúÐÔÄÜ¡£Ò»ÖֲⶨÁòº¬Á¿µÄ·½·¨Êǽ«¸ÖÑùÖÐÁòת»¯Îª¶þÑõ»¯ÁòÆøÌ壬ÔÙÓòâÁò×°ÖýøÐвⶨ¡£Ä³²â¶¨Áòº¬Á¿µÄÁ÷³ÌÈçÏ£º
£¨1£©ÆøÌåaµÄÖ÷Òª³É·ÖÓÐCO2¡¢______¡¢______¡£
£¨2£©Èô¸ÖÑùÖÐÁòÒÔFeSµÄÐÎʽ´æÔÚ£¬ìÑÉÕ×°ÖÃÖз¢ÉúµÄ»¯Ñ§·´Ó¦Îª3FeS£«5O2 1______ £«3______¡£___________
£¨3£©ÎüÊÕ×°ÖÃÖУ¬H2O2Ñõ»¯SO2µÄ»¯Ñ§·½³ÌʽÊÇ_________________¡£
£¨4£©ÓÃNaOHÈÜÒºÖкÍÉú³ÉµÄÈÜÒºb£¬ÏûºÄz mLNaOHÈÜÒº£¬ÈôÏûºÄ1 mLNaOHÈÜÒºÏ൱ÓÚÁòµÄÖÊÁ¿Îªy g£¬Ôò¸Ã¸ÖÑùÖÐÁòµÄÖÊÁ¿·ÖÊýΪ______¡£
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