ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿¸´ºÏÑõ»¯×ê¿ÉÓÃÓÚÉú²úºÏ½ð´ÅÐÔ²ÄÁÏ¡¢´ß»¯¼ÁµÈ¡£Îª³ä·ÖÀûÓÃ×ÊÔ´£¬ÔÚʵÑéÊÒÖÐ̽¾¿²ÉÓú¬îÜ·ÏÔü£¨º¬
µÈ£©À´ÖƱ¸¸´ºÏÑõ»¯îÜ£¬¾ßÌåÁ÷³ÌÈçÏ£º
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©µÄµç×ÓʽΪ_______________________¡£
£¨2£©¡°¼îÖóˮϴ¡±µÄÄ¿µÄΪ______________________________________________¡£
£¨3£©¡°Ëá½þ¡±Ê±£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________________________________£»ÂËÔü1ÖÐÖ÷Òª³É·ÖµÄ»¯Ñ§Ê½Îª_______________________¡£
£¨4£©¼ìÑ顰ˮϴҺ¡±Öк¬ÓеIJÙ×÷ºÍÏÖÏóΪ______________________________________________¡£ÉÏÊöÁ÷³ÌÖжà´ÎÓ¦ÓùýÂ˲Ù×÷¸Ã²Ù×÷ËùÓÃÖ÷Òª²£Á§ÒÇÆ÷ÓÐ_____________________¡£
£¨5£©ÏÂÁÐ×°ÖÃÖУ¬ÊʺϽøÐС°±ºÉÕ¡±²Ù×÷µÄÊÇ_____________________£¨ÌîÑ¡Ïî×Öĸ£©¡£
A. B.
C.
D.
£¨6£©³ÆÁ¿£¬³ä·Ö¡°±ºÉÕ¡±ºóµÃ¸´ºÏÑõ»¯îÜ
£¬Ôò¸´ºÏÑõ»¯×êµÄ»¯Ñ§Ê½Îª________________£»¡°±ºÉÕ¡±¹ý³ÌÖÐÉú³ÉµÄÆøÌåÄܲÎÓë´óÆøÑ»·£¬Ôò¸Ã¹ý³Ì_______________£¨Ìî¡°ÓС±»ò¡°ÎÞ¡±£©ÑõÆø²Î¼Ó·´Ó¦¡£
¡¾´ð°¸¡¿ ³ýÈ¥º¬îÜ·ÏÔü±íÃæµÄÓÍÎÛ Co2O3+4H++
=2Co2++
+2H2O CaSO4 Óýྻ²¬Ë¿ÕºÈ¡´ý¼ìÒºÔھƾ«µÆ»ðÑæÉÏ×ÆÉÕ£¬Í¸¹ýÀ¶É«îܲ£Á§¹Û²ì»ðÑæ³Ê×ÏÉ« ÉÕ±¡¢Â©¶·¡¢²£Á§°ô C Co2O7 ÓÐ
¡¾½âÎö¡¿
¸ù¾ÝÁ÷³ÌʾÒâͼ·ÖÎö¿ÉÖª£¬º¬îÜ·ÏÔüÓÃÈȵĴ¿¼îÈÜҺϴȥ±íÃæµÄÓÍÎÛ£¬¼ÓË®K2OÈܽ⣬ͬˮϴҺÁ÷³ö£¬Ï´µÓºó¼ÓÈëÁòËáÈÜÒººÍNa2SO3½øÐÐËá½þ£¬Ëá½þʱ·¢Éú·´Ó¦Co2O3+4H++=2Co2++
+2H2O£¬Í¬Ê±CaOºÍH2SO4·´Ó¦Éú³ÉCaSO4£¬µÃµ½º¬ÓÐCaSO4µÄÂËÔü1ºÍÖ÷Òªº¬CoSO4µÄÂËÒº1£¬ÔÙ¼ÓÈëNH4HCO3·¢Éú·´Ó¦CoSO4+NH4HCO3=CoCO3+NH4HSO4µÃµ½CoCO3µÄ³Áµí£¬³Áµí¾±ºÉÕºóµÃµ½¸´ºÏÑõ»¯îÜ(CoxOy)£¬¾Ý´Ë·ÖÎö½â´ð¡£
(1)K2OÊÇÀë×Ó»¯ºÏÎK+ºÍO2-ÐγÉÀë×Ó¼ü£¬Æäµç×ÓʽΪ£¬¹Ê´ð°¸Îª£º
£»
(2)ÈȵĴ¿¼îÈÜÒº¾ßÓнÏÇ¿µÄ¼îÐÔ£¬¿ÉʹÉúÎïÓÍÎÛË®½âΪ¿ÉÈÜÓÚË®µÄÎïÖÊ£¬±ãÓÚˮϴ³ýÈ¥£¬¹Ê´ð°¸Îª£º³ýÈ¥º¬îÜ·ÏÔü±íÃæµÄÓÍÎÛ£»
(3)¡°Ëá½þ¡±Ê±£¬Co2O3ÔÚËáÐÔÌõ¼þϱ»»¹Ô£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪCo2O3+4H++=2Co2++
+2H2O£¬Í¬Ê±£¬CaOºÍH2SO4·´Ó¦Éú³ÉÈܽâ¶ÈСµÄCaSO4£¬¹Ê´ð°¸Îª£ºCo2O3+4H++
=2Co2++
+2H2O£»CaSO4£»
(4)¼ìÑéK+¿ÉÓÃÑæÉ«·´Ó¦£¬²Ù×÷ºÍÏÖÏóΪÓýྻ²¬Ë¿ÕºÈ¡´ý¼ìÒºÔھƾ«µÆ»ðÑæÉÏ×ÆÉÕ£¬Í¸¹ýÀ¶É«îܲ£Á§¹Û²ì»ðÑæ³Ê×ÏÉ«£¬¹ýÂËËùÓõÄÖ÷Òª²£Á§ÒÇÆ÷ÓÐÉÕ±¡¢Â©¶·ºÍ²£Á§°ô£¬¹Ê´ð°¸Îª£ºÓýྻ²¬Ë¿ÕºÈ¡´ý¼ìÒºÔھƾ«µÆ»ðÑæÉÏ×ÆÉÕ£¬Í¸¹ýÀ¶É«îܲ£Á§¹Û²ì»ðÑæ³Ê×ÏÉ«£»ÉÕ±¡¢Â©¶·¡¢²£Á§°ô£»
(5)¡°±ºÉÕ¡±²Ù×÷Ó¦ÔÚÛáÛöÖнøÐУ¬CÑ¡ÏîÕýÈ·£¬¹Ê´ð°¸Îª£ºC£»
(6)5.95gCoCO3µÄÎïÖʵÄÁ¿Îª£¬Ôò¸´ºÏÑõ»¯îÜÖк¬îÜÔªËصÄÖÊÁ¿Îª0.05mol¡Á59g¡¤mol-1=2.95g£¬ÔòÑõÔªËصÄÖÊÁ¿Îª4.07g-2.95g=1.12g£¬¼´0.07mol£¬Òò´Ë¸´ºÏÑõ»¯îÜÖÐCoºÍOµÄ¸öÊý±ÈΪ5£º7£¬Ôò¸´ºÏÑõ»¯îܵĻ¯Ñ§Ê½ÎªCo2O7£¬CoCO3ת»¯ÎªCo5O7ʱCoÔªËصĻ¯ºÏ¼ÛÉý¸ß£¬ÔòÓÐÑõÆø²Î¼Ó·´Ó¦£¬¹Ê´ð°¸Îª£ºCo2O7£»ÓС£
![](http://thumb2018.1010pic.com/images/loading.gif)
¡¾ÌâÄ¿¡¿°ÑζÈΪ20 ¡æ£¬Å¨¶ÈΪ1.0 mol/LµÄH2SO4ÈÜÒººÍ2.2 mol/LµÄ¼îÈÜÒº¸÷50 mL»ìºÏ[ÈÜÒºÃܶȾùΪ1g/mL£¬±ÈÈÈÈÝΪ4.18 kJ/(kg¡¤¡æ)]ÇáÇá½Á¶¯£¬²âµÃËá¼î»ìºÏÒºµÄζȱ仯Êý¾ÝÈçÏ£º
·´Ó¦Îï | ÆðʼζÈt1 ¡æ | ÖÕֹζÈt2 ¡æ |
H2SO4£«NaOH | 20 | 33.6 |
H2SO4£«NH3¡¤H2O | 20 | 32.6 |
£¨1£©·´Ó¦NH3¡¤H2O(aq)NH4+ (aq)£«OH£(aq)µÄìʱäÔ¼____¡£
£¨2£©¼ÆËãÉÏÊöÁ½×éʵÑé²â³öµÄÖкÍÈÈ£º¦¤H1£½___kJ/mol£»¦¤H2£½__kJ/mol¡£
£¨3£©Óɱ¾Ìâ½áÂÛ¿ÉÔ¤²â½«µÚ1×éÖеÄ1 mol/LµÄH2SO4ÈÜÒº»»³É2mol/LµÄCH3COOHÈÜÒº½øÐÐʵÑ飬²âµÃµÄÖкÍÈÈÊýÖµ__(Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)56.848¡£