ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿A¡¢B¡¢C¡¢D¡¢E¡¢F¡¢G¡¢HÊÇÔªËØÖÜÆÚ±íÇ°ËÄÖÜÆÚ³£¼ûÔªËØ£¬ÇÒÔ×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÏà¹ØÐÅÏ¢ÈçÏÂ±í£º
ÔªËØ | Ïà¹ØÐÅÏ¢ |
A | Ô×ÓºËÍâÓÐ6ÖÖ²»Í¬Ô˶¯×´Ì¬µÄµç×Ó |
C | »ù̬Ô×ÓÖÐsµç×Ó×ÜÊýÓëpµç×Ó×ÜÊýÏàµÈ |
D | Ô×Ӱ뾶ÔÚͬÖÜÆÚÔªËØÖÐ×î´ó |
E | »ù̬Ô×Ó¼Û²ãµç×ÓÅŲ¼Ê½Îª3s23p1 |
F | »ù̬Ô×ÓµÄ×îÍâ²ãp¹ìµÀÓÐÁ½¸öµç×ÓµÄ×ÔÐý·½ÏòÓëÆäËûµç×ÓµÄ×ÔÐý·½ÏòÏà·´ |
G | »ù̬Ô×ÓºËÍâÓÐ7¸öÄܼ¶ÇÒÄÜÁ¿×î¸ßµÄÄܼ¶ÉÏÓÐ6¸öµç×Ó |
H | ÊÇÎÒ¹úʹÓÃ×îÔçµÄºÏ½ðÖеÄ×îÖ÷ÒªÔªËØ |
ÇëÓû¯Ñ§ÓÃÓïÌî¿Õ£º
£¨1£©AÔªËØÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃ______£¬BÔªËغÍCÔªËصĵÚÒ»µçÀëÄܱȽϣ¬½Ï´óµÄÊÇ___________£¬CÔªËغÍFÔªËصĵ縺ÐԱȽϣ¬½ÏСµÄÊÇ_____________¡£
£¨2£©BÔªËØÓëÓîÖæÖк¬Á¿×î·á¸»µÄÔªËØÐγɵÄ×î¼òµ¥»¯ºÏÎïµÄ·Ö×Ó¹¹ÐÍΪ____________£¬BÔªËØËùÐγɵĵ¥ÖÊ·Ö×ÓÖмüÓë¦ÐÊýÄ¿Ö®±ÈΪ_________¡£
£¨3£©GÔªËصĵͼÛÑôÀë×ӵĽṹʾÒâͼÊÇ_________________£¬FÔªËØÔ×ӵļ۵ç×ÓÅŲ¼Í¼ÊÇ__________£¬HÔªËصĻù̬Ô×ÓºËÍâµç×ÓÅŲ¼Ê½ÊÇ________________________________________¡£
£¨4£©GµÄ¸ß¼ÛÑôÀë×ÓµÄÈÜÒºÓëHµ¥ÖÊ·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________________;ÓëEÔªËسɶԽǹØϵµÄijԪËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¾ßÓÐÁ½ÐÔ£¬¸ÃÁ½ÐÔÎïÖÊÓëDÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________________________________¡£
¡¾´ð°¸¡¿ µÚ¶þÖÜÆÚIVA ×å N Cl Èý½Ç׶ÐÎ 1£º2 ÂÔ 1s22s22p63s23p63d104s1 »ò
2Fe3+ + Cu = Cu2+ + 2Fe2+ Be(OH)2+2NaOH=Na2BeO2+2H2O
¡¾½âÎö¡¿ÊÔÌâ·ÖÎö£ºAµÄÔ×ÓºËÍâÓÐ6ÖÖ²»Í¬Ô˶¯×´Ì¬µÄµç×Ó£¬¼´ºËÍâ¹²ÓÐ6¸öµç×Ó£¬AΪ̼ԪËØ¡£CµÄ»ù̬Ô×ÓÖÐsµç×Ó×ÜÊýÓëpµç×Ó×ÜÊýÏàµÈ£¬ÔòºËÍâµç×ÓÅŲ¼Îª1s22s22p4£¬CÊÇÑõÔªËØ¡£ÓÉÓÚA¡¢B¡¢CÔ×ÓÐòÊýÒÀ´ÎµÝÔö£¬¹ÊBÊǵªÔªËØ¡£DµÄÔ×Ӱ뾶ÔÚͬÖÜÆÚÔªËØÖÐ×î´ó£¬¹ÊDӦΪµÚÈýÖÜÆÚÔªËØ£¬ÊÇÄÆÔªËØ¡£EµÄ»ù̬Ô×Ó×îÍâ²ãµç×ÓÅŲ¼Ê½Îª3s23p1£¬ÊÇÂÁÔªËØ¡£FµÄ»ù̬Ô×ÓµÄ×îÍâ²ãp¹ìµÀÓÐÁ½¸öµç×ÓµÄ×ÔÐý·½ÏòÓëÆäËüµç×ÓµÄ×ÔÐý·½ÏòÏà·´£¬ËµÃ÷p¹ìµÀÉÏÓÐ5¸öµç×Ó£¬ÊÇÂÈÔªËØ¡£GµÄ»ù̬Ô×ÓºËÍâÓÐ7¸öÄܼ¶ÇÒÄÜÁ¿×î¸ßµÄÄܼ¶ÉÏÓÐ6¸öµç×Ó£¬ÔòÆäµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p64s23d6£¬ÊÇÌúÔªËØ¡£HÊÇÎÒ¹úʹÓÃ×îÔçµÄºÏ½ðÖеÄ×îÖ÷ÒªÔªËØ£¬ÊÇÍÔªËØ¡£
![](http://thumb2018.1010pic.com/images/loading.gif)
¡¾ÌâÄ¿¡¿£¨1£©ÄÉÃײÄÁÏÒ»Ö±ÊÇÈËÃÇÑо¿µÄÖØÒª¿ÎÌ⣬ÀýÈçÄÉÃ×¼¶Fe·Û±íÃæ»ý´ó£¬¾ßÓг¬Ç¿µÄ´ÅÐÔ£¬¸ßЧ´ß»¯ÐÔµÈÓÅÁ¼µÄÐÔÖÊ¡£ÊµÑéÊÒ²ÉÓÃÆøÏ໹Է¨ÖƱ¸ÄÉÃ×¼¶Fe£¬ÆäÁ÷³ÌÈçͼËùʾ£º
¢Ù ÄÉÃ×¼¶FeºÍÏ¡ÑÎËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________¡£
¢Ú ÄÉÃ×¼¶Fe·ÛÔÚ¿ÕÆøÖÐÒ××Ôȼ³ÉºÚÉ«¹ÌÌ壬µ«ÊÇÉú»îÖеÄÌúË¿»òÌú·ÛÔÚ¿ÕÆøÖмÓÈÈÒ²²»ÄÜȼÉÕ£¬ÆäÔÒòÊÇ_____________________¡£
¢Û ÈçºÎ½«FeC12¡¤nH2O¹ÌÌå¼ÓÈÈÍÑË®ÖƵÃÎÞË®FeCl2? ___________¡£
£¨2£©²éÔÄ×ÊÁÏ£ºÔÚ²»Í¬Î¶ÈÏ£¬ÄÉÃ×¼¶Fe·ÛÓëË®ÕôÆø·´Ó¦µÄ¹ÌÌå²úÎﲻͬ£ºÎ¶ȵÍÓÚ570¡æ ʱÉú³ÉFeO,¸ßÓÚ570¡æʱÉú³ÉFe3O4¡£¼×ͬѧÓÃÈçͼ¼××°ÖÃËùʾ½øÐÐÄÉÃ×¼¶Fe·ÛÓëË®ÕôÆø·´Ó¦µÄʵÑé¡£
¢Ù ¸Ã×°ÖÃÖÐÄÉÃ×¼¶Fe·ÛÓëË®ÕôÆø·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ__________¡£
ÒÒͬѧÓÃͼÒÒËùʾµÄ×°ÖýøÐÐÄÉÃ×¼¶Fe·ÛÓëË®ÕôÆøµÄ·´Ó¦²¢ÑéÖ¤²úÎ
¢Ú ×°ÖÃCµÄ×÷ÓÃÊÇ________________¡£
¢Û ÒÒͬѧΪ̽¾¿ÊµÑé½áÊøºóÊÔ¹ÜÄڵĹÌÌåÎïÖʳɷ֣¬½øÐÐÁËÏÂÁÐʵÑ飺
ʵÑé²½Öè | ʵÑé²Ù×÷ | ʵÑéÏÖÏó |
I | ½«·´Ó¦ºóµÃµ½µÄºÚÉ«·ÛÄ©X£¨¼Ù¶¨Îª¾ùÔȵģ©£¬È¡³öÉÙÁ¿·ÅÈëÁíÒ»ÊÔ¹ÜÖУ¬¼ÓÈëÉÙÁ¿ÑÎËᣬ΢ÈÈ | ºÚÉ«·ÛÄ©Öð½¥Èܽ⣬ÈÜÒº³ÊdzÂÌÉ«£»ÓÐÉÙÁ¿ÆøÅݲúÉú |
II | ÏòʵÑéI µÃµ½µÄÈÜÒºÖеμӼ¸µÎKSCNÈÜÒº£¬Õñµ´ | ÈÜҺûÓгöÏÖѪºìÉ« |
ÒÒͬѧÈÏΪ¸ÃÌõ¼þÏ·´Ó¦µÄ¹ÌÌå²úÎïΪFeO¡£±ûͬѧÈÏΪ¸Ã½áÂÛ²»ÕýÈ·£¬ËûµÄÀíÓÉÊÇ___________£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
¢Ü ¶¡Í¬Ñ§³ÆÈ¡2.80gFe·Û£¬ÓÃÒÒ×°Ö÷´Ó¦Ò»¶Îʱ¼äºó£¬Í£Ö¹¼ÓÈÈ¡£½«ÊÔ¹ÜÄڵĹÌÌåÎïÖÊÔÚ¸ÉÔïÆ÷ÖÐÀäÈ´ºó£¬³ÆµÃÖÊÁ¿Îª3.44g¡£È»ºó½«ÀäÈ´ºóµÄ¹ÌÌåÎïÖÊÓë×ãÁ¿FeCl3ÈÜÒº³ä·Ö·´Ó¦£¬ÏûºÄFeCl30.04mol¡£¶¡Í¬Ñ§ÊµÑéµÄ¹ÌÌå²úÎïΪ_________¡£
¡¾ÌâÄ¿¡¿¹¤ÒµÉÏÀûÓ÷ÏÌúм(º¬ÉÙÁ¿Ñõ»¯ÂÁ¡¢Ñõ»¯ÌúµÈ)Éú²ú¼îʽÁòËáÌú[Fe(OH)SO4]µÄ¹¤ÒÕÁ÷³ÌÈçÏ£º
ÒÑÖª£º²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ³ÁµíʱÈÜÒºµÄpHÈçÏÂ±í£º
³ÁµíÎï | Fe(OH)3 | Fe(OH)2 | Al(OH)3 |
¿ªÊ¼³Áµí | 2.3 | 7.5 | 3.4 |
ÍêÈ«³Áµí | 3.2 | 9.7 | 4.4 |
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©·´Ó¦¢ñÖз¢ÉúµÄÑõ»¯»¹Ô·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ ¡£
£¨2£©¼ÓÈëÉÙÁ¿NaHCO3µÄÄ¿µÄÊǵ÷½ÚpH£¬Ê¹ÈÜÒºÖеÄ________(Ìî¡°Fe3£«¡±¡¢¡°Fe2£«¡±»ò¡°Al3£«¡±)³Áµí¡£¸Ã¹¤ÒÕÁ÷³ÌÖС°½Á°è¡±µÄ×÷ÓÃÊÇ______________________¡£
£¨3£©·´Ó¦¢òµÄÀë×Ó·½³ÌʽΪ ___¡£ÔÚʵ¼ÊÉú²úÖУ¬·´Ó¦¢ò³£Í¬Ê±Í¨ÈëO2ÒÔ¼õÉÙNaNO2µÄÓÃÁ¿£¬O2ÓëNaNO2ÔÚ·´Ó¦Öоù×÷____ ____¡£Èô²ÎÓë·´Ó¦µÄO2ÓÐ11.2 L(±ê×¼×´¿ö)£¬ÔòÏ൱ÓÚ½ÚÔ¼NaNO2µÄÎïÖʵÄÁ¿Îª________¡£
£¨4£©¼îʽÁòËáÌúÈÜÓÚË®ºó²úÉúµÄFe(OH)2£«Àë×Ó£¬¿É²¿·ÖË®½âÉú³ÉFe2(OH)42+¾ÛºÏÀë×Ó¡£¸ÃË®½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________¡£
£¨5£©ÔÚÒ½Ò©Éϳ£ÓÃÁòËáÑÇÌúÓëÁòËá¡¢ÏõËáµÄ»ìºÏÒº·´Ó¦ÖƱ¸¼îʽÁòËáÌú¡£¸ù¾ÝÎÒ¹úÖÊÁ¿±ê×¼£¬²úÆ·Öв»µÃº¬ÓÐFe2£«¼°NO3-¡£Îª¼ìÑéËùµÃ²úÆ·ÖÐÊÇ·ñº¬ÓÐFe2£«£¬Ó¦Ê¹ÓõÄÊÔ¼ÁΪ________(Ìî×Öĸ)¡£
A. ÂÈË® B. KSCNÈÜÒº C. NaOHÈÜÒº D. ËáÐÔKMnO4ÈÜÒº