ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿¸ù¾Ý»¯Ñ§·´Ó¦ÓëÄÜÁ¿×ª»¯µÄÏà¹Ø֪ʶ£¬ÊԻشðÏÂÁÐÎÊÌâ:
£¨1£©ÒÑÖª£¬ºÚÁױȰ×Á×Îȶ¨£¬½á¹¹ÓëʯīÏàËÆ£¬ÏÂͼÄÜÕýÈ·±íʾ¸Ã·´ÕýÖÐÄÜÁ¿±ä»¯µÄÊÇ___£¨ÌîÐòºÅ£©¡£
£¨2£©ÒÑÖª£ºÇâÑõȼÁϵç³ØµÄ×Ü·´Ó¦·½³ÌʽΪ2H2O+O2=2H2O¡£ÔÚ¼îÐÔÌõ¼þÏ£¬Í¨ÈëÇâÆøÒ»¶ËµÄµç¼«·´Ó¦Ê½Îª_________________¡£µç·ÖÐÿתÒÆ0.2molµç×Ó£¬±ê×¼×´¿öÏÂÏûºÄH2µÄÌå»ýÊÇ______L¡£
£¨3£©ÒÑÖª¶Ï¿ª1mol N¡ÔN¼üÐèÒª946kJµÄÄÜÁ¿£¬¶Ï¿ª1mol H¡ªH¼üÐèÒª436kJµÄÄÜÁ¿£¬Éú³É1moN¡ªH¼ü·Å³ö391kJµÄÄÜÁ¿£¬ÊÔ¼ÆËãÉú³É2mol NH3ʱ»á______£¨Ìî¡°·Å³ö¡±»ò¡°ÎüÊÕ¡±£©______kJÄÜÁ¿¡£µ±ÔÚÏàͬµÄÌõ¼þÏÂÏòÈÝÆ÷ÖгäÈë1molN2ºÍ3molH2ʱ£¬ËüÃÇ·´Ó¦¶ÔÓ¦µÄÈÈÁ¿______£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©ÄãËù¼ÆËã³öµÄÖµ£¬ÔÒòÊÇ______¡£
¡¾´ð°¸¡¿£¨1£©A
£¨2£©H2-2e-+2OH- =2H2O £»2.24£»
£¨3£©·Å³ö£»92£»Ð¡ÓÚ£»¿ÉÄæ·´Ó¦£¬·´Ó¦Îï²»ÄÜÍêȫת»¯ÎªÉú³ÉÎï¡£
¡¾½âÎö¡¿
ÊÔÌâ·ÖÎö£º£¨1£©ÎïÖʺ¬ÓеÄÄÜÁ¿Ô½µÍ£¬ÎïÖʵÄÎȶ¨ÐÔ¾ÍԽǿ¡£ÓÉÓÚºÚÁױȰ×Á×Îȶ¨£¬ËùÒÔ°×Á×µÄÄÜÁ¿±Èü\Á׵ĸߣ¬°×Á×ת»¯Îªü\Á׵ķ´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬ÄÜÕýÈ·±íʾ¸Ã·´ÕýÖÐÄÜÁ¿±ä»¯µÄÊÇAͼÏñ£»£¨2£©ÇâÑõȼÁϵç³ØµÄ×Ü·´Ó¦·½³ÌʽΪ2H2O+O2=2H2O¡£ÔÚ¼îÐÔÌõ¼þÏ£¬Í¨ÈëÇâÆøÒ»¶ËÊǸº¼«£¬¸º¼«µÄµç¼«·´Ó¦Ê½ÎªH2-2e-+2OH- =2H2O¡£¸ù¾Ý×Ü·´Ó¦·½³Ìʽ¿ÉÖª£ºÃ¿ÓÐ2molÇâÆø·¢Éú·´Ó¦£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿ÊÇ4mol£¬Ôòµç·ÖÐÿתÒÆ0.2molµç×Ó£¬·´Ó¦ÏûºÄ±ê×¼×´¿öÏÂH2µÄÎïÖʵÄÁ¿ÊÇ0.1mol£¬ÆäÌå»ýÊÇV=0.1mol¡Á 22.4L/mol=2.24L£»£¨3£©ÒÑÖª¶Ï¿ª1mol N¡ÔN¼üÐèÒª946kJµÄÄÜÁ¿£¬¶Ï¿ª1mol H¡ªH¼üÐèÒª436kJµÄÄÜÁ¿£¬Éú³É1moN¡ªH¼ü·Å³ö391kJµÄÄÜÁ¿£¬Ôò¸ù¾Ý·´Ó¦·½³ÌʽN2(g)+3H2(g)=2NH3(g)¿ÉÖª£ºÉú³É2mol NH3ʱH=946kJ/mol+3¡Á436kJ/mol-6¡Á391kJ/mol=-92kJ/mol£¬¹Ê»á·Å³ö92kkJÄÜÁ¿¡£µ±ÔÚÏàͬµÄÌõ¼þÏÂÏòÈÝÆ÷ÖгäÈë1molN2ºÍ3molH2ʱ£¬ÓÉÓڸ÷´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬·´Ó¦Îï²»ÄÜÍêȫת»¯ÎªÉú³ÉÎËùÒÔËüÃÇ·´Ó¦¶ÔÓ¦µÄÈÈÁ¿Ð¡ÓÚ¼ÆËã³öµÄÖµ¡£
![](http://thumb2018.1010pic.com/images/loading.gif)
¡¾ÌâÄ¿¡¿ÊµÑéÊÒÖƱ¸»·¼ºÍªµÄ·´Ó¦ÔÀí£º
Æä·´Ó¦µÄ×°ÖÃʾÒâͼÈçÏ£¨¼Ð³Ö×°ÖᢼÓÈÈ×°ÖÃÂÔÈ¥£©£º
»·¼º´¼¡¢»·¼ºÍª¡¢±¥ºÍʳÑÎË®ºÍË®µÄ²¿·ÖÎïÀíÐÔÖʼûÏÂ±í£¨×¢£ºÀ¨ºÅÖеÄÊý¾Ý±íʾ¸ÃÓлúÎïÓëË®ÐγɵľßÓй̶¨×é³ÉµÄ»ìºÏÎïµÄ·Ðµã£©
ÎïÖÊ | ·Ðµã£¨¡æ£© | Ãܶȣ¨g¡¤cm£3, 20¡æ£© | ÈܽâÐÔ |
»·¼º´¼ | 161.1£¨97.8£© | 0.962 | ÄÜÈÜÓÚË® |
»·¼ºÍª | 155.6£¨95£© | 0.948 | ΢ÈÜÓÚË® |
±¥ºÍʳÑÎË® | 108.0 | 1.330 | |
Ë® | 100.0 | 0.998 |
£¨1£©ÊµÑéÖÐͨ¹ý×°ÖÃB½«ËáÐÔNa2Cr2O7ÈÜÒº¼Óµ½Ê¢Óл·¼º´¼µÄAÖУ¬ÔÚ55¡«60¡æ½øÐз´Ó¦¡£·´Ó¦Íê³Éºó£¬¼ÓÈëÊÊÁ¿Ë®£¬ÕôÁó£¬ÊÕ¼¯95 ~ 100¡æµÄÁó·Ö£¬µÃµ½Ö÷Òªº¬»·¼ºÍª´ÖÆ·ºÍË®µÄ»ìºÏÎï¡£
¢Ù ÒÇÆ÷BµÄÃû³ÆÊÇ ¡£
¢Ú ÕôÁó²Ù×÷ʱ£¬Ò»¶Îʱ¼äºó·¢ÏÖδͨÀäÄýË®£¬Ó¦²ÉÈ¡µÄÕýÈ··½·¨ÊÇ ¡£
¢Û ÕôÁó²»ÄÜ·ÖÀë»·¼ºÍªºÍË®µÄÔÒòÊÇ ¡£
£¨2£©»·¼ºÍªµÄÌá´¿ÐèÒª¾¹ýÒÔÏÂһϵÁеIJÙ×÷£º
a£®ÕôÁó£¬ÊÕ¼¯151~ 156¡æÁó·Ö£¬µÃµ½¾«Æ·
b£®¹ýÂË
c£®ÔÚÊÕ¼¯µ½µÄ´ÖÆ·ÖмÓNaCl¹ÌÌåÖÁ±¥ºÍ£¬¾²Ö㬷ÖÒº
d£®¼ÓÈëÎÞË®MgSO4¹ÌÌ壬³ýÈ¥ÓлúÎïÖÐÉÙÁ¿Ë®
¢Ù ÉÏÊö²Ù×÷µÄÕýȷ˳ÐòÊÇ £¨ÌîÐòºÅ£©¡£
¢Ú ÔÚÉÏÊö²Ù×÷cÖУ¬¼ÓÈëNaCl¹ÌÌåµÄ×÷ÓÃÊÇ ¡£
£¨3£©»·¼ºÍªÊÇÒ»ÖÖ³£ÓõÄÝÍÈ¡¼Á£¬ÒÔ»·¼ºÍªÎª³É·ÖÖ®Ò»µÄÝÍÈ¡Òº¶Ô½ðÊôÀë×ÓµÄÝÍÈ¡ÂÊÓëpHµÄ¹ØϵÈçÉÏͼ¡£ÏÖÓÐijîÜ¿óʯµÄÑÎËá½þ³öÒº, ½þ³öÒºº¬ÓеÄÑôÀë×ÓÖ÷ÒªÓÐH+¡¢Co2+¡¢Mn2+¡¢Al3+£¬³ýÈ¥Al3+ ºó£¬ÔÙÀûÓÃÝÍÈ¡·¨·ÖÀë³öMnCl2ÒԵõ½½ÏΪ´¿¾»µÄCoCl2ÈÜÒº£¬pH·¶Î§¿ØÖÆÔÚ ¡£
a£®2.0¡«2.5 b£®3.0¡«3.5 c£®4.0¡«4.5
£¨4£©ÔÚʵÑéÖÐÔÁÏÓÃÁ¿£º20mL»·¼º´¼¡¢ËáÐÔNa2Cr2O7ÈÜÒº10mL£¬×îÖճƵòúÆ·ÖÊÁ¿Îª14.7g£¬ÔòËùµÃ»·¼ºÍªµÄ²úÂÊΪ £¨°Ù·ÖÊý¾«È·µ½0.1£©¡£
£¨5£©ÊµÑé²úÉúµÄº¬¸õ·ÏÒºÒ×ÎÛȾ»·¾³£¬¿É½«Cr2O72£×ª»¯ÎªCr3£«£¬ÔÙת»¯ÎªCr£¨OH£©3³Áµí³ýÈ¥£¬ÎªÊ¹Cr3£«Å¨¶ÈСÓÚ1¡Á10-5mol/L£¬Ó¦µ÷½ÚÈÜÒºµÄpH=___________¡££¨ÒÑÖª¸ÃÌõ¼þÏÂCr£¨OH£©3µÄKspÊÇ6.4¡Á10-31£¬ lg2=0.3 lg5=0.7£©