ÌâÄ¿ÄÚÈÝ
£¨2£©Î¶ȼƵÄ×÷ÓÃÊÇ_____________________£¬Ëé´ÉƬµÄ×÷ÓÃÊÇ______________________£»
£¨3£©Ð´³öʵÑéÖвúÉúÒÒÏ©µÄ»¯Ñ§·½³Ìʽ£º_______________________________£»
£¨4£©¼×ͬѧÈÏΪ£ºäåË®ÍÊÉ«µÄÏÖÏó²»ÄÜÖ¤Ã÷ÒÒÏ©¾ßÓв»±¥ºÍÐÔ£¬ÆäÔÒòÊÇÉÕÆ¿ÖÐÒºÌå³ÊרºÚÉ«¶ø²úÉú
__________ÆøÌå¡£ÒÒͬѧ¾¹ý×Ðϸ¹Û²ìºóÈÏΪ£ºÊÔ¹ÜÖÐÁíÒ»¸öÏÖÏó¿ÉÖ¤Ã÷ÒÒÏ©¾ßÓв»±¥ºÍÐÔ£¬Õâ¸öÏÖÏóÊÇ_______________¡£±ûͬѧΪÑéÖ¤ÕâÒ»·´Ó¦ÊǼӳɶø²»ÊÇÈ¡´ú£¬Ìá³öÁ˽«ÔÓÖÊÆøÌåÎüÊպ󣬿ÉÓÃpHÊÔÖ½À´²âÊÔ·´Ó¦ºóÈÜÒºµÄËáÐÔ£¬ÀíÓÉÊÇ____________________________________£»
£¨5£©´¦ÀíÉÏÊöʵÑéÖÐÉÕÆ¿ÄÚ·ÏÒºµÄÕýÈ·²Ù×÷ÊÇ_______________¡£
A£®·ÏÒºÖ±½Óµ¹ÈëÏÂË®µÀ
B£®·ÏÒºµ¹Èë¿Õ·ÏÒº¸×ÖÐ
C£®½«Ë®µ¹ÈëÉÕÆ¿ÖÐ
D£®·ÏÒºµ¹ÈëÊ¢ÓÐË®µÄËÜÁÏͰÖУ¬¾´¦ÀíºóÔÙµ¹ÈëÏÂË®µÀ
£¨2£©¿ØÖÆ·´Ó¦Î¶ÈÔÚ170¡æ£»·ÀÖ¹±©·Ð
£¨3£©
£¨4£©O2£»Óв»ÈÜÓÚË®µÄÓÍ×´ÎïÉú³É£»ÈçÈô·¢ÉúÈ¡´ú·´Ó¦£¬±Ø¶¨Éú³ÉHBr£¬ÈÜÒºËáÐÔ½«»áÃ÷ÏÔÔöÇ¿£¬¹Ê¿ÉÓÃpHÊÔÖ½ÑéÖ¤
£¨5£©D
£¨A£©ÈçÏÂͼËùʾ£¬½«¼×¡¢ÒÒÁ½¸ö×°Óв»Í¬ÎïÖʵÄÕëͲÓõ¼¹ÜÁ¬½ÓÆðÀ´£¬½«ÒÒÕëͲÄÚµÄÎïÖÊѹµ½¼×ÕëͲÄÚ£¬½øÐÐϱíËùÁеIJ»Í¬ÊµÑé£¨ÆøÌåÔÚͬÎÂͬѹϲⶨ£©¡£
![]()
ʵÑéÐòºÅ | ¼×ÕëͲÄÚÎïÖÊ | ÒÒÕëͲÄÚÎïÖÊ | ¼×ÕëͲµÄÏÖÏó |
1 | 10 mL FeSO4ÈÜÒº | 10 mL NH3 | Éú³É°×É«³Áµí£¬ºó±äÉ« |
2 | 20 mL H2S | 10 mL SO2 |
|
3 | 30 mL NO2£¨Ö÷Òª£© | 10 mL H2O(l) | Ê£ÓÐÎÞÉ«ÆøÌ壬»îÈû×Ô¶¯ÏòÄÚѹËõ |
4 | 15 mL Cl2 | 40 mL NH3 |
|
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÊµÑé1ÖУ¬³Áµí×îÖÕ±äΪ___________É«£¬Ð´³ö³Áµí±äÉ«µÄ»¯Ñ§·½³Ìʽ_______________¡£
£¨2£©ÊµÑé2¼×ÕëͲÄÚµÄÏÖÏóÊÇ£ºÓÐ________Éú³É£¬»îÈû___________ÒÆ¶¯£¨Ìî¡°ÏòÍ⡱¡°ÏòÄÚ¡±»ò¡°²»¡±£©¡£·´Ó¦ºó¼×ÕëͲÄÚÓÐÉÙÁ¿µÄ²ÐÁôÆøÌ壬ÕýÈ·µÄ´¦Àí·½·¨Êǽ«ÆäͨÈë__________ÈÜÒºÖС£
£¨3£©ÊµÑé3ÖУ¬¼×ÖеÄ30 mLÆøÌåÊÇNO2ºÍN2O4µÄ»ìºÏÆøÌ壬ÄÇô¼×ÖÐ×îºóÊ£ÓàµÄÎÞÉ«ÆøÌåÊÇ__________£¬Ð´³öNO2ÓëH2O·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________________¡£
£¨4£©ÊµÑé4ÖУ¬ÒÑÖª£º3Cl2+2NH3
N2+6HCl¡£¼×ÕëͲ³ý»îÈûÓÐÒÆ¶¯¡¢ÕëͲÄÚÓа×Ñ̲úÉúÍâ£¬ÆøÌåµÄÑÕÉ«±ä»¯Îª___________£¬×îºóÕëͲÄÚÊ£ÓàÆøÌåµÄÌå»ýԼΪ______________mL¡£
£¨B£©Ä³ÊµÑéС×éÓÃÏÂÁÐ×°ÖýøÐÐÒÒ´¼´ß»¯Ñõ»¯µÄʵÑé¡£
![]()
£¨1£©ÊµÑé¹ý³ÌÖÐÍÍø³öÏÖºìÉ«ºÍºÚÉ«½»ÌæµÄÏÖÏó£¬Çëд³öÏàÓ¦µÄ»¯Ñ§·´Ó¦·½³Ìʽ
_____________________________________________________________________
_____________________________________________________________________¡£
ÔÚ²»¶Ï¹ÄÈë¿ÕÆøµÄÇé¿öÏ£¬Ï¨Ãð¾Æ¾«µÆ£¬·´Ó¦ÈÔÄܼÌÐø½øÐУ¬ËµÃ÷¸ÃÒÒ´¼Ñõ»¯·´Ó¦ÊÇ________·´Ó¦¡£
£¨2£©¼×ºÍÒÒÁ½¸öˮԡ×÷Óò»Ïàͬ¡£
¼×µÄ×÷ÓÃÊÇ____________________£»ÒÒµÄ×÷ÓÃÊÇ_____________________¡£
£¨3£©·´Ó¦½øÐÐÒ»¶Îʱ¼äºó£¬¸ÉÔïÊÔ¹ÜaÖÐÄÜÊÕ¼¯µ½²»Í¬µÄÎïÖÊ£¬ËüÃÇÊÇ____________________________¡£¼¯ÆøÆ¿ÖÐÊÕ¼¯µ½µÄÆøÌåµÄÖ÷Òª³É·ÖÊÇ______________¡£
£¨4£©ÈôÊÔ¹ÜaÖÐÊÕ¼¯µ½µÄÒºÌåÓÃ×ÏɫʯÈïÊÔÖ½¼ìÑ飬ÊÔÖ½ÏÔºìÉ«£¬ËµÃ÷ÒºÌåÖл¹º¬ÓÐ__________¡£Òª³ýÈ¥¸ÃÎïÖÊ£¬¿ÉÏÈÔÚ»ìºÏÒºÖмÓÈë______________£¨Ìîд×Öĸ£©¡£
a.ÂÈ»¯ÄÆÈÜÒº b.±½
c.̼ËáÇâÄÆÈÜÒº d.ËÄÂÈ»¯Ì¼
È»ºó£¬ÔÙͨ¹ý_____________£¨ÌîʵÑé²Ù×÷Ãû³Æ£©¼´¿É³ýÈ¥¡£
¼×´¼À´Ô´·á¸»£®¼Û¸ñµÍÁ®£®ÔËÊäÖü´æ·½±ã£¬ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤ÔÁÏ£¬ÓÐ×ÅÖØÒªµÄÓÃ;ºÍÓ¦ÓÃǰ¾°¡£
£¨1£©¹¤ÒµÉú²ú¼×´¼µÄ³£Ó÷½·¨ÊÇ£ºCO(g)+2H2(g)
CH3OH(g) ¡÷H = ¡ª90.8kJ/mol¡£ÈôÔÚºãκãÈݵÄÈÝÆ÷ÄÚ½øÐз´Ó¦CO(g)+2H2(g)
CH3OH(g)£¬ÏÂÁбíʾ¸Ã·´Ó¦´ïµ½Æ½ºâ״̬µÄ±êÖ¾ÓÐ £¨Ìî×ÖĸÐòºÅ£©¡£
A£®ÈÝÆ÷ÖлìºÏÆøÌåµÄÃܶȲ»±ä»¯
B£®CO°Ù·Öº¬Á¿±£³Ö²»±ä
C£®ÈÝÆ÷ÖлìºÏÆøÌåµÄѹǿ²»±ä»¯
D£®ÓÐ1¸öH¡ªH¼üÉú³ÉµÄͬʱÓÐ 3¸öC¡ªH¼üÉú³É
£¨2£©ÖƼ״¼ËùÐèÒªµÄH2£¬¿ÉÓÃÏÂÁз´Ó¦ÖÆÈ¡£ºH2O(g)+CO(g)
H2(g)+ CO2(g)
¡÷H£¼0£¬Ä³Î¶Èϸ÷´Ó¦µÄƽºâ³£ÊýK = 1¡£ÈôÆðʼʱc(CO)=1mol•L-1£¬c(H2O)=2mol•L£1£¬ÊԻشðÏÂÁÐÎÊÌ⣺
¢Ù¸ÃζÈÏ£¬·´Ó¦½øÐÐÒ»½×¶Îʱ¼äºó£¬²âµÃH2µÄŨ¶ÈΪ0.5mol•L-1£¬Ôò´Ëʱ¸Ã·´Ó¦
v(Õý) v(Äæ)£¨Ìî¡°£¾¡±£®¡°£¼¡±»ò¡°£½¡±£©£»
¢ÚÈô·´Ó¦Î¶Ȳ»±ä£¬´ïµ½Æ½ºâºó£¬H2OµÄת»¯ÂÊΪ ¡£
£¨3£©Ä³ÊµÑéС×éÉè¼ÆÁËÈçÏÂͼËùʾµÄ¼×´¼È¼ÁÏµç³Ø×°Öá£
| ||||||||||
| ||||||||||
| ||||||||||
¢Ù¸Ãµç³Ø¹¤×÷ʱ£¬OH¡ª Ïò ¼«Òƶ¯£¨Ìî¡°a¡±»ò¡°b¡±£©£»
¢Ú¹¤×÷Ò»¶Îʱ¼äºó£¬²âµÃÈÜÒºµÄpH¼õС£¬¸Ãµç³Ø¸º¼«µç¼«·´Ó¦Ê½Îª ¡£