ÌâÄ¿ÄÚÈÝ

ÓÐЩ»¯Ñ§ÊµÑé¾ßÓÐÒ»¶¨µÄΣÏÕÐÔ£¬±ØÐëÑϸñ°´ÕÕ²Ù×÷¹æ¶¨½øÐÐʵÑé¡£ÏÂÃæµÄʵÑé²Ù×÷ÕýÈ·µÄÊÇ¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­£¨  £©

A. ÖÆC2H4ʱ£¬ÔÚ´óÊÔ¹ÜÀïµ¹Èë¾Æ¾«ºÍŨH2SO4µÄ»ìºÏÒº£¬¼ÓÈÈÇ°¼ÓÈëËé´ÉƬ£¬ÒÔ·ÀÖ¹±©·Ð

B. ÅäÖÆÒø°±ÈÜҺʱ£¬°±Ë®±ØÐë¹ýÁ¿£¬µ«¾ÃÖõÄÒø°±ÈÜÒº²»ÄÜʹÓã¬Ó¦¸ÃËæÅäËæÓÃ

C. ÔÚÓÃпÁ£ºÍÑÎËá·´Ó¦ÖÆÈ¡ÇâÆøµÄʵÑéÖУ¬µ±Ð¿Á£·´Ó¦ÍêÈ«ºó£¬´ò¿ª·´Ó¦Æ÷µÄÈû×Óºó×°ÈëпÁ££¬ÈûÉÏÈû×Ó£¬È»ºóµãȼÇâÆø

D. ÖÆÈ¡»òʹÓÃCl2¡¢H2S¡¢NO2¡¢COµÈÓж¾ÆøÌåʱ£¬Ò»°ãÔÚͨ·ç³÷ÖнøÐУ¬¶àÓàµÄβÆø±ØÐëÊÕ¼¯»òÎüÊÕ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¢ñ£®Ñغ£µØÇøÓÐ×ŷḻµÄº£Ë®×ÊÔ´£¬º£Ë®ÖÐÖ÷Òªº¬ÓÐNa£«¡¢K£«¡¢Ca2£«¡¢Mg2£«¡¢Cl£­¡¢SO42£­¡¢Br£­¡¢CO32£­¡¢HCO3£­µÈÀë×Ó¡£ºÏÀíÀûÓÃ×ÊÔ´ºÍ±£»¤»·¾³ÊǿɳÖÐø·¢Õ¹µÄÖØÒª±£Ö¤¡£

£¨1£©º£Ë®¾­¹ý´¦Àíºó¿ÉÒԵõ½ÎÞË®ÂÈ»¯Ã¾£¬ÎÞË®ÂÈ»¯Ã¾Êǹ¤ÒµÖÆȡþµÄÔ­ÁÏ¡£ÊÔд³öµç½âÈÛÈÚÂÈ»¯Ã¾ÖÆÈ¡½ðÊôþµÄ»¯Ñ§·´Ó¦·½³Ìʽ???????????????????????????????? ¡£

£¨2£©Ä³»¯¹¤³§Éú²ú¹ý³ÌÖлá²úÉúº¬ÓÐCu2£«ºÍPb2£«µÄÎÛË®¡£ÅÅ·ÅÇ°ÄâÓóÁµí·¨³ýÈ¥ÕâÁ½ÖÖÀë×Ó£¬¸ù¾ÝÏÂÁÐÊý¾Ý£¬ÄãÈÏΪͶÈë?????????? £¨Ñ¡Ìî¡°Na2S¡±»ò¡°NaOH¡±£©Ð§¹û¸üºÃ¡£

ÄÑÈܵç½âÖÊ

Cu(OH)2

CuS

Pb(OH)2

PbS

Ksp

4£®8¡Á10£­20

6£®3¡Á10£­36

1£®2¡Á10£­15

1£®0¡Á10£­28

 

£¨3£©»ðÁ¦·¢µçÔÚÎÒ¹úµÄÄÜÔ´ÀûÓÃÖÐÕ¼½Ï´ó±ÈÖØ£¬µ«ÊÇÅŷųöµÄSO2»áÔì³ÉһϵÁл·¾³ºÍÉú̬ÎÊÌâ¡£ÀûÓú£Ë®ÍÑÁòÊÇÒ»ÖÖÓÐЧµÄ·½·¨£¬Æ乤ÒÕÁ÷³ÌÈçÏÂͼËùʾ£º

¢ÙÌìÈ»º£Ë®µÄpH¡Ö8£¬ÊÔÓÃÀë×Ó·½³Ìʽ½âÊÍÌìÈ»º£Ë®³ÊÈõ¼îÐÔµÄÔ­Òò????????? £¨ÈÎдһ¸ö£©¡£

¢ÚijÑо¿Ð¡×éΪ̽¾¿Ìá¸ßº¬ÁòÑÌÆøÖÐSO2ÎüÊÕЧÂʵĴëÊ©£¬½øÐÐÁËÌìÈ»º£Ë®ÎüÊÕº¬ÁòÑÌÆøµÄÄ£ÄâʵÑ飬ʵÑé½á¹ûÈçͼËùʾ¡£

ÇëÄã¸ù¾ÝͼʾʵÑé½á¹û£¬¾ÍÈçºÎÌá¸ßÒ»¶¨Å¨¶Èº¬ÁòÑÌÆøÖÐSO2µÄÎüÊÕЧÂÊ£¬Ìá³öÒ»ÌõºÏÀí»¯½¨Ò飺??? ¡£

¢ÛÌìÈ»º£Ë®ÎüÊÕÁ˺¬ÁòÑÌÆøºó»áÈÜÓÐH2SO3¡¢HSO3£­µÈ·Ö×Ó»òÀë×Ó£¬Ê¹ÓÃÑõÆø½«ÆäÑõ»¯µÄ»¯Ñ§Ô­ÀíÊÇ?????????????????? £¨ÈÎдһ¸ö»¯Ñ§·½³Ìʽ»òÀë×Ó·½³Ìʽ£©¡£Ñõ»¯ºóµÄ¡°º£Ë®¡±ÐèÒªÒýÈë´óÁ¿µÄÌìÈ»º£Ë®ÓëÖ®»ìºÏºó²ÅÄÜÅÅ·Å£¬¸Ã²Ù×÷µÄÖ÷ҪĿµÄÊÇ???????????????????????? ¡£

¢ò£®ÄÜÔ´ÊÇÈËÀàÉú´æºÍ·¢Õ¹µÄÖØÒªÖ§Öù¡£Ñо¿»¯Ñ§·´Ó¦¹ý³ÌÖеÄÄÜÁ¿±ä»¯ÔÚÄÜÔ´½ôȱµÄ½ñÌì¾ßÓÐÖØÒªµÄÀíÂÛÒâÒå¡£ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ

¢Ù2H2(g)+O2(g)£½2H2O(l)???? H£½£­570kJ/mol£»

¢ÚH2(g)+1/2O2(g)£½H2O(g)??? H£½£­242kJ/mol£»

¢ÛC(s)+1/2O2(g)£½CO(g)???? H£½¡ª110£®5kJ/moL£»

¢ÜC(s)+O2(g)£½CO2(g)??????? H£½¡ª393£®5kJ/moL£»

¢ÝCO2(g)+2H2O(g)£½2CH4(g)+2 O2(g)? H£½+890kJ/moL

»Ø´ðÏÂÁÐÎÊÌâ

£¨1£©ÉÏÊö·´Ó¦ÖÐÊôÓÚÎüÈÈ·´Ó¦µÄÊÇ??????????????? ¡£

£¨2£©H2µÄȼÉÕÈÈΪ¡÷H£½??????????????? ¡£

£¨3£©¸Ç˹¶¨ÂÉÔÚÉú²úºÍ¿ÆѧÑо¿ÖÐÓкÜÖØÒªµÄÒâÒå¡£ÓÐЩ·´Ó¦µÄ·´Ó¦ÈÈËäÈ»ÄÑÖ±½Ó²â¶¨£¬µ«¿Éͨ¹ý¼ä½ÓµÄ·½·¨ÇóµÃ¡£ÒÑÖªC(s) + H2O(g)£½H2(g)+ CO(g)???? H£½akJ/moL£»Ôòa£½???????? £»¸Ã·´Ó¦µÄìØS???????? 0(Ñ¡Ìî¡°£¾¡±¡¢¡°£½¡±¡¢¡°£¼¡±)£»ÒÑÖª×ÔÓÉÄÜG£½H¡ªTS£¬µ±G£¼0ʱ¿É×Ô·¢½øÐС£Ôò¸Ã·´Ó¦ÔÚʲôÌõ¼þÏ¿É×Ô·¢½øÐÐ__________________¡£

£¨4£©CO·ÖÎöÒÇÒÔȼÁϵç³ØΪ¹¤×÷Ô­Àí£¬Æä×°ÖÃÈçͼËùʾ£¬¸Ãµç³ØÖеç½âÖÊΪÑõ»¯îÆ£­Ñõ»¯ÄÆ£¬ÆäÖÐO2-¿ÉÒÔÔÚ¹ÌÌå½éÖÊNASICONÖÐ×ÔÓÉÒƶ¯¡£ÏÂÁÐ˵·¨´íÎóµÄÊÇ????? ¡£

A£®¸º¼«µÄµç¼«·´Ó¦Ê½Îª£ºCO+O2¡ª¨D2e-£½CO2

B£®¹¤×÷ʱµç¼«b×÷Õý¼«£¬O2¡ªÓɵ缫aÁ÷Ïòµç¼«b

C£®¹¤×÷ʱµç×ÓÓɵ缫aͨ¹ý´«¸ÐÆ÷Á÷Ïòµç¼«b

D£®´«¸ÐÆ÷ÖÐͨ¹ýµÄµçÁ÷Ô½´ó£¬Î²ÆøÖÐCOµÄº¬Á¿Ô½¸ß

 

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø