ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿¶ÌÖÜÆÚÖ÷×åÔªËØX¡¢Y¡¢Z¡¢MµÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬XÔ­×ÓºËÍâ×îÍâ²ãµç×ÓÊýÊÇÆäµç×Ó²ãÊýµÄ2±¶£¬X¡¢YµÄºËÍâ×îÍâ²ãµç×ÓÊýÖ®±ÈΪ2£º3¡£½ðÊôµ¥ÖÊZÔÚY2ÖÐȼÉÕÉú³ÉµÄ»¯ºÏÎï¿ÉÓëË®·¢ÉúÑõ»¯»¹Ô­·´Ó¦¡£µÄ×îÍâ²ãΪ8µç×ӽṹ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺.

£¨1£©XµÄ×î¼òµ¥Ç⻯ÎïµÄ·Ö×ÓʽΪ__________¡£

£¨2£©ZµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÊôÓÚ_________£¨Ìî¡°Ëᡱ»ò¡°¼î¡±£©¡£

£¨3£©ÈÈÎȶ¨ÐÔ£ºXµÄ×î¼òµ¥Ç⻯Îï±ÈYµÄ×î¼òµ¥Ç⻯Îï_______£¨Ìî¡°Ç¿¡±»ò¡°Èõ¡±£©¡£

£¨4£©½ðÊôµ¥ÖÊZÔÚY2ÖÐȼÉÕÉú³É»¯ºÏÎ1mol¸Ã»¯ºÏÎïÓëË®·´Ó¦Ê±×ªÒƵç×ÓÊýΪ_______mol¡£

£¨5£©Y¡¢MÁ½ÔªËØÖ®¼äÐγɵĻ¯ºÏÎ³£ÓÃ×÷Ë®µÄÏû¶¾¼ÁµÄÊÇ_______£¨Ìî·Ö×Óʽ£©¡£

¡¾´ð°¸¡¿CH4 ¼î Èõ 1 ClO2

¡¾½âÎö¡¿

XÔ­×ÓºËÍâ×îÍâ²ãµç×ÓÊýÊÇÆäµç×Ó²ãÊýµÄ2±¶£¬XÊÇCÔªËØ£»X¡¢YµÄºËÍâ×îÍâ²ãµç×ÓÊýÖ®±ÈΪ2£º3£¬ËµÃ÷YµÄ×îÍâ²ãµç×ÓÊýΪ6£¬YÊÇOÔªËØ£»½ðÊôµ¥ÖÊZÔÚY2ÖÐȼÉÕÉú³ÉµÄ»¯ºÏÎï¿ÉÓëË®·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬ËµÃ÷ZÊÇNaÔªËØ£»µÄ×îÍâ²ãΪ8µç×ӽṹ£¬ËµÃ÷MÊÇClÔªËØ¡£

£¨1£©¸ù¾Ý·ÖÎö£¬XÊÇCÔªËØ£¬CµÄ×î¼òµ¥Ç⻯ÎïµÄ·Ö×ÓʽΪCH4£»

£¨2£©¸ù¾Ý·ÖÎö£¬ZÊÇNaÔªËØ£¬NaµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÊÇNaOH£¬ÊÇÇ¿¼î£»

£¨3£©XÊÇCÔªËØ£¬YÊÇOÔªËØ£¬·Ç½ðÊôÐÔԽǿ£¬Ç⻯ÎïÔ½Îȶ¨£¬·Ç½ðÊôÐÔC£¼O£¬¹ÊCH4µÄÈÈÎȶ¨ÐÔÈõÓÚH2O£»

£¨4£©NaÔÚO2ÖÐȼÉÕÉú³ÉµÄNa2O2ÓëË®·´Ó¦2Na2O2+2H2O=4NaOH+O2 ¡ü£¬Na2O2ÖÐ-1¼ÛµÄO·¢ÉúÆ绯£¬±äΪ0¼ÛºÍ-2¼Û£¬¹Ê1mol Na2O2ÓëË®·´Ó¦Ê±×ªÒƵç×ÓÊýΪ1mol£»

£¨5£©ÂȵÄÑõ»¯ÎïÖУ¬ClO2³£×÷ΪˮµÄÏû¶¾¼Á¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ï±íÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬±íÖеÄÿ¸ö×Öĸ´ú±íÒ»ÖÖÔªËØ£¬Çë¸ù¾ÝÒªÇó»Ø´ðÎÊÌâ¡£

×å

ÖÜÆÚ

¢ñA

0

1

a

¢òA

¢óA

¢ôA

¢õA

¢öA

¢÷A

2

b

c

d

3

e

f

g

£¨1£©ÔªËØgÔÚÔªËØÖÜÆÚ±íµÄλÖÃΪ____________________¡£

£¨2£©bºÍgÁ½ÖÖÔªËصÄÔ­×Ӱ뾶´óС¹Øϵ£ºb______g£¨Ìî¡°>¡±»ò¡°<¡±£©.

£¨3£©ÓÉÔ­×Ó¸öÊý±ÈΪ1£º1£º1µÄa¡¢b¡¢cÈýÖÖÔªËØ×é³ÉµÄ¹²¼Û»¯ºÏÎïX£¬¹²ÐγÉ4¶Ô¹²Óõç×Ó¶Ô£¬ÔòXµÄ½á¹¹Ê½Îª______________¡£

£¨4£©fµÄ×î¸ß¼ÛÑõ»¯ÎïÓëeµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÔÚÈÜÒºÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________¡£

£¨5£©A¡¢B¡¢D¡¢EÊÇÓÉÉÏÊö²¿·ÖÔªËØ×é³ÉµÄ»¯ºÏÎËüÃÇÖ®¼äµÄת»¯¹ØϵÈçͼËùʾ£¨²¿·Ö²úÎïÒÑÂÔÈ¥£©¡£A¡¢B¡¢DµÄÑæÉ«·´Ó¦¾ù³Ê»ÆÉ«£¬Ë®ÈÜÒº¾ùΪ¼îÐÔ¡£Çë»Ø´ð£º

¢ÙEµÄµç×ÓʽΪ_______________£¬BµÄ»¯Ñ§Ê½Îª____________________¡£

¢ÚAÖеĻ¯Ñ§¼üÀàÐÍΪ____________________

¢Û×ÔÈ»½çÖдæÔÚB¡¢DºÍH2O°´Ò»¶¨±ÈÀý½á¾§¶ø³ÉµÄ¹ÌÌ塣ȡһ¶¨Á¿¸Ã¹ÌÌåÈÜÓÚË®Åä³É100mLÈÜÒº£¬²âµÃÈÜÒºÖнðÊôÑôÀë×ÓµÄŨ¶ÈΪ0.5mo/L¡£ÈôÈ¡ÏàͬÖÊÁ¿µÄ¹ÌÌå¼ÓÈÈÖÁÖÊÁ¿²»ÔÙ·¢Éú±ä»¯£¬Ê£Óà¹ÌÌåµÄÖÊÁ¿Îª___________g¡£

¡¾ÌâÄ¿¡¿¢ñ.£¨1£©298Kʱ£¬0.5 mol C2H4 (g)ÍêȫȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍҺ̬ˮ£¬·Å³ö705.5kJµÄÈÈÁ¿¡£Çëд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ______________________¡£

£¨2£©Na2SO3¾ßÓл¹Ô­ÐÔ£¬ÆäË®ÈÜÒº¿ÉÒÔÎüÊÕCl2(g)£¬¼õÉÙ»·¾³ÎÛȾ¡£

ÒÑÖª·´Ó¦£º Na2SO3(aq)+Cl2(g)+H2O(l)£½Na2SO4(aq)+2HCl(aq) ¦¤H1=a kJ¡¤mol1

Cl2(g)+H2O(l)£½HCl(aq)+HClO(aq) ¦¤H2=b kJ¡¤mol1

ÊÔд³öNa2SO3(aq)ÓëHClO(aq)·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º________________________¡£

¢ò.ºì·¯ÄÆ£¨Na2Cr2O7¡¤2H2O£©ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬¹¤ÒµÉÏÓøõÌú¿ó£¨Ö÷Òª³É·ÖÊÇFeO¡¤Cr2O3£©ÖƱ¸ºì·¯ÄƵĹý³ÌÖлᷢÉúÈçÏ·´Ó¦£º4FeO(s)£«4Cr2O3(s)£«8Na2CO3(s)£«7O2(g) 8Na2CrO4(s)£«2Fe2O3(s)£«8CO2(g)¡¡¦¤H<0

£¨1£©Çëд³öÉÏÊö·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽ£ºK£½_____________________¡£

£¨2£©Èçͼ1Ëùʾ£¬ÔÚ0~2minÄÚCO2µÄƽ¾ù·´Ó¦ËÙÂÊΪ____________________¡£

£¨3£©Í¼1¡¢Í¼2±íʾÉÏÊö·´Ó¦ÔÚ2minʱ´ïµ½Æ½ºâ¡¢ÔÚ4minʱÒò¸Ä±äij¸öÌõ¼þ¶ø·¢Éú±ä»¯µÄÇúÏß¡£ÓÉͼ1Åжϣ¬·´Ó¦½øÐÐÖÁ4minʱ£¬ÇúÏß·¢Éú±ä»¯µÄÔ­ÒòÊÇ______________£¨ÓÃÎÄ×Ö±í´ï£©£»ÓÉͼ2Åжϣ¬4minµ½6minµÄÇúÏ߱仯µÄÔ­Òò¿ÉÄÜÊÇ________£¨ÌîдÐòºÅ£©¡£

a.Éý¸ßÎÂ¶È b£®¼Ó´ß»¯¼Á c£®Í¨ÈëO2 d£®ËõСÈÝÆ÷Ìå»ý

£¨4£©¹¤ÒµÉÏ¿ÉÓÃÉÏÊö·´Ó¦Öеĸ±²úÎïCO2À´Éú²ú¼×´¼£ºCO2(g)£«3H2(g) CH3OH(g)£«H2O(g)¡£

ÒÑÖª¸Ã·´Ó¦ÄÜ×Ô·¢½øÐУ¬ÔòÏÂÁÐͼÏñÕýÈ·µÄÊÇ____________¡£

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

¾«Ó¢¼Ò½ÌÍø