ÉÂÎ÷Ê¡Î÷°²Êнì¸ßÈýÄ£Ä⣨һ£©Àí×Û»¯Ñ§ÊÔÌâWord°æº¬´ð°¸ - ͼÎÄ

·¢²¼Ê±¼ä : ÐÇÆÚÈý ÎÄÕÂÉÂÎ÷Ê¡Î÷°²Êнì¸ßÈýÄ£Ä⣨һ£©Àí×Û»¯Ñ§ÊÔÌâWord°æº¬´ð°¸ - ͼÎĸüÐÂÍê±Ï¿ªÊ¼ÔĶÁ

£¨3£©Ó÷ϵç³ØµÄпƤ¿ÉÓÃÓÚ»ØÊÕÖÆ×÷ZnSO4¡¤7H2O¡£¹ý³ÌÖУ¬Ðè³ýȥпƤÖеÄÉÙÁ¿ÔÓÖÊÌú£¬Æä·½·¨ÊÇ£º³£ÎÂÏ£¬¼ÓÈëÏ¡H2SO4ºÍH2O2£¬ÌúÈܽâ±äΪFe£¬¼Ó¼îµ÷½ÚpHΪ4£¬Ê¹ÈÜÒºÖеÄFeת»¯ÎªFe(OH)3³Áµí£¬´ËʱÈÜÒºÖÐc(Fe)=_________¡£¼ÌÐø¼Ó¼îµ÷½ÚpHΪ____ʱ£¬Ð¿¿ªÊ¼³Áµí£¨¼Ù¶¨ZnŨ¶ÈΪ0.1mol/L£©¡£²¿·ÖÄÑÈܵĵç½âÖÊÈܶȻý³£Êý£¨Ksp£©ÈçÏÂ±í£º

2+

3+

3+

3+

28.ijºìÉ«¹ÌÌå·ÛÄ©ÑùÆ·¿ÉÄܺ¬ÓÐFe2O3ºÍCu2OÖеÄÒ»ÖÖ»òÁ½ÖÖ£¬Ä³»¯Ñ§ÐËȤС×é¶ÔÆä×é³É

½øÐÐ̽¾¿¡£Íê³ÉÏÂÁпոñ¡£ ¢ÙÌá³ö¼ÙÉ裺

¼ÙÉè1£ºÖ»´æÔÚFe2O3£»¼ÙÉè2£º_________£»¼ÙÉè3£º¼È´æÔÚFe2O3Ò²´æÔÚCu2O¡£ ¢Ú²éÕÒ×ÊÁÏ£ºCu2OÔÚËáÐÔÈÜÒºÖлᷢÉú·´Ó¦£ºCu2O+2H=Cu+Cu+H2O¡£ ¢ÛʵÑé·½°¸Éè¼ÆÓë·ÖÎö£º

·½°¸Ò»£º²½ÖèÒ»£ºÈ¡ÉÙÁ¿ÑùÆ·ÓÚÉÕ±­ÖУ¬¼ÓÈë¹ýÁ¿Å¨ÏõËᣬ²úÉúÒ»ÖÖºì×ØÉ«µÄÆøÌå¡£Óɴ˿ɵóö¼ÙÉè____²»³ÉÁ¢£¬Ð´³ö²úÉúÉÏÊöÆøÌåµÄ»¯Ñ§·½³Ìʽ___________________¡£ ²½Öè¶þ£ºÈ¡ÉÙÁ¿²½ÖèÒ»ÈÜÒºÖÃÓÚÊÔ¹ÜÖеμÓ_______£¬Õñµ´£¬Èô________£¬Ôò˵Ã÷¼ÙÉè3³ÉÁ¢¡£ ·½°¸¶þ£º

È¡ÉÙÁ¿ÑùÆ·ÓÚÉÕ±­ÖУ¬¼ÓÈë¹ýÁ¿Ï¡ÁòËᣬÈô¹ÌÌåÈ«²¿Èܽ⣬˵Ã÷¼ÙÉè£ß²»³ÉÁ¢¡£ ·½°¸Èý£º

+

2+

ͬѧÃÇÉè¼ÆÁËÈçÏÂʵÑé·½°¸²â¶¨¸ÃÑùÆ·ÖÐFe2O3µÄÖÊÁ¿·ÖÊý£¨×°ÖÃÆøÃÜÐÔÁ¼ºÃ£¬¼ÙÉèÑùÆ·ÍêÈ«·´Ó¦£©:

²½ÖèÒ»£ºÈ¡ÑùÆ·²¢³ÆÁ¿¸ÃÑùÆ·µÄÖÊÁ¿Îªm1£» ²½Öè¶þ£º²â³ö·´Ó¦Ç°¹ã¿ÚÆ¿ºÍÆ¿ÄÚÎïÖÊ×ÜÖÊÁ¿m2£» ²½ÖèÈý£º²â³ö·´Ó¦ºó¹ã¿ÚÆ¿ºÍÆ¿ÄÚÎïÖÊ×ÜÖÊÁ¿m3£» ²½ÖèËÄ£º¼ÆËãµÃ³ö¿óÎïÑùÆ·ÖÐFe2O3µÄÖÊÁ¿·ÖÊý¡£

ÌÖÂÛ·ÖÎö£º¸ÃʵÑé·½°¸________£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©²â³ö¿óÎïÖÐFe2O3µÄÖÊÁ¿·ÖÊý¡£²»¸Ä±ä×°ÖúÍÒ©Æ·£¬Í¨¹ý¼ÆËãµÃ³ö¿óÎïÖÐFe2O3µÄÖÊÁ¿·ÖÊý£¬Ä㻹¿ÉÒÔͨ¹ý²â¶¨_______¡£Èô²âµÃm1Ϊ3.04g£¬m3=m2£«1.76g£¬ÔòÔ­ÑùÆ·ÖÐFe2O3µÄÖÊÁ¿·ÖÊýΪ_____£¨½á¹û±£ÁôËÄλÓÐЧÊý×Ö£©¡£

35£®¡¾»¯Ñ§¨DÑ¡ÐÞ3£ºÎïÖʽṹÓëÐÔÖÊ¡¿£¨15·Ö£©

A¡¢B¡¢C¡¢D¡¢EÊÇÔªËØÖÜÆÚ±íÖÐÇ°36ºÅÔªËØ£¬ºËµçºÉÊýÒÀ´ÎÔö´ó£¬AÓëBÄÜÐγÉÖÖÀà·±¶àµÄ»¯ºÏÎDÔ­×ÓÖгɶԵç×ÓÊýµÈÓÚδ³É¶Ôµç×ÓÊýµÄ3±¶£»EÖÐËùÓеç×ÓÕýºÃ³äÂúK¡¢L¡¢MÈý¸öµç×Ӳ㡣

£¨1£©»ù̬CÔ­×ÓºËÍâÓÐ_____ÖÖÔ˶¯×´Ì¬²»Í¬µÄµç×Ó£¬ÔªËØCµÄÆø̬Ç⻯ÎïµÄ¿Õ¼ä¹¹ÐÍΪ____¡£

£¨2£©B¡¢C¡¢DÈýÖÖÔªËصĵÚÒ»µçÀëÄÜÓÉ´óµ½Ð¡µÄ˳ÐòΪ_________¡££¨ÓÃÔªËØ·ûºÅ±íʾ£© £¨3£©AÓëBÐγɵĻ¯ºÏÎïB2A2ÖÐBÔ­×ÓµÄÔÓ»¯·½Ê½Îª____£¬·Ö×ÓÖк¬ÓеĦҼüºÍ¦Ð¼ü¸öÊý·Ö±ðÊÇ______¡¢_______¡£

£¨4£©DÓëÄÆÔªËØÐγɵÄÔ­×ÓÊýÖ®±ÈΪl:1µÄÎïÖÊÖоßÓеĻ¯Ñ§¼üÀàÐÍΪ______¡£ £¨5£©EÓëCµÄ¼òµ¥Àë×ÓÐγɾ§ÌåµÄ¾§°û½á¹¹Èçͼ1Ëùʾ£¬Í¼Öа×Çò±íʾ_______¡£ £¨6£©EµÄµ¥Öʾ§ÌåµÄ¾§°û½á¹¹Èçͼ2Ëùʾ£¬Æä¿Õ¼äÀûÓÃÂÊΪ_____£¨Ô²ÖÜÂÊÓæбíʾ£¬

£©

+

+

36£®¡¾»¯Ñ§¨DÑ¡ÐÞ5£ºÓлú»¯Ñ§»ù´¡¡¿£¨15·Ö£©

ÒÔÈâ¹ðËáÒÒõ¥MΪԭÁÏ£¬¾­¹ýÏà¹Ø»¯Ñ§·´Ó¦ºÏ³ÉµÄ¿¹°©Ò©¶ÔÖÎÁÆÈéÏÙ°©ÓÐ×ÅÏÔÖøµÄÁÆЧ¡£ÒÑÖªMÄÜ·¢ÉúÈçÏÂת»¯£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©AµÄ½á¹¹¼òʽΪ________£¬EÖк¬ÓйÙÄÜÍŵÄÃû³ÆÊÇ________¡£ £¨2£©Ð´³ö·´Ó¦¢ÛºÍ¢ÞµÄ»¯Ñ§·½³Ìʽ£º________¡¢________¡£

£¨3£©ÔÚºÏÊʵĴ߻¯¼ÁÌõ¼þÏ£¬ÓÉE¿ÉÒÔÖƱ¸¸ß·Ö×Ó»¯ºÏÎïH£¬HµÄ½á¹¹¼òʽΪ________£¬ÓÉEµ¼HµÄ·´Ó¦ÀàÐÍΪ________¡£

£¨4£©·´Ó¦¢Ù¡«¢ÞÖУ¬ÊôÓÚÈ¡´ú·´Ó¦µÄÊÇ_______________¡£

£¨5£©IÊÇBµÄͬ·ÖÒì¹¹Ì壬ÇҺ˴Ź²ÕñÇâÆ×ÖÐÖ»ÓÐÒ»¸öÎüÊշ壬IµÄ½á¹¹¼òʽΪ______¡£ £¨6£©1molAÓëÇâÆøÍêÈ«·´Ó¦£¬ÐèÒªÇâÆø_______L£¨±ê×¼×´¿öÏ£©¡£

£¨7£©AµÄͬ·ÖÒì¹¹ÌåÓжàÖÖ£¬ÆäÖÐÊôÓÚ·¼Ïã×廯ºÏÎ¼ÈÄÜʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«£¬ÓÖÄÜÓë̼ËáÇâÄÆÈÜÒº·´Ó¦µÄͬ·ÖÒì¹¹ÌåÓÐ_______ÖÖ£¨²»°üº¬A£©¡£

7-13£ºBDCAC BA

26£®£¨14·Ö£©(l) CH4(g)+2NO2(g)= N2(g)+ CO2(g)+ 4H2O(l) ¡÷H=-955kJ¡¤mol(2·Ö£©

£¨2£©¢Ù2NO+2CO=N2+2CO2£¨2·Ö£©¢Ú0.005mol L min£¨1·Ö£© 50%£¨1·Ö£© 5£¨2·Ö£© ¢Ûc£¨2·Ö£© ¢Üa b£¨2·Ö£©

£¨3£©CH3OCH3-12e+16OH=2CO3+11H2O£¨2·Ö£©

27£®£¨15·Ö£©£¨1£©¢ÙÔö´ó½Ó´¥Ãæ»ý£¬¼Ó¿ì·´Ó¦ËÙÂÊ¡££¨1·Ö£© ¢Ú¹ýÂË£¨1·Ö£© ÒýÁ÷£¨1·Ö£©

¢Û³ýȥ̿·Û£¨2·Ö£© £¨2£©¢Ù2MnOOH+6HCl=2MnCl2+Cl2¡ü+4H2O£¨2·Ö£©

¢Ú2MnOOH+H2C2O4+2H2SO4=2MnSO4+2CO2¡ü+4H2O£¨2·Ö£©¹¤ÒÕÁ÷³Ì¼òµ¥£ºÉú³ÉCO2ºÍH2O²»Ó°ÏìMnSO4´¿¶È£»·´Ó¦¹ý³ÌÎÞÓж¾Óк¦ÎïÖÊÉú³É£¬²»Ôì³É¶þ´ÎÎÛȾ£»·ÏÎï×ÊÔ´»¯µÈ£¨´ð1µã¼´¿É£©£¨2·Ö£©

£¨3£©2.6¡Á10£¨2·Ö£©6 £¨2·Ö£©

28£®£¨14·Ö£©Ö»´æÔÚCu2O£¨l·Ö£©1£¨1·Ö£©Cu2O+6HNO3(Ũ)=2Cu(NO3)2+2NO2¡ü+3H2O£¨2·Ö£©

KSCNÈÜÒº£¨1·Ö£© ÈÜÒºÏÔºìÉ«£¨1·Ö£© 2£¨2·Ö£© ÄÜ£¨2·Ö£©²âÁ¿·´Ó¦ºó¹ÌÌåµÄÖÊÁ¿»ò¹ã¿ÚÆ¿ÖгÁµíµÄÖÊÁ¿¡££¨ÆäËûºÏÀí´ð°¸Ò²¸ø·Ö£©£¨2·Ö£© 52.63%£¨2·Ö£© 35£®¡¾Ñ¡ÐÞÈý¡ª¡ªÎïÖʽṹÓëÐÔÖÊ¡¿£¨15·Ö£©

£¨1£©7£¨2·Ö£© Èý½Ç׶ÐΣ¨1·Ö£© £¨2£©N>O>C£¨2·Ö£© £¨3£©SP£¨2·Ö£© 3£¨1·Ö£©2£¨1·Ö£©

£¨4£©Àë×Ó¼ü£¨1·Ö£© ·Ç¼«ÐÔ¹²¼Û¼ü£¨1·Ö£© £¨5£©N£¨2·Ö£© £¨6£©36£®¡¾Ñ¡ÐÞÎ塪¡ªÓлú»¯Ñ§»ù´¡¡¿£¨15·Ö£©

£¨1£©

£¨2·Ö£©£¬ôÇ»ù ôÈ»ù£¨2·Ö£©

3+

-9

--2--1

-1

-l

2? 6£¨2£©2CH3CH2OH+O2

2CH3CHO+2H2O£¨2·Ö£©

£¨2·Ö£©

£¨3£©£¨¸÷1·Ö£© £¨4£©¢Ù¢Û¢Ü¢Ý£¨2·Ö£©

£¨5£©CH3-O-CH3£¨1·Ö£© £¨6£©89.6£¨1·Ö£© £¨7£©4£¨1·Ö£©

ÁªÏµºÏͬ·¶ÎÄ¿Í·þ£ºxxxxx#qq.com(#Ì滻Ϊ@)