ÌâÄ¿ÄÚÈÝ

4£®Ä³ÌþAÏà¶ÔÓÚÇâÆøµÄÃܶÈΪ35£¬Ò»¶¨Á¿AÔÚ×ãÁ¿µÄÑõÆøÖÐÍêȫȼÉÕ£¬µÃµ½µÈÎïÖʵÄÁ¿µÄCO2ºÍH2O£¬AÄÜ·¢ÉúÈçÏÂת»¯£¨Í¼ÖеÄÎÞ»ú²úÎï¾ùÒÑÂÔÈ¥£©£®ÆäÖÐFÄÜ·¢ÉúÒø¾µ·´Ó¦£¬E²»ÄÜ·¢ÉúÒø¾µ·´Ó¦£»G¡¢H»¥ÎªÍ¬·ÖÒì¹¹Ì壬GÖÐÖ»º¬1¸ö¼×»ù£®
ÒÑÖª£º

£¨1£©ÓÃϵͳÃüÃû·¨¶ÔAÃüÃû2-¼×»ù-1-¶¡Ï©£®
£¨2£©CÓëHIO4µÄ·´Ó¦ÀàÐÍÊÇÑõ»¯·´Ó¦£®
£¨3£©Ð´³öFºÍGµÄ½á¹¹¼òʽ£ºFHCHO¡¢GCH2=C£¨COOH£©CH2CH3£®
£¨4£©Ð´³öB¡úCºÍD¡úIµÄ»¯Ñ§·½³Ìʽ£®
B¡úCCH2BrCBr£¨CH3£©CH2CH3+2NaOH$¡ú_{¡÷}^{H_{2}O}$CH2OHCOH£¨CH3£©CH2CH3+2NaBr£»
D¡úI2HOOCCOH£¨CH3£©CH2CH3$¡ú_{¡÷}^{ŨÁòËá}$+2H2O£®
£¨5£©IÓжàÖÖͬ·ÖÒì¹¹Ì壬д³öÂú×ãÒÔÏÂÌõ¼þµÄËùÓеÄͬ·ÖÒì¹¹Ì壺¢ÙÁùÔª»·£»¢Ú»·ÉϵÄÒ»Ïõ»ùÈ¡´úÎïÖ»ÓÐÒ»ÖÖ£»¢Û1 mol¸ÃÎïÖÊÓë×ãÁ¿µÄNaHCO3×÷Ó÷ųö2 molÆøÌ壻¢ÜºË´Å¹²ÕñÇâÆ×Ϊ3×é·å£®·ûºÏÌõ¼þµÄ½á¹¹¼òʽ£¨²»¿¼ÂÇÁ¢ÌåÒì¹¹£©£®

·ÖÎö ijÌþAÏà¶ÔÓÚÇâÆøµÄÃܶÈΪ35£¬¸ÃÌþµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª70£¬Ò»¶¨Á¿AÔÚ×ãÁ¿µÄÑõÆøÖÐÍêȫȼÉÕ£¬µÃµ½µÈÎïÖʵÄÁ¿µÄCO2ºÍH2O£¬ËµÃ÷CHÔ­×Ó¸öÊý±ÈΪ1£º2£¬ÔòAµÄ·Ö×ÓʽΪC5H10£¬¾ÝÒÑÖªÐÅÏ¢ºÍFÄÜ·¢ÉúÒø¾µ·´Ó¦£¬E²»ÄÜ·¢ÉúÒø¾µ·´Ó¦¿ÉÖª£¬CÄÜÑõ»¯Éú³ÉD£¬DµÄ·Ö×ÓʽΪC5H10O3£¬ÇÒÄÜ·¢Éú·Ö×ÓÄÚõ¥»¯·´Ó¦Éú³Éõ¥I£¬ËùÒÔAΪCH2=C£¨CH3£©CH2CH3£¬ÔòBΪCH2BrCBr£¨CH3£©CH2CH3£¬CΪCH2OHCOH£¨CH3£©CH2CH3£¬EΪCH3COCH2CH3£¬FΪHCHO£¬CÑõ»¯Éú³ÉD£¬DΪHOOCCOH£¨CH3£©CH2CH3£¬¾ÝGHµÄ·Ö×Óʽ¿ÉÖª£¬D¡úGHµÄ·´Ó¦ÊÇ´¼ôÇ»ùµÄÏûÈ¥·´Ó¦£¬GÖÐÖ»º¬1¸ö¼×»ù£¬GΪCH2=C£¨COOH£©CH2CH3£¬D·¢Éú·Ö×ÓÄÚõ¥»¯·´Ó¦Éú³Éõ¥I£¬IΪ£¬¾Ý´Ë·ÖÎö£®

½â´ð ½â£ºÄ³ÌþAÏà¶ÔÓÚÇâÆøµÄÃܶÈΪ35£¬¸ÃÌþµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª70£¬Ò»¶¨Á¿AÔÚ×ãÁ¿µÄÑõÆøÖÐÍêȫȼÉÕ£¬µÃµ½µÈÎïÖʵÄÁ¿µÄCO2ºÍH2O£¬ËµÃ÷CHÔ­×Ó¸öÊý±ÈΪ1£º2£¬ÔòAµÄ·Ö×ÓʽΪC5H10£¬¾ÝÒÑÖªÐÅÏ¢ºÍFÄÜ·¢ÉúÒø¾µ·´Ó¦£¬E²»ÄÜ·¢ÉúÒø¾µ·´Ó¦¿ÉÖª£¬CÄÜÑõ»¯Éú³ÉD£¬DµÄ·Ö×ÓʽΪC5H10O3£¬ÇÒÄÜ·¢Éú·Ö×ÓÄÚõ¥»¯·´Ó¦Éú³Éõ¥I£¬ËùÒÔAΪCH2=C£¨CH3£©CH2CH3£¬ÔòBΪCH2BrCBr£¨CH3£©CH2CH3£¬CΪCH2OHCOH£¨CH3£©CH2CH3£¬EΪCH3COCH2CH3£¬FΪHCHO£¬CÑõ»¯Éú³ÉD£¬DΪHOOCCOH£¨CH3£©CH2CH3£¬¾ÝGHµÄ·Ö×Óʽ¿ÉÖª£¬D¡úGHµÄ·´Ó¦ÊÇ´¼ôÇ»ùµÄÏûÈ¥·´Ó¦£¬GÖÐÖ»º¬1¸ö¼×»ù£¬GΪCH2=C£¨COOH£©CH2CH3£¬D·¢Éú·Ö×ÓÄÚõ¥»¯·´Ó¦Éú³Éõ¥I£¬IΪ£¬
£¨1£©AΪCH2=C£¨CH3£©CH2CH3£¬ÆäÃû³ÆΪ2-¼×»ù-1-¶¡Ï©£¬¹Ê´ð°¸Îª£º2-¼×»ù-1-¶¡Ï©£»
£¨2£©CΪCH2OHCOH£¨CH3£©CH2CH3£¬¾ÝÐÅÏ¢CÓëHIO4µÄ·´Ó¦ÀàÐÍÊÇÑõ»¯·´Ó¦£¬¹Ê´ð°¸Îª£ºÑõ»¯·´Ó¦£»
£¨3£©FΪHCHO£¬GΪCH2=C£¨COOH£©CH2CH3£¬¹Ê´ð°¸Îª£ºHCHO£»CH2=C£¨COOH£©CH2CH3£»
£¨4£©BΪCH2BrCBr£¨CH3£©CH2CH3£¬ÆäÔÚNaOHÈÜÒºÖÐË®½âÉú³ÉC£¬CΪCH2OHCOH£¨CH3£©CH2CH3£¬·´Ó¦·½³ÌʽΪCH2BrCBr£¨CH3£©CH2CH3+2NaOH$¡ú_{¡÷}^{H_{2}O}$CH2OHCOH£¨CH3£©CH2CH3+2NaBr£¬DΪHOOCCOH£¨CH3£©CH2CH3£¬D·¢Éú·Ö×ÓÄÚõ¥»¯·´Ó¦Éú³Éõ¥I£¬IΪ£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2HOOCCOH£¨CH3£©CH2CH3$¡ú_{¡÷}^{ŨÁòËá}$+2H2O£¬¹Ê´ð°¸Îª£ºCH2BrCBr£¨CH3£©CH2CH3+2NaOH$¡ú_{¡÷}^{H_{2}O}$CH2OHCOH£¨CH3£©CH2CH3+2NaBr£»2HOOCCOH£¨CH3£©CH2CH3$¡ú_{¡÷}^{ŨÁòËá}$+2H2O£»
£¨5£©IΪ£¬Æäͬ·ÖÒì¹¹Ìå1 mol¸ÃÎïÖÊÓë×ãÁ¿µÄNaHCO3×÷Ó÷ųö2 molÆøÌ壬˵Ã÷·Ö×ÓÖк¬ÓÐ2¸öôÈ»ù£¬½áºÏÆäËûÐÅÏ¢¿ÉÖª£¬Æäͬ·ÖÒì¹¹ÌåΪ£¬¹Ê´ð°¸Îª£º£®

µãÆÀ ±¾Ì⿼²éÓлúÎïµÄÍƶϣ¬ÌâÄ¿ÄѶÈÖеȣ¬±¾Ìâ×¢Òâ°ÑÎÕÓлúÎï·Ö×ÓʽµÄÈ·¶¨£¬ÍƶÏʱעÒâ°ÑÎÕÍ»ÆÆ¿Ú£¬°ÑÎÕÓлúÎï¹ÙÄÜÍŵÄÐÔÖÊÒÔ¼°×ª»¯£¬ÌرðÊÇÌâ¸øÐÅÏ¢£¬ÔÚ½â´ðʱҪעÒâÉóÌ⣮

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
3£®ÎÞ»úÑÎXÔÚÒ½Ò©¡¢»¯¹¤¡¢Å©ÒµµÈÁìÓò¶¼Óй㷺ÓÃ;£¬¹¤ÒµÉÏÒ»°ãͨ¹ýµ¥Öʼ×ÓëCa£¨OH£©2×ÇÒº¹²Èȵķ½·¨À´ÖƱ¸£®¾­²â¶¨£¬XÖк¬ÓÐÁ½ÖÖÔªËØ£¬Ïà¶ÔÖÊÁ¿Îª168£»È¡8.40gXÓë×ãÁ¿ÑÎËá·´Ó¦£¬²úÎïÖÐÓÐ0.15molµ¥Öʼ׺Í1.12LÆøÌåÒÒ£¨ÒÑÕÛËã³É±ê×¼×´¿ö£©£®ÒÑÖªÆøÌåÒÒÔÚ±ê×¼×´¿öϵÄÃܶÈԼΪ1.52g•L-1£¬ÔÚO2ÖÐȼÉյõ½ÎÞÉ«¡¢´Ì¼¤ÐÔÆøζÆøÌå±û£¬±ûÊÇÒ»ÖÖ³£¼ûµÄ´óÆøÎÛȾÎÇë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©XµÄ»¯Ñ§Ê½ÎªCaS4£¬Òҵĵç×ÓʽΪ£®
£¨2£©XÓë×ãÁ¿ÑÎËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCaS4+2HCl=CaCl2+3S¡ý+H2S¡ü£®
£¨3£©ÆøÌå±ûÓëµâË®·´Ó¦µÄÀë×Ó·½³ÌʽΪSO2+I2+2H2O=4H++2I-+SO42-£®
£¨4£©Ca£¨OH£©2×ÇÒºÓë¼×¹²ÈÈÊÇ·ñÖ»Éú³ÉXºÍË®£¿²¢ËµÃ÷ÀíÓÉ·ñ£¬¸ù¾ÝÑõ»¯»¹Ô­·´Ó¦¹æÂÉ£¬²úÎïÖÐÂÈÔªËصļÛ̬²»¿ÉÄÜÖ»½µµÍ²»Éý¸ß£®
£¨5£©ÊµÑéÊÒÓùýÁ¿NaOHÈÜÒºÎüÊÕ±ûʱ£¬Éú³É²úÎﶡ£¬¶¡ÖпÉÄÜ»ìÓÐ×é³ÉÔªËØÍêÈ«ÏàͬµÄÁíÒ»ÖÖÑÎÎ죬ÇáÉè¼ÆʵÑé·½°¸ÑéÖ¤ÎìÊÇ·ñ´æÔÚ£¿È¡ÉÙÁ¿ÊµÑé½áÊøºóµÄÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬ÏȼÓÈë×ãÁ¿ÑÎËáËữ£¬ÔٵμӼ¸µÎBaCl2ÈÜÒº£¬Èô³öÏÖ°×É«³Áµí£¬Ö¤Ã÷´æÔÚNa2SO4£¬·´Ö®ÔòÎÞ£®
9£®Åð¼°Æ仯ºÏÎïÔÚÏÖ´ú¹¤Òµ¡¢Éú»îºÍ¹ú·ÀÖÐÓÐÖØÒªÓ¦ÓüÛÖµ£®
£¨1£©ÅðÔ­×ӵĵç×ÓÅŲ¼Ê½ÊÇ1s22s22p1£®
£¨2£©×î¼òµ¥µÄÅðÍéÊÇB2H6£¨ÒÒÅðÍ飩£¬½á¹¹¼ûͼ£¬ÆäÖÐBÔ­×ÓµÄÔÓ»¯·½Ê½Îªsp3ÔÓ»¯£®
£¨3£©BF3ºÍBCl3¶¼ÓÐÇ¿ÁÒ½ÓÊܹµç×Ó¶ÔµÄÇãÏò£¬ÈçÈý·ú»¯ÅðÆøÌåÓë°±ÆøÏàÓö£¬Á¢¼´Éú³É°×É«¹ÌÌ壬д³ö¸Ã°×É«¹ÌÌå½á¹¹Ê½£¬²¢±ê×¢³öÆäÖеÄÅäλ¼ü£®
£¨4£©½üÄêÀ´£¬ÈËÃǿ϶¨ÅðÊÇÈ˺Ͷ¯Îï·úÖж¾µÄÖØÒª½â¶¾¼Á£®ÅðÔÚÌåÄÚ¿ÉÓë·úÐγÉÎȶ¨µÄÅäºÏÎïBF4-
¡¢£¬²¢ÒԺͷúÏàͬµÄ;¾¶²Î¼ÓÌåÄÚ´úл£¬µ«¶¾ÐԱȷúС£¬ÇÒÒ×ËæÄòÅųö£¬¹ÊÈÏΪÅð¶Ô·ú»¯Îï¾ßÓнⶾ×÷Óã®
£¨5£©¾­½á¹¹Ñо¿Ö¤Ã÷£¬ÅðËᾧÌåÖÐB£¨OH£©3µ¥Ôª½á¹¹Èçͼ£¨1£©Ëùʾ£®¸÷µ¥ÔªÖеÄÑõÔ­×Óͨ¹ýO-H¡­OÇâ¼üÁ¬½á³É²ã×´½á¹¹Èçͼ£¨2£©Ëùʾ£®²ãÓë²ãÖ®¼äÒÔ΢ÈõµÄ·Ö×Ó¼äÁ¦Ïà½áºÏ¹¹³ÉÕû¸öÅðËᾧÌ壮
¢ÙH3BO3ÊÇÒ»ÔªÈõËᣬд³öËüÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽH3BO3+H2O?[B£¨OH£©4]-+H+£¬
¢Ú¸ù¾Ý½á¹¹ÅжÏÏÂÁÐ˵·¨ÕýÈ·µÄÊÇbcdg£®
a£®ÅðËᾧÌåÊôÓÚÔ­×Ó¾§Ìå       b£®ÅðËᾧÌåÓÐÁÛƬ״µÄÍâ²ã
c£®ÅðËᾧÌåÊÇ·Ö×Ó¾§Ìå         d£®ÅðËᾧÌåÓл¬Äå¸Ð£¬¿É×÷È󻬼Á
e£®ÔÚB£¨OH£©3µ¥ÔªÖУ¬BÔ­×ÓÒÔsp3ÔÓ»¯¹ìµÀºÍÑõÔ­×Ó½áºÏ¶ø³É
f£®H3BO3·Ö×ÓµÄÎȶ¨ÐÔÓëÇâ¼üÓйØ
g£®º¬1mol H3BO3µÄ¾§ÌåÖÐÓÐ3molÇâ¼ü
h£®·Ö×ÓÖÐÅðÔ­×Ó×îÍâ²ãΪ8e-Îȶ¨½á¹¹
¢ÛÅðËá³£ÎÂÏÂΪ°×ɫƬ״¾§Ì壬ÈÜÓÚË®£¨273KʱÈܽâ¶ÈΪ6.35£©£¬ÔÚÈÈË®ÖÐÈܽâ¶ÈÃ÷ÏÔÔö´ó£¨373KʱΪ27.6£©£®Çë·ÖÎöÆäÖÐÔ­ÒòÊÜÈÈʱÅðËᾧÌåÖеĴóÁ¿Çâ¼üÓв¿·Ö¶ÏÁÑËùÖ£¬¾§ÌåÄÚÇâ¼ü²»ÀûÓÚÎïÖÊÈܽ⣮

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

¾«Ó¢¼Ò½ÌÍø