ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÄαØÂå¶ûÊÇÒ»ÖÖÓÃÓÚѪ¹ÜÀ©ÕŵĽµÑªÑ¹Ò©ÎһÖֺϳÉÄαØÂå¶ûÖмäÌåGµÄ²¿·ÖÁ÷³ÌÈçÏ£º

ÒÑÖª£ºÒÒËáôûµÄ½á¹¹¼òʽΪ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©AµÄÃû³ÆÊÇ______£»BÖÐËùº¬¹ÙÄÜÍŵÄÃû³ÆÊÇ______¡£

£¨2£©·´Ó¦¢ÝµÄ»¯Ñ§·½³ÌʽΪ______£¬¸Ã·´Ó¦µÄ·´Ó¦ÀàÐÍÊÇ______¡£

£¨3£©GµÄ·Ö×ÓʽΪ______¡£

£¨4£©Ð´³öÂú×ãÏÂÁÐÌõ¼þµÄEµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º______¡¢______¡£

¢ñ£®±½»·ÉÏÖ»ÓÐÈý¸öÈ¡´ú»ù

¢òºË´Å¹²ÕñÇâÆ×ͼÖÐÖ»ÓÐ4×éÎüÊÕ·å

¢ó.1mol¸ÃÎïÖÊÓë×ãÁ¿NaHCO3ÈÜÒº·´Ó¦Éú³É2molCO2

£¨5£©¸ù¾ÝÒÑÓÐ֪ʶ²¢½áºÏÏà¹ØÐÅÏ¢£¬Ð´³öÒÔΪԭÁÏÖƱ¸µÄºÏ³É·ÏßÁ÷³Ìͼ£¨ÎÞ»úÊÔ¼ÁÈÎÑ¡£©______£¬ºÏ³É·ÏßÁ÷³ÌͼʾÀýÈçÏ£ºCH3CH2BrCH3CH2OHCH3COOCH2CH3

¡¾´ð°¸¡¿¶Ô·ú±½·Ó£¨4-·ú±½·Ó£© ·úÔ­×Ó¡¢õ¥»ù ÏûÈ¥·´Ó¦ C10H9O3F

¡¾½âÎö¡¿

(1)·úÔ­×ÓÓëôÇ»ùÔÚ±½»·µÄ¶ÔλÉÏ£¬Ãû³ÆΪ£º¶Ô·ú±½·Ó£¨4-·ú±½·Ó£©£»BÖк¬ÓеĹÙÄÜÍÅΪõ¥»ù¡¢·úÔ­×Ó£»

£¨2£©Í¨¹ý¶Ô±ÈE¡¢G£¬EÖеĴ¼ôÇ»ù±äΪÇâÔ­×Ó£¬Ôò¢ÝΪ´¼ôÇ»ùµÄÏûÈ¥·´Ó¦£»

£¨3£©·Ö×ÓGÖк¬Óб½»ù¡¢ôÈ»ù£¬Ö§Á´»¹ÓÐ3¸ö̼ԭ×Ó£¬Æä·Ö×ÓʽΪ£ºC10H9O3F£»

£¨4£©EµÄ·Ö×ÓʽΪ£ºC10H9O4F£¬ 1mol¸ÃÎïÖÊÓë×ãÁ¿ NaHCO3ÈÜÒº·´Ó¦Éú³É2 mol CO2£¬Æäͬ·ÖÒì¹¹ÌåÖк¬ÓÐ2¸öôÈ»ù£¬ºË´Å¹²ÕñÇâÆ×ͼÖÐÖ»ÓÐ4×éÎüÊշ壬±½»·ÉÏÖ»ÓÐÈý¸öÈ¡´ú»ù£¬ÔòÈ¡´ú»ùΪ¶Ô³Æ½á¹¹£¬´ð°¸Îª£º¡¢£»

£¨5£©ÒÑÖª±½·ÓÓëäåË®·´Ó¦Éú³É2£¬4£¬6-Èýäå±½·Ó£¬ÔòÒýÈëäåÔ­×ÓÇ°Ó¦ÏȽøÐÐÌâ¸ÉÖТڵķ´Ó¦£»

(1)·úÔ­×ÓÓëôÇ»ùÔÚ±½»·µÄ¶ÔλÉÏ£¬Ãû³ÆΪ£º¶Ô·ú±½·Ó£¨4-·ú±½·Ó£©£»BÖк¬ÓеĹÙÄÜÍÅΪõ¥»ù¡¢·úÔ­×Ó£»

£¨2£©Í¨¹ý¶Ô±ÈE¡¢G£¬EÖеĴ¼ôÇ»ù±äΪÇâÔ­×Ó£¬Ôò¢ÝΪ´¼ôÇ»ùµÄÏûÈ¥·´Ó¦£¬»¯Ñ§·½³ÌʽΪ£º£»·´Ó¦ÀàÐÍΪ£ºÏûÈ¥·´Ó¦£»

£¨3£©·Ö×ÓGÖк¬Óб½»ù¡¢ôÈ»ù£¬Ö§Á´»¹ÓÐ3¸ö̼ԭ×Ó£¬Æä·Ö×ÓʽΪ£ºC10H9O3F£»

£¨4£©EµÄ·Ö×ÓʽΪ£ºC10H9O4F£¬ 1mol¸ÃÎïÖÊÓë×ãÁ¿ NaHCO3ÈÜÒº·´Ó¦Éú³É2 mol CO2£¬Æäͬ·ÖÒì¹¹ÌåÖк¬ÓÐ2¸öôÈ»ù£¬ºË´Å¹²ÕñÇâÆ×ͼÖÐÖ»ÓÐ4×éÎüÊշ壬±½»·ÉÏÖ»ÓÐÈý¸öÈ¡´ú»ù£¬ÔòÈ¡´ú»ùΪ¶Ô³Æ½á¹¹£¬´ð°¸Îª£º¡¢£»

£¨5£©ÒÑÖª±½·ÓÓëäåË®·´Ó¦Éú³É2£¬4£¬6-Èýäå±½·Ó£¬ÔòÒýÈëäåÔ­×ÓÇ°Ó¦ÏȽøÐÐÌâ¸ÉÖТڵķ´Ó¦£¬¹Ê·´Ó¦Á÷³ÌΪ£º

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿îéËáÄÆ( NaBiO3 )ÊÇÒ»ÖÖÄÑÈÜÓÚË®µÄÇ¿Ñõ»¯¼Á£¬ÔÚ¸ÖÌú¹¤ÒµÖг£ÓÃ×÷ÃÌÔªËصķÖÎö²â¶¨¡£Ä³Ñо¿Ð¡×éÓø¡Ñ¡¹ýµÄ»Ôîé¿ó(Ö÷Òª³É·ÖÊÇBi2S3£¬»¹º¬ÉÙÁ¿Bi2O3£¬SiO2µÈÔÓÖÊ)ÖƱ¸îéËáÄÆ£¬ÆäÁ÷³ÌÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ΪÁËÌá¸ß¡°½þÈ¡¡±ÖÐÔ­ÁϵĽþ³öÂÊ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊÇ_________________(дһÖÖ¼´¿É) ¡£

(2)¡°½þÈ¡¡±Ê±Í¨³£¼ÓÈëFeCl3ÈÜÒººÍŨÑÎËᣬÏòÆäÖмÓÈë¹ýÁ¿Å¨ÑÎËáµÄÄ¿µÄÊÇ_____£¬¡°ÂËÔü¡±µÄ³É·ÝÊÇ____________(Ìѧʽ)¡£

(3)¡°³Áµí¡±·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________________________£»

(4)¡°³Áµí¡±²Ù×÷ʱ£¬Óð±Ë®µ÷½ÚpHÖÁ6£¬Í¨¹ý¼ÆËã˵Ã÷£¬´ËʱÈÜÒºÖеÄBi3+ÊÇ·ñÍêÈ«³Áµí£º____________________(ÒÑÖª£ºBi(OH)3µÄÈܶȻýKsp=3¡Á10-32) ¡£

(5)¡°±ºÉÕ¡±³ýÁ˲ÉÓÃ×î¼ÑµÄÖÊÁ¿±È¡¢ºÏÊʵÄζÈÍ⣬ÄãÈÏΪ»¹ÐèÒª¿ØÖƵÄÌõ¼þÊÇ________¡£

(6)ÒÑÖª£¬ÔÚËáÐÔÈÜÒºÖÐNaBiO3½«Mn2+Ñõ»¯ÎªMnO4-¡£Çë³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º__________¡£

(7)ij»Ôîé¿óÖÐîéÔªËصÄÖÊÁ¿·ÖÊýΪ20.90%£¬Èô100¶Ö¸Ã»Ôîé¿óÍêÈ«ÓÃÓÚÉú²ú£¬¹²µÃµ½25.00¶ÖNaBiO3£¬Ôò²úÂÊÊÇ___________¡£

¡¾ÌâÄ¿¡¿£¨1£©¹ãÖÝÑÇÔ˻ᡰ³±Á÷¡±»ð¾æÄÚÐÜÐÜ´ó»ðÀ´Ô´ÓÚ±ûÍéµÄȼÉÕ£¬±ûÍéÊÇÒ»ÖÖÓÅÁ¼µÄȼÁÏ¡£ÊԻشðÏÂÁÐÎÊÌ⣺

¢ÙÈçͼÊÇÒ»¶¨Á¿±ûÍéÍêȫȼÉÕÉú³ÉCO2ºÍ1 mol H2O(l)¹ý³ÌÖеÄÄÜÁ¿±ä»¯Í¼£¬ÇëÔÚͼÖеÄÀ¨ºÅÄÚÌîÈë¡°£«¡±»ò¡°£­¡±_____¡£

¢Úд³ö±íʾ±ûÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ£º___________________________________¡£

¢Û¶þ¼×ÃÑ(CH3OCH3)ÊÇÒ»ÖÖÐÂÐÍȼÁÏ£¬Ó¦ÓÃÇ°¾°¹ãÀ«¡£1 mol¶þ¼×ÃÑÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ·Å³ö1 455 kJÈÈÁ¿¡£Èô1 mol±ûÍéºÍ¶þ¼×ÃѵĻìºÏÆøÌåÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ¹²·Å³ö1835 kJÈÈÁ¿£¬Ôò»ìºÏÆøÌåÖУ¬±ûÍéºÍ¶þ¼×ÃѵÄÎïÖʵÄÁ¿Ö®±ÈΪ_______¡£

£¨2£©¿Æѧ¼Ò¸Ç˹ÔøÌá³ö£º¡°²»¹Ü»¯Ñ§¹ý³ÌÊÇÒ»²½Íê³É»ò·Ö¼¸²½Íê³É£¬Õâ¸ö×ܹý³ÌµÄÈÈЧӦÊÇÏàͬµÄ¡£¡±ÀûÓøÇ˹¶¨ÂɿɲâijЩÌرð·´Ó¦µÄÈÈЧӦ¡£

¢ÙP4(s£¬°×Á×)£«5O2(g)===P4O10(s)¡¡¦¤H1£½£­2 983.2 kJ¡¤mol£­1

¢ÚP(s£¬ºìÁ×)£«5/4O2(g)===1/4P4O10(s)¡¡¦¤H2£½£­738.5 kJ¡¤mol£­1

Ôò°×Á×ת»¯ÎªºìÁ×µÄÈÈ»¯Ñ§·½³ÌʽΪ_________________________________________¡£ÏàͬµÄ×´¿öÏ£¬ÄÜÁ¿½ÏµÍµÄÊÇ________£»°×Á×µÄÎȶ¨ÐԱȺìÁ×________(Ìî¡°¸ß¡±»ò¡°µÍ¡±)¡£

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

¾«Ó¢¼Ò½ÌÍø