2020½ì¸ß¿¼»¯Ñ§¶þÂÖ¸´Ï°»¯¹¤¹¤ÒÕÁ÷³ÌרÌâ¾í ÏÂÔر¾ÎÄ

1.3¡Á1022£­1£­£­

(5)FeÇ¡ºÃ³ÁµíÍêȫʱ£¬c(PO4)£½£½1.3¡Á1017mol¡¤L1£¬c3(Mg2£­5 mol¡¤L

1.0¡Á10

3£«

3£­

£«

£­

3

)¡¤c2(PO34)£½0.01¡Á(1.3¡Á10

£­

£­17

)2£½1.7¡Á10

£­40

µí

(6)2FePO4£«Li2CO3£«H2C2O4=====2LiFePO4£«3CO2¡ü£«H2O¡ü

3£®×îÐÂÑо¿±íÃ÷As2O3ÔÚÒ½Ò©ÁìÓòÓÐÖØÒªÓ¦Óá£Ä³Ð¡×é´Ó¹¤Òµ·ÏÁÏÖÐÌáÈ¡As2O3£¬Éè¼ÆÁ÷³ÌÈçÏ£º

¸ßÎÂ

ÒÑÖª£ºH3AsO3ΪÈõËᣬÈÈÎȶ¨ÐԲ

(1)д³öÒ»Ìõ¡°¼î½þ¡±Ê±Ìá¸ß½þÈ¡Âʵķ½·¨£º________________________£»¡°¼î½þ¡±ÖÐH3AsO3ת»¯³ÉNa3AsO3µÄÀë×Ó·½³ÌʽΪ__________________________________¡£

(2)¡°Ñõ»¯¡±Ê±Ïò»ìºÏÒºÖÐͨÈëO2ʱ¶ÔÌåϵ¼Óѹ£¬Ä¿µÄÊÇ________________________¡£ (3)¡°³ÁÉ顱¹ý³ÌÖÐÓÐÈçÏ·´Ó¦£º ¢ÙCa(OH)2(s)

£«

£­

Ca2(aq)£«2OH(aq) ¦¤H£¼0

£­

£«£­

¢Ú5Ca2£«OH£«3AsO34Ca5(OH)(AsO4)3¡ý ¦¤H£¾0

³ÁÉéÂÊÓëζȹØϵÈçͼ¡£³ÁÉé×î¼ÑζÈΪ____________________________________£¬¸ßÓÚ85 ¡æʱ£¬³ÁÉéÂÊϽµµÄÔ­ÒòÊÇ________________________________¡£

(4)¡°»¹Ô­¡±¹ý³ÌÖлñµÃH3AsO3µÄ»¯Ñ§·½³ÌʽΪ_____________________________£»¡°²Ù×÷A¡±Îª________¡¢¹ýÂË£»ÂËÒº¢òµÄÖ÷ÒªÈÜÖÊÊÇ________¡£

(5)Èôÿ²½¾ùÍêÈ«·´Ó¦£¬¡°Ñõ»¯¡±ºÍ¡°»¹Ô­¡±Ê±ÏûºÄÏàͬÌõ¼þÏÂO2ºÍSO2µÄÌå»ý·Ö±ðΪx L¡¢y L£¬Ôò·ÏË®ÖÐn(H3AsO3)¡Ãn(H3AsO4)£½________(д³öº¬x¡¢yµÄ¼ÆËãʽ)¡£

½âÎö£º¹¤Òµ·ÏÁÏ(º¬H3AsO3¡¢H3AsO4)¼ÓÈëÇâÑõ»¯ÄÆÈÜÒº¡°¼î½þ¡±£¬ÓëÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉNa3AsO3¡¢Na3AsO4ÈÜÒº£¬Í¨ÈëÑõÆø£¬½«Na3AsO3Ñõ»¯ÎªNa3AsO4£¬ÏòÈÜÒºÖмÓÈëʯ»ÒÈ飬µÃµ½Ca5(OH)(AsO4)3³Áµí£¬½«Ca5(OH)(AsO4)3ÓÃÁòËáËữµÃµ½H3AsO4£¬Í¨Èë¶þÑõ»¯ÁòÓëH3AsO4ÈÜÒº»ìºÏ£ºH3AsO4£«H2O£«SO2===H3AsO3£«H2SO4£¬»¹Ô­ºó¼ÓÈÈÈÜÒº£¬H3AsO3·Ö½âΪAs2O3£¬¾Ý´Ë·ÖÎö½â´ð¡£

(1)·´Ó¦Îï½Ó´¥Ê±¼äÔ½³¤£¬·´Ó¦Ô½³ä·Ö£¬ËùÒÔ¡°¼î½þ¡±Ê±¿É²ÉÓýÁ°è»ò¶à´Î½þȡʹÆä³ä·Ö·´Ó¦£¬Ìá¸ßÔ­ÁϽþÈ¡ÂÊ£»H3AsO4ÓëÇâÑõ»¯ÄÆ·´Ó¦ÎªËá¼îÖкͷ´Ó¦£¬Àë×Ó·½³ÌʽΪH3AsO3£«3OH===AsO33£«3H2O¡£

(2)¡°Ñõ»¯¡±Ê±Ïò»ìºÏÒºÖÐͨÈëO2ʱ¶ÔÌåϵ¼Óѹ£¬¿ÉÒÔÔö´óO2µÄÈܽâ¶È£¬¼Ó¿ì·´Ó¦ËÙÂÊ¡£

(3)¡°³ÁÉ顱Êǽ«ÉéÔªËØת»¯ÎªCa5(OH)(AsO4)3³Áµí£¬·¢ÉúµÄÖ÷Òª·´Ó¦ÓТÙCa(OH)2(s)

Ca2(aq)£«2OH(aq) ¦¤H<0¡¢¢Ú5Ca2£«OH£«3AsO34

£«

£­

£«

£­

£«

£­

£­

£­

£­

Ca5(OH)(AsO4)3

¦¤H>0£¬¸ßÓÚ85 ¡æ£¬Ëæ×ÅζÈÉý¸ß£¬·´Ó¦¢ÙÖÐc(Ca2)¡¢c(OH)¼õС£¬Î¶ȶÔƽºâÒƶ¯µÄÓ°Ïì·´Ó¦¢Ù´óÓÚ·´Ó¦¢Ú£¬Ê¹·´Ó¦¢ÚÖÐƽºâÄæÏòÒƶ¯£¬³ÁÉéÂÊϽµ¡£

(4)¡°»¹Ô­¡±¹ý³ÌÖжþÑõ»¯Áò½«H3AsO4»¹Ô­ÎªH3AsO3£¬×ÔÉí±»Ñõ»¯Éú³ÉÁòËᣬ»¯Ñ§·½³ÌʽΪH3AsO4£«H2O£«SO2===H3AsO3£«H2SO4£»»¹Ô­ºó¼ÓÈÈÈÜÒº£¬H3AsO3·Ö½âΪAs2O3£¬ËùÒÔ¡°²Ù×÷A¡±Îª¼ÓÈÈ¡¢¹ýÂË£»ÓÉ»¯Ñ§·½³Ìʽ¿ÉÖª£¬ÂËÒº¢òµÄÖ÷ÒªÈÜÖÊÊÇH2SO4¡£

(5)¡°Ñõ»¯¡±¹ý³ÌÖн«Na3AsO3Ñõ»¯ÎªNa3AsO4£¬¹ØϵʽΪ2H3AsO3¡«2Na3AsO3¡«O2£»¡°»¹Ô­¡±¹ý³ÌÖжþÑõ»¯Áò½«H3AsO4»¹Ô­ÎªH3AsO3£¬¹ØϵʽΪH3AsO4¡«SO2£»ÒÑÖªÏûºÄÏà

ͬÌõ¼þÏÂO2ºÍSO2µÄÌå»ý·Ö±ðΪx L¡¢y L£¬Ìå»ý±ÈµÈÓÚÎïÖʵÄÁ¿Ö®±È£¬¼´n(O2)¡Ãn(SO2)£½x¡Ãy£¬ÓɹØϵʽ¿ÉÖª£¬n(H3AsO3)£½2n(O2)£½2x£¬Ôò·ÏË®ÖÐÔ­ÓеÄn(H3AsO4)£½y£­2x£¬Ôò·ÏË®ÖÐn(H3AsO3)¡Ãn(H3AsO4)£½2x¡Ã(y£­2x)¡£

´ð°¸£º(1)½Á°è(»ò¶à´Î½þÈ¡»òÆäËûºÏÀí´ð°¸) H3AsO3£«3OH===AsO33£«3H2O (2)Ôö´óO2µÄÈܽâ¶È£¬¼Ó¿ì·´Ó¦ËÙÂÊ (3)85 ¡æ Ëæ×ÅζÈÉý¸ß£¬·´Ó¦¢ÙÖÐc(Ca2)¡¢c(OH)¼õС£¬Î¶ȶÔƽºâÒƶ¯µÄÓ°Ïì·´Ó¦¢Ù´óÓÚ·´Ó¦¢Ú£¬Ê¹·´Ó¦¢ÚÖÐƽºâÄæÏòÒƶ¯£¬³ÁÉéÂÊϽµ (4)H3AsO4£«H2O£«SO2===H3AsO3£«H2SO4 ¼ÓÈÈ H2SO4 (5)2x¡Ã(y£­2x)

4£®(2019¡¤¸ß¿¼ÃûУÁª¿¼ÐÅÏ¢ÓÅ»¯¾í)ÅðÇ⻯ÄÆ(NaBH4)¹ã·ºÓÃÓÚ»¯¹¤Éú²ú£¬³£ÎÂÏÂÄÜÓëË®·´Ó¦£¬Ò×ÈÜÓÚÒì±û°·(·ÐµãΪ33 ¡æ)¡£¹¤ÒµÉÏ¿ÉÓÃÅðþ¿ó(Ö÷Òª³É·ÖΪMg2B2O5¡¤H2O£¬º¬ÉÙÁ¿ÔÓÖÊFe3O4)ÖÆÈ¡NaBH4£¬Æ乤ÒÕÁ÷³ÌÈçÏ£º

£«

£­

£­

£­

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)NaBH4µÄµç×ÓʽΪ________¡£

(2)¼îÈÜʱMg2B2O5·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_________________________________¡£ (3)ÂËÔüµÄ³É·ÖÊÇ________¡£

(4)¸ßκϳÉÖУ¬¼ÓÁÏ֮ǰÐ轫·´Ó¦Æ÷¼ÓÈÈÖÁ100 ¡æÒÔÉϲ¢Í¨Èëë²Æø£¬¸Ã²Ù×÷µÄÄ¿µÄÊÇ________________________________________________________________________£»Ô­ÁÏÖеĽðÊôÄÆͨ³£±£´æÔÚ________ÖУ¬ÊµÑéÊÒÈ¡ÓÃÉÙÁ¿½ðÊôÄÆÓõ½µÄʵÑéÓÃÆ·ÓÐ________¡¢²£Á§Æ¬ºÍСµ¶¡£

(5)²Ù×÷2µÄÃû³ÆΪ________¡£Á÷³ÌÖпÉÑ­»·ÀûÓõÄÎïÖÊÊÇ________¡£

(6)ÔÚ¼îÐÔÌõ¼þÏ£¬ÓöèÐԵ缫µç½âNaBO2ÈÜÒºÒ²¿ÉÖƵÃNaBH4£¬×°ÖÃÈçͼËùʾ£¬µç½â×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________________________¡£

(7)NaBH4³£ÓÃ×÷»¹Ô­¼Á£¬H2Ò²Êdz£¼ûµÄ»¹Ô­¼Á¡£ÓëÏàͬÑõ»¯¼Á·´Ó¦Ê±£¬1 g NaBH4µÄ»¹Ô­ÄÜÁ¦Ï൱ÓÚ________g H2µÄ»¹Ô­ÄÜÁ¦(BÔªËØ»¯ºÏ¼Û²»±ä£¬Ñõ»¯²úÎïÖÐÇâÔªËØ»¯ºÏ¼ÛÏàͬ£¬¼ÆËã½á¹û±£ÁôÁ½Î»Ð¡Êý)¡£

½âÎö£º(1)NaBH4µÄµç×ÓʽΪNa[H¡¤B¡¤¡¤,¡¤¡¤H,H¡¤H]¡£(2)ÓÉÁ÷³Ìͼ¿ÉÖª£¬¼îÈÜʱMg2B2O5

ÓëNaOH·´Ó¦×ª»¯ÎªNaBO2£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇMg2B2O5£«2NaOH£«H2O===2NaBO2£«2Mg(OH)2¡£(3)Åðþ¿óÖÐFe3O4²»ÈÜÓÚŨNaOHÈÜÒº£¬ÒÔÂËÔüµÄÐÎʽÎö³ö£¬ÁíÍ⣬¼îÈÜʱÉú³ÉµÄMg(OH)2Ò²ÒÔÂËÔüµÄÐÎʽÎö³ö£¬ËùÒÔÂËÔüµÄ³É·ÖΪMg(OH)2ºÍFe3O4¡£(4)NaµÄÐÔÖÊ»îÆã¬ÄÜÓëË®ºÍ¿ÕÆøÖеÄÑõÆø·´Ó¦£¬½áºÏÌâ¸ÉÖÐÅðÇ⻯ÄÆÔÚ³£ÎÂÏÂÄÜÓëË®·´Ó¦£¬ËùÒÔ¼ÓÁÏ֮ǰÐëÏÈÅųýË®ÕôÆøºÍ¿ÕÆø(»òÑõÆø)µÄ¸ÉÈÅ¡£(5)NaBH4Ò×ÈÜÓÚÒì±û°·ÇÒÒì±û°·µÄ·Ðµã½ÏµÍ£¬½áºÏÁ÷³Ìͼ¿ÉÖª£¬²Ù×÷2ÊÇÕôÁóNaBH4µÄÒì±û°·ÈÜÒº£¬»ØÊÕÒì±û°·£¬Òì±û°·¿ÉÑ­»·ÀûÓá£(6)ÓÉ¡°ÔÚ¼îÐÔÌõ¼þÏ£¬ÓöèÐԵ缫µç½âNaBO2ÈÜÒºÒ²¿ÉÖƵÃNaBH4¡±¿ÉÖª£¬µç½â×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪBO2£«2H2O=====BH4£«2O2¡ü¡£(7)NaBH4ºÍÇâÆø×ö»¹Ô­¼ÁʱÑõ»¯²úÎïÖÐÇâÔªËض¼Îª£«1¼Û£¬Ôò1 g NaBH4µÄ»¹Ô­ÄÜÁ¦Ï൱ÓÚ0.21 g H2µÄ»¹Ô­ÄÜÁ¦¡£

´ð°¸£º(1)

£­

£«

¡¤

¡¤

£­

µç½â

£­

(2)Mg2B2O5£«2NaOH£«H2O===2NaBO2£«2Mg(OH)2 (3)Fe3O4ºÍMg(OH)2

(4)³ýÈ¥·´Ó¦Æ÷ÖеÄË®ÕôÆøºÍ¿ÕÆø úÓÍ Ä÷×Ó¡¢ÂËÖ½ (5)ÕôÁó Òì±û°·

(6)BO2£«2H2O=====BH4£«2O2¡ü (7)0.21

£­

µç½â

£­