¡¾ÌâÄ¿¡¿2019Äê3ÔÂ21ÈÕÊǵڶþÊ®Æ߽조ÊÀ½çË®ÈÕ¡±£¬±£»¤Ë®×ÊÔ´£¬ºÏÀíÀûÓ÷ÏË®½ÚÊ¡Ë®×ÊÔ´£¬¼ÓÇ¿·ÏË®µÄ»ØÊÕÀûÓÃÒѱ»Ô½À´Ô½¶àµÄÈËËù¹Ø×¢¡£ÒÑÖª£ºÄ³ÎÞÉ«·ÏË®ÖпÉÄܺ¬ÓÐH+¡¢NH4+¡¢Fe3+¡¢Al3+¡¢Mg2+¡¢Na+¡¢NO3-¡¢CO32-¡¢SO42-Öеļ¸ÖÖ£¬Îª·ÖÎöÆä³É·Ö£¬·Ö±ðÈ¡·ÏË®ÑùÆ·100£¬½øÐÐÁËÈý×éʵÑ飬Æä²Ù×÷ºÍÓйØͼÏñÈçÏÂͼËùʾ£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)¸ù¾ÝÉÏÊö3×éʵÑé¿ÉÒÔ·ÖÎö·ÏË®ÖÐÒ»¶¨²»´æÔÚµÄÒõÀë×ÓÊÇ_______£¬Ò»¶¨´æÔÚµÄÑôÀë×ÓÊÇ____________________¡£

(2)д³öʵÑé¢ÛͼÏñÖгÁµí´ïµ½×î´óÁ¿ÇÒÖÊÁ¿²»ÔÙ·¢Éú±ä»¯½×¶Î·¢Éú·´Ó¦µÄÀë×Ó·´Ó¦·½³Ìʽ£º_________________________¡£

(3)·ÖÎöͼÏñ£¬ÔÚÔ­ÈÜÒºÖÐc(NH4+)Óëc(Al3+)µÄ±ÈֵΪ__________£¬ËùµÃ³ÁµíµÄ×î´óÖÊÁ¿ÊÇ__________g¡£

(4)Èôͨ¹ýʵÑéÈ·¶¨Ô­·ÏË®ÖÐc(Na+)=0.14mol/L£¬ÊÔÅжÏÔ­·ÏË®ÖÐNO3-ÊÇ·ñ´æÔÚ?__________(Ìî¡°´æÔÚ¡±¡°²»´æÔÚ¡±»ò¡°²»È·¶¨¡±)¡£Èô´æÔÚ£¬c(NO3-)=__________mol/L¡£(Èô²»´æÔÚ»ò²»È·¶¨Ôò´Ë¿Õ²»Ìî)¡£

¡¾´ð°¸¡¿CO32- Na+¡¢H+¡¢Al3+¡¢NH4+ NH4++OH-=NH3¡¤H2O 1£º1 0.546 ´æÔÚ 0.36

¡¾½âÎö¡¿

(1)¸ù¾ÝʵÑé¢ÙÑæÉ«·´Ó¦Îª»ÆÉ«È·¶¨´æÔÚNa+£¬¸ù¾ÝʵÑé¢Ú¼ÓÈëÏ¡HClËữµÄBaCl2ÈÜÒº£¬²úÉú°×É«³Áµí£¬È·¶¨´æÔÚSO42-£»¸ù¾ÝʵÑé¢Û¼ÓÈëNaOHÈÜÒº£¬¿ªÊ¼ÎÞ³Áµí£¬ËµÃ÷º¬ÓÐH+£¬ÓÉÓÚH+ÓëCO32-²»ÄÜ´óÁ¿¹²´æ£¬Ö¤Ã÷²»º¬ÓÐCO32-£»ºóÀ´²úÉú°×É«³Áµí£¬Ö¤Ã÷²»º¬ÓвúÉúºìºÖÉ«³ÁµíµÄFe3+£»È»ºó³Áµí²»Ôٱ仯£¬×îºó³ÁµíÍêÈ«Ïûʧ£¬Ö¤Ã÷º¬ÓÐAl3+£¬ÎÞMg2+£¬¿É¼ûͨ¹ý¢Û¿ÉÖ¤Ã÷º¬ÓÐH+¡¢Al3+¡¢NH4+£¬ÎÞCO32-¡¢Mg2+¡¢Fe3+¡£È»ºó¸ù¾Ý·´Ó¦¹ý³ÌÖÐÏûºÄµÄNaOHµÄÎïÖʵÄÁ¿È·¶¨º¬ÓÐÀë×ÓµÄÎïÖʵÄÁ¿µÄ¶àÉÙ£¬½áºÏÈÜÒºµçÖÐÐÔÈ·¶¨ÆäËüÀë×ӵĴæÔÚ¼°ÆäŨ¶È´óС¡£

(1)¸ù¾ÝʵÑé¢ÙÈ·¶¨´æÔÚNa+£¬¸ù¾ÝʵÑé¢ÚÈ·¶¨´æÔÚSO42-£¬¸ù¾ÝʵÑé¢ÛÈ·¶¨ÓÐH+¡¢Al3+¡¢NH4+£¬Ã»ÓÐFe3+¡¢Mg2+£¬ÒòΪCO32-ÓëH+¡¢Al3+²»Äܹ²´æ£¬ËùÒÔÒ»¶¨²»´æÔÚCO32-£»¿É¼û¸Ã·ÏË®ÖÐÒ»¶¨²»´æÔÚµÄÒõÀë×ÓÊÇCO32-£»Ò»¶¨´æÔÚµÄÑôÀë×ÓÊÇNa+¡¢H+¡¢Al3+¡¢NH4+£»

(2)ʵÑé¢ÛͼÏóÖгÁµí´ïµ½×î´óÁ¿ÇÒÖÊÁ¿²»ÔÙ·¢Éú±ä»¯½×¶Î£¬ÊÇNH4+ÓëOH-Ö®¼äµÄÀë×Ó·´Ó¦£¬Àë×Ó·½³ÌʽΪ£ºNH4++OH-=NH3¡¤H2O£»

(3)¸ù¾ÝͼÏ󣬴ӿªÊ¼ÐγɳÁµíµ½³Áµí´ïµ½×î´óÖµ£¬·¢Éú·´Ó¦Al3++3OH-=Al(OH)3¡ý£¬n(Al3+)==0.007mol£¬n(NH4+)=0.042mol-0.035mol=0.007mol£»ÈÜÒºµÄÌå»ýÏàͬ£¬ËùÒÔÔ­ÈÜÒºÖÐc(NH4+)Óëc(Al3+)µÄ±ÈֵΪ1£º1£»ËùµÃ³ÁµíµÄ×î´óÖÊÁ¿m[Al(OH)3]=0.007mol¡Á78g/mol=0.546g£»

(4)¸ù¾ÝÁòËá±µ³ÁµíÖÊÁ¿ÊÇ2.33g£¬n(SO42-)=n(BaSO4)==0.01mol£¬¸ù¾ÝµçºÉÊغ㣬ÒõÀë×ÓµçºÉ×ÜÎïÖʵÄÁ¿Îª0.01mol¡Á2=0.02mol£¬ÑôÀë×ÓµçºÉ×ÜÎïÖʵÄÁ¿Îªn(H+)+n(Al3+)+ n(NH4+)+n(Na+)=0.014mol+0.007¡Á3mol+0.007mol+0.14mol/L¡Á0.1L=0.056mol£¬ÕýµçºÉ×ÜÊý´óÓÚ¸ºµçºÉ×ÜÊý£¬ËùÒÔÔ­·ÏË®ÖдæÔÚNO3-£¬c(NO3-)==0.36mol/L¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨1£©ÓÐÒ»ÖÖ½ðÊôµ¥ÖÊA£¬ÑæÉ«·´Ó¦ÑÕÉ«³Ê»ÆÉ«£¬ÄÜ·¢ÉúÏÂͼËùʾ±ä»¯£º

ÉÏͼÖе­»ÆÉ«¹ÌÌåBÊÇ_____________Ìѧʽ£©

£¨2£©ÄƵĻ¯ºÏÎïÖУ¬¿ÉÓÃÓÚºôÎüÃæ¾ß×÷ΪO2À´Ô´µÄÊÇ______

£¨3£©Ð´³ö£¨1£©ÖÐCÈÜÒºÓëÏõËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³Ð£¿ÎÍâС×éΪ²â¶¨Ä³Ì¼ËáÄƺÍ̼ËáÇâÄÆ»ìºÏÎïÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊý£¬¼×¡¢ÒÒÁ½×éͬѧ·Ö±ð½øÐÐÁËÏÂÁÐÏà¹ØʵÑ飮

·½°¸¢ñ£®¼××éͬѧÓÃÖÊÁ¿·¨£¬°´ÈçÏÂͼËùʾµÄʵÑéÁ÷³Ì½øÐÐʵÑ飺

£¨1£©ÊµÑéʱ£¬Õô·¢½á¾§²Ù×÷ÖУ¬³ýÁ˾ƾ«µÆÍ⣬»¹ÒªÓõ½µÄÒÇÆ÷ÊÇ_______

£¨2£©ÓÐͬѧÈÏΪ¡°¼ÓÈëÊÊÁ¿ÑÎËᡱ²»ºÃ²Ù¿Ø£¬Ó¦¸ÄΪ¡°¹ýÁ¿ÑÎËᡱ£¬±ãÓÚ²Ù×÷ÇÒ²»Ó°Ïì²â¶¨µÄ׼ȷÐÔ£¬ÄãÈÏΪ¶Ô»ò´í_______£¬ÎªÊ²Ã´___________________

£¨3£©ÈôʵÑéÖвâµÃÑùÆ·ÖÊÁ¿Îª46.4g£¬¹ÌÌåÖÊÁ¿Îª40.95g£¬Ôò̼ËáÄƵÄÖÊÁ¿·ÖÊýΪ_______£®£¨±£Áô3λÓÐЧÊý×Ö£©

£¨4£©Õô·¢½á¾§¹ý³ÌÖÐÈôÓйÌÌå·É½¦£¬²âµÃ̼ËáÄƵÄÖÊÁ¿·ÖÊý____________£¨ÌîÆ«´ó ƫС ÎÞÓ°Ï죩£®

·½°¸¢ò£ºÒÒ×éͬѧµÄÖ÷ҪʵÑéÁ÷³ÌͼÈçÏ£º

°´ÈçÏÂͼËùʾװÖýøÐÐʵÑ飺

£¨5£©ÔÚCÖÐ×°¼îʯ»ÒÀ´ÎüÊÕ¾»»¯ºóµÄÆøÌ壮D×°ÖõÄ×÷ÓÃÊÇ_____________________£®

£¨6£©ÓеÄͬѧÈÏΪΪÁ˼õÉÙʵÑéÎó²î£¬ÔÚ·´Ó¦Ç°ºó¶¼Í¨ÈëN2£¬·´Ó¦ºóͨÈëN2µÄÄ¿µÄÊÇ______________________________£®

·½°¸¢ó£ºÆøÌå·ÖÎö·¨

£¨7£©°ÑÒ»¶¨Á¿ÑùÆ·Óë×ãÁ¿Ï¡ÁòËá·´Ó¦ºó£¬ÓÃÈçͼװÖòâÁ¿²úÉúCO2ÆøÌåµÄÌå»ý£¬BÈÜÒº×îºÃ²ÉÓÃ_________£¨ÒÔÏÂÑ¡ÏîÖÐÑ¡Ôñ£©Ê¹²âÁ¿Îó²î½ÏС£®

A£®±¥ºÍ̼ËáÄÆÈÜÒº

B£®±¥ºÍ̼ËáÇâÄÆÈÜÒº

C£®±¥ºÍÇâÑõ»¯ÄÆÈÜÒº

D£®±¥ºÍÁòËáÍ­ÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏòµÈÁ¿µÄNaOHÈÜÒºÖзֱðͨÈëCO2ÆøÌå¡£ÒòCO2µÄͨÈëÁ¿²»Í¬£¬µÃµ½×é·Ö²»Í¬µÄÈÜÒºM¡£ÈôÏòMÖÐÖðµÎ¼ÓÈëÑÎËᣬ²úÉúµÄÆøÌåÌå»ýV(CO2)Óë¼ÓÈëÑÎËáµÄÌå»ýV(HCl)¹ØϵÈçͼ£¬(×¢£º¢Ù¼ÙÉèCO2È«²¿Òݳö£»¢ÚͼÖÐOa<ab£¬¢ÛͼÖÐOa=ab£¬¢ÜͼÖÐOa>ab)¡£ÆäÖÐMÖÐÖ»ÓÐ1ÖÖÈÜÖʵÄÊÇ£¨ £©

A.Ö»ÓТÙB.Ö»ÓТÛC.¢Ú¢ÜD.¢Ù¢Û

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¸ù¾ÝÏÂÁÐʵÑé×°ÖÃͼ»Ø´ðÎÊÌ⣺

£¨1£©Å¨H2SO4ºÍľ̿ÔÚ¼ÓÈÈʱ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ__£¬Èç¹ûÓÐ0.2molµç×ÓתÒÆ£¬ÔòÔÚ±ê×¼×´¿öϲúÉúÆøÌå__¡£

£¨2£©ÈôÓÃͼʾÖеÄ×°ÖüìÑéÉÏÊö·´Ó¦µÄÈ«²¿²úÎд³öÓйØÒÇÆ÷ÖÐÓ¦¼ÓÈëµÄÊÔ¼ÁºÍ×÷ÓãºÎÞË®CuSO4ÊÔ¼Á×÷ÓÃÊÇ__£¬BÖмÓÈëµÄÊÔ¼ÁÊÇ__£¬×÷ÓÃÊÇ__£¬×ãÁ¿KMnO4ÈÜÒº×÷ÓÃÊÇ___£¬DÖмÓÈëµÄÊÔ¼ÁÊÇ__£¬×÷ÓÃÊÇ__£¬NaOHÈÜÒºµÄ×÷ÓÃÊÇ__¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁл¯Ñ§·½³Ìʽ»òµçÀë·½³ÌʽÖУ¬²»ÄÜÕýÈ·±í´ï·´Ó¦ÑÕÉ«±ä»¯µÄÊÇ

A.½«°±ÆøͨÈëµÎÓзÓ̪ÊÔÒºµÄË®ÖУ¬ÈÜÒº±äºì£ºNH3£«H2ONH3¡¤H2ONH4£«£«OH£­

B.ÉÙÁ¿FeCl3ÈÜÒºµÎÈë·ÐË®ÖбäΪºìºÖÉ«ÒºÌ壺FeCl3£«3H2OFe(OH)3(½ºÌå)£«3HCl

C.ÏòCuCl2ÈÜÒºÖмÓÈë×ãÁ¿µÄÌú·Û£¬ÈÜÒºÓÉÀ¶É«±äΪdzÂÌÉ«£ºFe£«CuCl2£½Cu£«FeCl2

D.ºôÎüÃæ¾ßʹÓúó£¬Na2O2Óɵ­»ÆÉ«Öð½¥±äΪ°×É«£º2Na2O2£½2Na2O£«O2¡ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÊµÑéÊÒÖƱ¸¹âÃô²ÄÁÏKaFeb(C2O4)c¡¤xH2OµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)¡°Èܽ⡱ʱ£¬ÎªÊ¹ËùÓõÄË®Öв»º¬O2£¬²ÉÓõIJÙ×÷·½·¨ÊÇ_________________¡£

(2)ÓÃH2C2O4(ÈõËá)¡°³ÁÌú¡±Ê±£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________¡£

(3)FeC2O4¡¤2H2OÈÈ·Ö½âÓëÆø·Õ¼°Î¶ÈÓйأ¬ÔÚN2Æø·ÕÖÐÈÈ·Ö½âʱ£¬¹ÌÌåµÄ²ÐÁôÂÊ(¹ÌÌåÑùÆ·µÄÊ£ÓàÖÊÁ¿/¹ÌÌåÑùÆ·µÄÆðʼÖÊÁ¿¡Á100£¥)ÓëζȵĹØϵÈçͼËùʾ£¬ÔòB¡úCµÄ±ä»¯ÖУ¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________¡£

(4)¡°Ñõ»¯¡±Ê±£¬Î¶Ȳ»Ò˳¬¹ý40¡æ£¬ÆäÔ­ÒòÊÇ_______¡£

(5)Ϊ²â¶¨²úÆ·KaFeb(C2O4)c¡¤xH2O(ÌúÔªËØΪ£«3¼Û)µÄ×é³É£¬³ÆÈ¡²úÆ·0.2455gÓÃÁòËáÈܽâºó£¬ÓÃ0.02000 mol¡¤L£­1µÄKMnO4±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ±ê×¼ÈÜÒº30.00 mL¡£ÔÚÉÏÊöµÎ¶¨¹ýC2O42£­µÄ±£ÁôÒºÖмÓÈë×ãÁ¿Ð¿·Û£¬¼ÓÈÈÖÁ»ÆÉ«Ïûʧ£¬¹ýÂËÏ´µÓ£¬ÂËÒº¼°Ï´µÓÒºÔÙÓÃ0.02000 mol¡¤L£­1µÄKMnO4±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ±ê×¼ÈÜÒº5.00 mL¡£Ôò¸Ã²úÆ·µÄ»¯Ñ§Ê½Îª____________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿°±ÔÚÈËÀàµÄÉú²úºÍÉú»îÖÐÓÐ׏㷺µÄÓ¦Óã¬Ä³»¯Ñ§ÐËȤС×éÀûÓÃͼһװÖÃ̽¾¿°±ÆøµÄÓйØÐÔÖÊ¡£

£¨1£©×°ÖÃAÖÐÉÕÆ¿ÄÚÊÔ¼Á¿ÉÑ¡Óà £¨ÌîÐòºÅ£©¡£BµÄ×÷ÓÃÊÇ

a£®¼îʯ»Ò b£®Å¨ÁòËá c£®Éúʯ»Ò d£®ÉÕ¼îÈÜÒº

£¨2£©Á¬½ÓºÃ×°Öò¢¼ìÑé×°ÖõÄÆøÃÜÐÔºó£¬×°ÈëÒ©Æ·£¬È»ºóÓ¦ÏÈ £¨ÌîI»ò¢ò£©£®

¢ñ£®´ò¿ªÐýÈûÖðµÎÏòÔ²µ×ÉÕÆ¿ÖмÓÈ백ˮ ¢ò£®¼ÓÈÈ×°ÖÃC

£¨3£©ÊµÑéÖй۲쵽CÖÐCuO·ÛÄ©±äºì£¬DÖÐÎÞË®ÁòËáÍ­±äÀ¶£¬²¢ÊÕ¼¯µ½Ò»ÖÖµ¥ÖÊÆøÌ壬Ôò¸Ã·´Ó¦Ïà¹Ø»¯Ñ§·½³ÌʽΪ ,£®¸Ã·´Ó¦Ö¤Ã÷°±Æø¾ßÓÐ ÐÔ£®

£¨4£©¸ÃʵÑéȱÉÙβÆøÎüÊÕ×°Öã¬Í¼¶þÖÐÄÜÓÃÀ´ÎüÊÕβÆøµÄ×°ÖÃÊÇ £¨Ìî×°ÖÃÐòºÅ£©£®

£¨5£©°±Æø¼«Ò×ÈÜÓÚË®£¬Èô±ê×¼×´¿öÏ£¬½«2.24LµÄ°±ÆøÈÜÓÚË®Åä³É0.5LÈÜÒº£¬ËùµÃÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ mol/L£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³»¯Ñ§»î¶¯Ð¡×éÉè¼ÆÒÔÏÂ×°ÖýøÐв»Í¬µÄʵÑé¡£ÆäÖÐaΪÓÃÓÚ¹ÄÈë¿ÕÆøµÄÆøÄÒ£¬b ΪÂÝÐý×´Í­Ë¿£¬cÖÐÊ¢ÓбùË®¡£

£¨1£©ÈôÓÃA×°ÖÃ×öÒÒ´¼ÓëÒÒËáµÄõ¥»¯·´Ó¦ÊµÑ飬Ôò»¹ÐèÁ¬½ÓµÄ×°ÖÃÊÇ____________(ÌîÐòºÅ)£¬¸Ã×°ÖÃÖÐÓ¦¼ÓÈëÊÔ¼Á____________¡£´ÓʵÑ鰲ȫ½Ç¶È¿¼ÂÇ£¬A×°ÖÃÊÔ¹ÜÖгý¼ÓÈë·´Ó¦ÒºÍ⣬»¹Ðè¼ÓÈëµÄ¹ÌÌåÎïÖÊÊÇ____________¡£

£¨2£©¸ÃС×éͬѧÓû×öÒÒ´¼Ñõ»¯³ÉÒÒÈ©µÄʵÑ飬ÔòӦѡÓõÄ×°ÖÃÊÇ____________(ÌîÐòºÅ)£¬ÔÙÓÃÖƵõÄÒÒÈ©ÈÜÒº½øÐÐÒø¾µ·´Ó¦£¬ÕýÈ·µÄ²Ù×÷˳ÐòÊÇ____________(ÌîÐòºÅ)¡£

¢ÙÏòÊÔ¹ÜÖеÎÈë3µÎÒÒÈ©ÈÜÒº

¢ÚÒ»±ßÕñµ´Ò»±ßµÎÈë2%µÄÏ¡°±Ë®£¬Ö±ÖÁ×î³õ²úÉúµÄ³ÁµíÇ¡ºÃÈܽâΪֹ

¢ÛÕñµ´ºó·ÅÈëÈÈË®ÖУ¬Ë®Ô¡¼ÓÈÈ

¢ÜÔڽྻµÄÊÔ¹ÜÖмÓÈë1 mL 2%µÄAgNO3ÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸