¡¾ÌâÄ¿¡¿îâ(Mo)¼°ÆäºÏ½ðÔÚÒ±½ð¡¢Å©Òµ¡¢µçÆø¡¢»¯¹¤¡¢»·±£ºÍÓµÈÖØÒªÁìÓòÓÐ׏㷺µÄÓ¦ÓúÍÁ¼ºÃµÄǰ¾°£¬³ÉΪ¹úÃñ¾­¼ÃÖÐÒ»ÖÖÖØÒªµÄÔ­ÁϺͲ»¿ÉÌæ´úµÄÕ½ÂÔÎïÖÊ¡£îâËáÄÆ¾§Ìå(Na2MoO4 ¡¤2H2O)ÊÇÒ»ÖÖÖØÒªµÄ½ðÊô»ºÊ´¼Á¡£Ä³¹¤³§ÀûÓÃî⾫¿ó(Ö÷Òª³É·Ö MoS2 )ÖÆ±¸îâËáÄÆ¾§ÌåºÍ½ðÊôîâµÄÁ÷³ÌÈçͼËùʾ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)Na2 MoO 4¡¤2H2O ÖÐ Mo µÄ»¯ºÏ¼ÛÊÇ ________¡£

(2)ÒÑÖª¡°±ºÉÕ¡±¹ý³ÌÖÐ MoS2 ±ä³É MoO3£¬ÔòÆøÌå 1 ÖжԴóÆøÓÐÎÛȾµÄÊÇ ________¡£

(3)¡°¼î½þ¡±¹ý³ÌÖÐÉú³É Na2MoO4 µÄ»¯Ñ§·½³ÌʽÊÇ ________£»¡°½á¾§¡±µÄîâËáÄÆ¾§ÌåÈÔº¬ÆäËûÔÓÖÊ£¬ÒªµÃµ½½Ï´¿µÄ¾§Ì壬»¹Ó¦²ÉÈ¡µÄ·½·¨ÊÇ ________¡£

(4)¡°ÂËÒº¡±µÄÖ÷Òª³É·ÖÊÇ ________¡£

(5)½«¹ýÂ˲Ù×÷µÃµ½µÄîâËá³Áµí½øÐиßαºÉÕ£¬ÊµÑéÊÒÄ£Äâ¸ßαºÉÕʱÓÃÓÚÊ¢·ÅîâËáµÄÒÇÆ÷ÊÇ ________¡£

(6)îâËá¸ßαºÉյIJúÎïÓë Al ÔÚ¸ßÎÂÏ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ ________¡£

(7)²âµÃ¼î½þÒºÖв¿·ÖÀë×ÓŨ¶ÈΪ£ºc(MoO42£­) =0.4 mol¡¤L£­1£¬c(SO42£­) =0.02 mol¡¤L£­1¡£½á¾§Ç°¼ÓÈëÇâÑõ»¯±µ¹ÌÌå³ýÈ¥SO42£­£¬²»¿¼ÂǼÓÈëÇâÑõ»¯±µ¹ÌÌåºóÈÜÒºÌå»ýµÄ±ä»¯£¬µ±BaMoO4¿ªÊ¼³Áµíʱ£¬SO42£­ µÄÈ¥³ýÂÊΪ ________ (±£ÁôÈýλÓÐЧÊý×Ö)¡£ [ÒÑÖª£º Ksp(BaSO4)=1.1¡Á10£­10£¬Ksp(BaMoO4)=4.0¡Á10£­8]

¡¾´ð°¸¡¿+6 ¶þÑõ»¯Áò»ò SO2 MoO3 + Na2CO3 =Na2MoO4 +CO2¡ü ÖØ½á¾§ NaCl¡¢HCl ÛáÛö 2Al+MoO3Mo+Al2O3 94.5%»ò0.945

¡¾½âÎö¡¿

î⾫¿ó£¨Ö÷Òª³É·ÖÊÇMoS2£©ÖƱ¸îâËáÄÆ¼°ÖƱ¸½ðÊôîâµÄÖ÷ÒªÁ÷³Ì£¬î⾫¿ó¿ÕÆøÖÐׯÉյõ½º¬¶þÑõ»¯ÁòµÄÎ²ÆøºÍ´Ö²úÆ·MoO3£¬¼ÓÈëNa2CO3ÈÜÒº·´Ó¦Éú³ÉNa2MoO4£¬½á¾§Îö³öµÃµ½¾§Ì壬Na2MoO4ÈÜÒº¼ÓÈëÑÎËáÉú³ÉîâËᣬ¸ßÎÂׯÉÕ·Ö½âÉú³ÉMoO3£¬MoO3ÔÚ¸ßÎÂÌõ¼þÏÂÓëÂÁ»¹Ô­Éú³Éµ¥ÖÊî⣬ÒԴ˽â´ð¸ÃÌâ¡£

£¨1£©Na2 MoO 4¡¤2H2OÖÐÄÆÔªËØ»¯ºÏ¼Û+1¼Û£¬ÑõÔªËØ»¯ºÏ¼Û2¼Û£¬ÉèîâÔªËØµÄ»¯ºÏ¼ÛÊÇx¼Û£¬¸ù¾Ý»¯ºÏÎïµÄ´úÊýºÍΪ0¿ÉÖª£¬2¡Á(+1)+x+(2)¡Á4=0£¬½âµÃx = +6£¬¼´¼ÆËãµÃµ½îâÔªËØµÄ»¯ºÏ¼ÛÊÇ+6¼Û£¬

¹Ê´ð°¸Îª£º+6£»

£¨2£©¡°±ºÉÕ¡±¹ý³ÌÖÐMoS2±ä³ÉMoO3,£¬S±»Ñõ»¯Éú³É¶Ô´óÆøÓÐÎÛȾµÄÆøÌ壬ӦΪSO2£¬

¹Ê´ð°¸Îª£º¶þÑõ»¯Áò»òSO2£»

£¨3£©¼ÓÈëNa2CO3ÈÜÒººÍMoO3·´Ó¦Éú³É¶þÑõ»¯Ì¼ºÍNa2MoO4£¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºMoO3 + Na2CO3 =Na2MoO4 +CO2 ¡ü£¬¡°½á¾§¡±µÄîâËáÄÆ¾§ÌåÈÔº¬ÆäËûÔÓÖÊ£¬ÒªµÃµ½½Ï´¿µÄ¾§Ì壬Ӧ½øÐÐÖØ½á¾§£¬

¹Ê´ð°¸Îª£ºMoO3 + Na2CO3 =Na2MoO4 +CO2 ¡ü£»Öؽᾧ£»

£¨4£©¼ÓÈë¹ýÁ¿ÑÎËᣬÉú³ÉH2MoO4ºÍNaCl£¬ÔòÂËÒºµÄÖ÷Òª³É·ÖΪNaCl¡¢HCl£¬

¹Ê´ð°¸Îª£ºNaCl¡¢HCl£»

£¨5£©×ÆÉÕ¹ÌÌ壬ӦÔÚÛáÛöÖнøÐУ¬

¹Ê´ð°¸Îª£ºÛáÛö£»

£¨6£©ÀûÓÃÂÁÈÈ·´Ó¦¿É»ØÊÕ½ðÊôî⣬MoO3·¢ÉúÂÁÈÈ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º4Al+2MoO3 2Mo+2Al2O3£¬

¹Ê´ð°¸Îª£º4Al+2MoO3 2Mo+2Al2O3£»

£¨7£©Ksp(BaMoO4) = 4.0¡Á108£¬îâËáÄÆÈÜÒºÖÐc(MoO42) = 0.4 molL1£¬BaMoO4¿ªÊ¼³Áµíʱ£¬ÈÜÒºÖбµÀë×ÓµÄŨ¶ÈΪ£ºc(Ba2+) = = 1.0¡Á107 mol/L£¬ÓÖKsp(BaSO4)=1.1¡Á1010£¬ÔòÈÜÒºÖÐÁòËá¸ùÀë×ÓµÄŨ¶ÈΪ£ºc(SO42) = =1.1¡Á103 mol/L£¬ÒòΪԭÈÜÒºÖÐÁòËá¸ùÀë×ÓµÄŨ¶ÈΪ0.02 molL1£¬¹ÊÆäÁòËá¸ùÀë×ÓµÄÈ¥³ýÂÊΪ£º(1)¡Á100%=15.5% = 94.5 %»ò0.945£¬

¹Ê´ð°¸Îª£º94.5 %»ò0.945¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨1£©13 g C2H2(g)ÍêȫȼÉÕÉú³ÉCO2ºÍH2O(l)ʱ£¬·Å³ö659 kJµÄÈÈÁ¿£¬Ð´³ö±íʾ¸ÃÎïÖÊȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ___________________________¡£

£¨2£©ÒÑÖª·´Ó¦£ºCl2£«2HBr===Br2£«2HCl¡£

¢ñ.µ±ÓÐ0.2 mol HClÉú³Éʱ·Å³ö8.1 kJµÄÈÈÁ¿¡£

¢ò.ÆäÄÜÁ¿±ä»¯Ê¾ÒâͼÈçͼ£º

Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ_____________________________________¡£ÓÉÉÏÊöÊý¾ÝÅж϶Ͽª1 mol H¡ªCl ¼üÓë¶Ï¿ª 1 mol H¡ªBr¼üËùÐèÄÜÁ¿Ïà²îԼΪ________kJ¡£

£¨3£©ÒÑÖª£º¢Ù2CO(g)£«O2(g) ===2CO2(g)¡¡¦¤H£½£­566 kJ¡¤mol£­1

¢ÚNa2O2(s)£«CO2(g) ===Na2CO3(s)£«1/2O2(g)¡¡¦¤H£½£­226 kJ¡¤mol£­1

ÔòCO(g)ÓëNa2O2(s)·´Ó¦·Å³ö509 kJÈÈÁ¿Ê±£¬µç×Ó×ªÒÆÊýĿΪ________¡£

£¨4£©ÒÑÖªCO ¡¢H2 ¡¢CH4 µÄȼÉÕÈÈ·Ö±ðΪ283 kJ¡¤mol£­1¡¢286 kJ¡¤mol£­1¡¢890 kJ¡¤mol£­1¡£Èô½«a mol CH4¡¢COºÍH2µÄ»ìºÏÆøÌåÍêȫȼÉÕ,Éú³ÉCO2ÆøÌåºÍҺ̬ˮ,ÇÒCO2ºÍË®µÄÎïÖʵÄÁ¿ÏàµÈʱ,Ôò·Å³öÈÈÁ¿(Q)µÄȡֵ·¶Î§ÊÇ____¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿100 ¡æÊ±£¬ÏòÈÝ»ýΪ2 LµÄÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨Á¿µÄXÆøÌ壬·¢ÉúÈçÏ·´Ó¦£ºX(g)£«2Y(g)Z(g)£¬·´Ó¦¹ý³ÌÖвⶨµÄ²¿·ÖÊý¾Ý¼ûÏÂ±í£ºÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ( )

·´Ó¦Ê±¼ä/min

n(X)/mol

n(Y)/mol

0

2.00

2.40

10

1.00

30

0.40

A. ζÈΪ200 ¡æÊ±£¬ÉÏÊö·´Ó¦Æ½ºâ³£ÊýΪ20£¬ÔòÕý·´Ó¦ÎªÎüÈÈ·´Ó¦

B. ÈôÃܱÕÈÝÆ÷Ìå»ý¿É±ä£¬ÆäËûÌõ¼þ²»±ä£¬ÔڴﵽƽºâºóËõСÈÝÆ÷Ìå»ýΪԭÀ´Ò»°ë£¬Ôòc(X)£¾1 mol/L

C. ±£³ÖÆäËûÌõ¼þ²»±ä£¬ÏòÈÝÆ÷ÖÐÔÙ³äÈë1.00 mol XÆøÌåºÍ1.20 mol YÆøÌ壬µ½´ïƽºâºó£¬Xת»¯ÂÊÔö´ó

D. ±£³ÖÆäËûÌõ¼þ²»±ä£¬ÈôÔÙÏòÈÝÆ÷ÖÐͨÈë0.10 mol XÆøÌ壬0.10 mol YºÍ0.10 mol Z£¬Ôòv(Õý)£¼v(Äæ)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿³£ÎÂÏÂÇâÑõ»¯Ð¿ÔÚ¼îÐÔÈÜÒºÖеı仯ÈçͼËùʾ£¬ºá×ø±êΪÈÜÒºµÄpH£¬×Ý×ø±êΪZn2£«»òZnOµÄÎïÖʵÄÁ¿Å¨¶ÈµÄ¶ÔÊý£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)

A. ÒÀͼÖÐÊý¾Ý¼ÆËã¿ÉµÃ¸ÃζÈÏÂZn(OH)2µÄÈܶȻý(Ksp)£½1¡Á10£­17

B. ÈÜÒºÖмÓÈë×ãÁ¿°±Ë®£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪZn2£«£«4OH£­===ZnO£«2H2O

C. ΪÌáÈ¡¹¤Òµ·ÏÒºÖеÄZn2£«£¬¿ÉÒÔ¿ØÖÆÈÜÒºµÄpHÔÚ13×óÓÒ

D. Zn2£«ÔÚÈÜÒºÖеĴæÔÚÐÎʽÓëAl3£«ÏàËÆ£¬¼îÐÔÈÜÒºÖÐÖ»ÒÔZn(OH)2ÐÎʽ´æÔÚ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿·¼Ïã×廯ºÏÎïXºÍY¶¼ÊÇ´ÓÕÁÄÔ¿ÆÖ²ÎïÖÐÌáÈ¡µÄÏãÁÏ£®X¿É°´Èçͼ·ÏߺϳÉY£®

ÒÑÖª£º¢ÙRCH=CHR¡ä RCHO+R¡äCHO

¢Ú²»º¬¦Á-ÇâÔ­×Ó(ÓëÈ©»ùÖ±½ÓÏàÁ¬µÄ̼ÉϵÄÇâ)µÄÈ©ÔÚŨ¼î×÷ÓÃÏÂÈ©·Ö×Ó×ÔÉíͬʱ·¢ÉúÑõ»¯Ó뻹ԭ·´Ó¦£¬Éú³ÉÏàÓ¦µÄôÈËá(ÔÚ¼îÈÜÒºÖÐÉú³ÉôÈËáÑÎ)ºÍ´¼µÄ·´Ó¦¡£

È磺2HCHOHCOOH+CH3OH

(1)XµÄ¹ÙÄÜÍÅÃû³ÆÎª_________________

(2)YµÄ½á¹¹¼òʽΪ___________________

(3)D+G¡úYµÄÓлú·´Ó¦ÀàÐÍΪ£º___________£®

(4)ÏÂÁÐÎïÖʲ»ÄÜÓëD·´Ó¦µÄÊÇ_______(Ñ¡ÌîÐòºÅ)£®

a£®½ðÊôÄÆ b£®ÇâäåËá c£®Ì¼ËáÄÆÈÜÒº d£®ÒÒËá

(5)д³öÏÂÁз´Ó¦·½³Ìʽ£ºX¡úGµÄµÚ¢Ù²½·´Ó¦__________________E¡úF£º______________

(6)GÓжàÖÖͬ·ÖÒì¹¹Ì壬д³öͬʱÂú×ãÏÂÁÐÌõ¼þµÄGµÄËùÓÐͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ(Òª¿¼ÂǿռäÁ¢ÌåÒì¹¹)

i£®ÄÜ·¢ÉúÒø¾µ·´Ó¦ ii£®ÄÜ·¢ÉúË®½â·´Ó¦ iii£®±½»·ÉÏÖ»ÓÐÒ»¸öÈ¡´ú»ù

(7)GµÄÒ»ÖÖͬ·ÖÒì¹¹ÌåHµÄ½á¹¹¼òʽΪ£¬Ð´³öÒÔΪÓлúÔ­ÁÏ£¬ÖÆÈ¡HµÄºÏ³É·Ïß(¿ÉÈÎÑ¡ÎÞ»úÔ­ÁÏ)__________________________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³¿ÎÍâ»î¶¯Ð¡×éÓÃÈçͼװÖýøÐÐʵÑ飬ÊԻشðÏÂÁÐÎÊÌâ¡£

£¨1£©Èô¿ªÊ¼Ê±¿ª¹ØKÓëaÁ¬½Ó£¬ÔòA¼«µÄµç¼«·´Ó¦Ê½Îª_____________________¡£

£¨2£©Èô¿ªÊ¼Ê±¿ª¹ØKÓëbÁ¬½Ó£¬ÔòB¼«µÄµç¼«·´Ó¦Ê½Îª____________________¡£

£¨3£©ÈôÓÃÂÁÌõºÍþÌõ·Ö±ð´úÌæÍ¼ÖÐʯīºÍÌúµç¼«£¬µç½âÖÊÈÜҺΪÇâÑõ»¯ÄÆÈÜÒº£¬Çëд³öÔ­µç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½_____________________________¡£

£¨4£©ÈôÓöþÑõ»¯Ç¦ºÍǦ×÷µç¼«£¬ÁòËáÈÜҺΪµç½âÖÊÈÜÒº¹¹³ÉǦÐîµç³Ø£¬Ôò³äµçʱÑô¼«µÄµç¼«·´Ó¦Ê½_________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚÒ»¶¨Ìõ¼þÏ£¬ÀûÓÃÖû»·´Ó¦X+Y=W+Z£¬¿ÉʵÏÖÎïÖÊÖ®¼äµÄת»¯

(1)ÈôXΪMg£¬WΪC£¬ÔòZΪ_________¡£

(2)ÈôXΪSiO2£¬WΪCO£¬¸Ã·´Ó¦ÔÚ¹¤ÒµµÄÓÃ;ÊÇ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³»¯Ñ§Ð¡×éͬѧÔÚѧϰCl2µÄʵÑéÊÒÖÆ·¨ºó£¬·¢ÏÖÓüÓÈÈŨÑÎËáÓëMnO2»ìºÏÎïµÄ·½·¨ÖÆCl2¼ÈÀË·ÑÄÜÔ´ÓÖ²»Ò׿ØÖÆÎ¶ȣ¬ËûÃÇÏÖÓû¶Ô¸ÃʵÑé½øÐиĽø²¢ÑéÖ¤Cl2µÄ²¿·ÖÐÔÖÊ£¬½øÐÐÁËÈçÏÂ̽¾¿»î¶¯¡£

²éÔÄ×ÊÁÏ£ºKMnO4ºÍKClO3µÄÑõ»¯ÐÔ¶¼±ÈMnO2Ç¿£¬ÔÚ²»¼ÓÈȵÄÌõ¼þϼ´¿ÉÓëŨÑÎËá·¢Éú·´Ó¦£¬´Ó¶øÊµÏÖÖÆÈ¡Cl2µÄÄ¿µÄ¡£

£¨1£©È·¶¨·´Ó¦Ô­Àí£º2KMnO4£«16HCl(Ũ)===______£«5Cl2¡ü£«8H2O¡£

£¨2£©Ñ¡ÔñʵÑé×°ÖãºÏÖÓÐÈçͼËùʾµÄ×°Ö㬿ÉÒÔÑ¡ÔñÆäÖеÄ____(ÌîÐòºÅ)½øÐÐʵÑé¡£

£¨3£©×°ÖõÄÕýÈ·Á¬½Ó˳ÐòΪ_______(ÓÃСд×ÖĸÐòºÅ±íʾ)¡£

£¨4£©¢ÙʵÑéʱCÖÐËù×°ÒºÌåÊÇ________¡£

¢ÚÔÚÖÆÂÈÆøÇ°£¬±ØÐë½øÐеÄÒ»Ïî²Ù×÷ÊÇ________¡£

¢ÛDÖÐËù×°ÒºÌåÊÇ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÎÒ¹úÅ®¿ÆÑ§¼ÒÍÀßÏßÏ·¢ÏÖÇàÝïËØ(ÇàÝïËØµÄ»¯Ñ§Ê½£ºC15H22O5)£¬ËüÊÇÒ»ÖÖÓÃÓÚÖÎÁÆÅ±¼²µÄÒ©Îï£¬Ôø¾­Íì¾ÈÁËÊý°ÙÍòÈ˵ÄÉúÃü£¬½üÈÕ»ñµÃŵ±´¶ûÉúÀíÓëҽѧ½±£¬³ÉΪÎÒ¹ú»ñµÃŵ±´¶û¿ÆÑ§½±µÄµÚÒ»ÈË¡£ÏÂÁйØÓÚÇàÝïËØµÄÐðÊö´íÎóµÄÊÇ(¡¡¡¡)

A. ÇàÝïËØµÄÒ»¸ö·Ö×ÓÖк¬ÓÐ42¸öÔ­×Ó

B. ÇàÝïËØÖÐÌ¼ÔªËØµÄÖÊÁ¿·ÖÊýԼΪ63.8%

C. ÇàÝïËØµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª282

D. 0.1 molÇàÝïËØµÄÖÊÁ¿Îª28.2

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸