ÍÊÇÉúÎïÌå±ØÐèµÄ΢Á¿ÔªËØ£¬Ò²ÊÇÈËÀà×îÔçʹÓõĽðÊôÖ®Ò»¡£ÍµÄÉú²úºÍʹÓöԹú¼ÆÃñÉú¸÷¸ö·½Ã涼²úÉúÁËÉîÔ¶µÄÓ°Ïì¡£
£¨1£©Ð´³öÍÓëÏ¡ÏõËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________________________________________________________________________________________________________________________________________¡£
£¨2£©ÎªÁ˱£»¤»·¾³ºÍ½ÚÔ¼×ÊÔ´£¬Í¨³£ÏÈÓÃH2O2ºÍÏ¡ÁòËáµÄ»ìºÏÈÜÒºÈܳö·Ï¾ÉÓ¡Ë¢µç·°åÖеÄÍ£¬×îÖÕʵÏÖ͵ĻØÊÕÀûÓá£Ð´³öÈܳö͵ÄÀë×Ó·½³Ìʽ£º________________________________________________________________________________________________________________________________________________¡£
£¨3£©¹¤ÒµÉÏÒÔ»ÆÍ¿óΪÔÁÏ£¬²ÉÓûð·¨ÈÛÁ¶¹¤ÒÕÉú²úÍ¡£¸Ã¹¤ÒÕµÄÖмä¹ý³Ì»á·¢Éú·´Ó¦£º2Cu2O£«Cu2S6Cu£«SO2¡ü£¬¸Ã·´Ó¦µÄÑõ»¯¼ÁÊÇ______________£»µ±Éú³É19.2 g Cuʱ£¬·´Ó¦ÖÐתÒƵĵç×ÓΪ__________mol¡£
£¨4£©ÍÔÚ³±ÊªµÄ¿ÕÆøÖÐÄÜ·¢ÉúÎüÑõ¸¯Ê´¶øÉúÐ⣬ÍÐâµÄÖ÷Òª³É·ÖΪCu2£¨OH£©2CO3£¨¼îʽ̼ËáÍ£©¡£ÊÔд³öÉÏÊö¹ý³ÌÖиº¼«µÄµç¼«·´Ó¦Ê½£º________________________________________________________________________________________________________________________________________________¡£
£¨5£©Ñо¿ÐÔѧϰС×éÓá°¼ä½ÓµâÁ¿·¨¡±²â¶¨Ä³ÊÔÑùCuSO4¡¤5H2O£¨²»º¬ÄÜÓëI£·´Ó¦µÄÑõ»¯ÐÔÔÓÖÊ£©µÄº¬Á¿¡£È¡a gÊÔÑùÅä³É100 mLÈÜÒº£¬Ã¿´ÎÈ¡25.00 mL£¬µÎ¼ÓKIÈÜÒººóÓа×É«µâ»¯Îï³ÁµíÉú³É¡£Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º___________________________¡£¼ÌÐøµÎ¼ÓKIÈÜÒºÖÁ³Áµí²»ÔÙ²úÉú£¬ÈÜÒºÖеÄI2ÓÃÁò´úÁòËáÄƱê×¼ÈÜÒºµÎ¶¨£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪI2£«2Na2S2O3=2NaI£«Na2S4O6£¬Æ½¾ùÏûºÄc mol/LµÄNa2S2O3ÈÜÒºV mL¡£ÔòÊÔÑùÖÐCuSO4¡¤5H2OµÄÖÊÁ¿·ÖÊýΪ______________¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
2013Äê10ÔÂÎÒÊÐÒǫ̀·ç·ÆÌØÔâÊܵ½ÖØ´óËðʧ£¬Êм²¿ØÖÐÐĽô¼±²É¹ºÏû¶¾Ò©Æ·£¬ÒÔÂú×ãÔÖºóÐèÒª¡£¸´·½¹ýÑõ»¯ÇâÏû¶¾¼Á¾ßÓиßЧ¡¢»·±£¡¢Î޴̼¤ÎÞ²ÐÁô£¬ÆäÖ÷Òª³É·ÖH2O2ÊÇÒ»ÖÖÎÞÉ«Õ³³íÒºÌ壬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÏÂÁз½³ÌÖÐH2O2ËùÌåÏÖµÄÐÔÖÊÓëÆä¿ÉÒÔ×÷ΪÏû¶¾¼ÁÍêÈ«Ò»ÖµÄÊÇ ¡£
A£®BaO2+2HClH2O2+BaCl2
B£®Ag2O+H2O2 =2Ag+O2+H2O
C£®2H2O22H2O+O2¡ü
D£®H2O2+NaCrO2+NaOH=Na2CrO4 +H2O
£¨2£©»ð¼ý·¢Éä³£ÒÔҺ̬ëÂ(N2H4)ΪȼÁÏ£¬ÒºÌ¬H2O2ΪÖúȼ¼Á¡£ÒÑÖª£º
N2H4£¨1£©+O2(g)=N2(g)+2H2O(g) ¡÷H=" -" 534 kJ¡¤mol£1
H2O2£¨1£©=H2O£¨1£©+1/2O2(g) ¡÷H=" -" 98.64 kJ¡¤mol£1
H2O£¨1£©=H2O(g) ¡÷H=+44kJ¡¤mol£l
Ôò·´Ó¦N2H4£¨1£©+2H2O2£¨1£©=N2(g)+4H2O(g)µÄ¡÷H= £¬
¸Ã·´Ó¦µÄ¡÷S= 0(Ìî¡°£¾¡±»ò¡°<¡±)¡£
£¨3£©H2O2ÊÇÒ»ÖÖ²»Îȶ¨Ò×·Ö½âµÄÎïÖÊ¡£
¢ÙÈçͼÊÇH2O2ÔÚûÓд߻¯¼Áʱ·´Ó¦½ø³ÌÓëÄÜÁ¿±ä»¯Í¼£¬ÇëÔÚͼÉÏ»³öʹÓô߻¯¼Á¼Ó¿ì·Ö½âËÙÂÊʱÄÜÁ¿Óë½ø³Ìͼ
¢ÚʵÑé֤ʵ£¬ÍùNa2CO3ÈÜÒºÖмÓÈëH2O2Ò²»áÓÐÆøÅݲúÉú¡£ÒÑÖª³£ÎÂʱH2CO3µÄµçÀë³£Êý·Ö±ðΪKal=4.3¡Ál0£7£¬Ka2 =" 5.0" ¡Ál0£11 ¡£Na2CO3ÈÜÒºÖÐCO32£µÚÒ»²½Ë®½â³£Êý±í´ïʽKhl= £¬³£ÎÂʱKhlµÄֵΪ ¡£ÈôÔÚNa2CO3ÈÜÒºÖÐͬʱ¼ÓÈëÉÙÁ¿Na2CO3¹ÌÌåÓëÊʵ±Éý¸ßÈÜҺζȣ¬ÔòKhlµÄÖµ
(Ìî±ä´ó¡¢±äС¡¢²»±ä»ò²»È·¶¨)¡£
£¨4£©Ä³ÎÄÏ×±¨µ¼Á˲»Í¬½ðÊôÀë×Ó¼°ÆäŨ¶È¶ÔË«ÑõË®Ñõ»¯½µ½âº£ÔåËáÄÆÈÜÒº·´Ó¦ËÙÂʵÄÓ°Ï죬ʵÑé½á¹ûÈçͼ1¡¢Í¼2Ëùʾ¡£
×¢£ºÒÔÉÏʵÑé¾ùÔÚζÈΪ20¡æ¡¢w(H2O2)=0£®25%¡¢pH=7£®12¡¢º£ÔåËáÄÆÈÜҺŨ¶ÈΪ8mg¡¤L-lµÄÌõ¼þϽøÐС£Í¼1ÖÐÇúÏßa£ºH2O2£»b£ºH2O2+Cu2+£»c£ºH2O2+Fe2+£»d£ºH2O2+Zn2+£»e£ºH2O2+Mn2+£»Í¼2ÖÐÇúÏßf£º·´Ó¦Ê±¼äΪ1h£»g£º·´Ó¦Ê±¼äΪ2h£»Á½Í¼ÖеÄ×Ý×ø±ê´ú±íº£ÔåËáÄÆÈÜÒºµÄÕ³¶È(º£ÔåËáÄÆŨ¶ÈÓëÈÜÒºÕ³¶ÈÕýÏà¹Ø)¡£
ÓÉÉÏÊöÐÅÏ¢¿ÉÖª£¬ÏÂÁÐÐðÊö´íÎóµÄÊÇ (ÌîÐòºÅ)¡£
A£®ÃÌÀë×ÓÄÜʹ¸Ã½µ½â·´Ó¦ËÙÂʼõ»º
B£®ÑÇÌúÀë×ӶԸýµ½â·´Ó¦µÄ´ß»¯Ð§ÂʱÈÍÀë×ÓµÍ
C£®º£ÔåËáÄÆÈÜÒºÕ³¶ÈµÄ±ä»¯¿ìÂý¿É·´Ó³³öÆä½µ½â·´Ó¦ËÙÂʵĿìÂý
D£®Ò»¶¨Ìõ¼þÏ£¬ÍÀë×ÓŨ¶ÈÒ»¶¨Ê±£¬·´Ó¦Ê±¼äÔ½³¤£¬º£ÔåËáÄÆÈÜҺŨ¶ÈԽС
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
Ðí¶àÁòµÄº¬ÑõËáÑÎÔÚÒ½Ò©¡¢»¯¹¤µÈ·½ÃæÓÐ×ÅÖØÒªµÄÓÃ;¡£
£¨1£©Öؾ§Ê¯£¨BaSO4£©³£×ö賦µÀÔìÓ°¼Á¡£
ÒÑÖª£º³£ÎÂÏ£¬Ksp£¨BaSO4£©£½1.1¡Á10£10¡£ÏòBaSO4Ðü×ÇÒºÖмÓÈëÁòËᣬµ±ÈÜÒºµÄpH£½2ʱ£¬ÈÜÒºÖÐc£¨Ba2£«£©£½ __¡£
£¨2£©ÁòËáÑÇÌú茶§Ìå[£¨NH4£©2Fe£¨SO4£©2¡¤6H2O]³£×ö·ÖÎö¼Á¡£
¢Ù¼ìÑ龧ÌåÖк¬ÓÐNH4+µÄ·½·¨Îª ¡£
¢ÚµÈÎïÖʵÄÁ¿Å¨¶ÈµÄËÄÖÖÏ¡ÈÜÒº£º
a£®£¨NH4£©2Fe£¨SO4£©2 b£®NH4HSO4
c£®£¨NH4£© 2SO4 d£®£¨NH4£©2SO3£¬
ÆäÖÐc£¨NH4+£©ÓÉ´óµ½Ð¡µÄ˳ÐòΪ __£¨ÌîÑ¡Ïî×Öĸ£©¡£
£¨3£©¹ý¶þÁòËá¼Ø£¨K2S2O8£©³£×öÇ¿Ñõ»¯¼Á£¬Na2S2O3³£×ö»¹Ô¼Á¡£
¢ÙK2S2O8ÈÜÒºÓëËáÐÔMnSO4ÈÜÒº»ìºÏ£¬ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬¿ÉÒԹ۲쵽ÈÜÒº±äΪ×ÏÉ«£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ __¡£
¢ÚÓò¬×öµç¼«£¬µç½âH2SO4ºÍK2SO4µÄ»ìºÏÈÜÒº¿ÉÒÔÖƱ¸K2S2O8£¬ÆäÑô¼«µÄµç¼«·´Ó¦Ê½Îª __£¬µç½â¹ý³ÌÖÐÒõ¼«¸½½üÈÜÒºµÄpH½« __£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©¡£
¢Û²úÆ·ÖÐK2S2O8µÄº¬Á¿¿ÉÓõâÁ¿·¨²â¶¨¡£²Ù×÷²½ÖèΪ³ÆÈ¡0.3000 g²úÆ·ÓÚµâÁ¿Æ¿ÖУ¬¼Ó50 mLË®Èܽ⣻¼ÓÈë4.000 g KI¹ÌÌ壨ÉÔ¹ýÁ¿£©£¬Õñµ´Ê¹Æä³ä·Ö·´Ó¦£»¼ÓÈëÊÊÁ¿´×ËáÈÜÒºËữ£¬ÒÔ __Ϊָʾ¼Á£¬ÓÃ0.1000 mol¡¤L£1 Na2S2O3±ê×¼ÒºµÎ¶¨ÖÁÖյ㣨ÒÑÖª£ºI2£«2S2O32-=2I££«S4O62-£©¡£Öظ´2´Î£¬²âµÃƽ¾ùÏûºÄ±ê×¼Òº21.00 mL¡£¸Ã²úÆ·ÖÐK2S2O8µÄÖÊÁ¿·ÖÊýΪ£¨ÔÓÖʲ»²Î¼Ó·´Ó¦£© __£¨ÁÐʽ²¢¼ÆË㣩¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
Á×ÊǵؿÇÖк¬Á¿½ÏΪ·á¸»µÄ·Ç½ðÊôÔªËØ£¬Ö÷ÒªÒÔÄÑÈÜÓÚË®µÄÁ×ËáÑÎÈçCa3(PO4)2µÈÐÎʽ´æÔÚ¡£ËüµÄµ¥Öʺͻ¯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Óá£
(1)°×Á×(P4)¿ÉÓÉCa3(PO4)2¡¢½¹Ì¿ºÍSiO2ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦»ñµÃ¡£Ïà¹ØÈÈ»¯Ñ§·½³ÌʽÈçÏ£º
2Ca3(PO4)2(s)£«10C(s)=6CaO(s)£«P4(s)£«10CO(g)¡¡ ¦¤H1£½£«3359.26 kJ¡¤mol£1
CaO(s)£«SiO2(s)=CaSiO3(s) ¦¤H1£½£89.61 kJ¡¤mol£1
2Ca3(PO4)2(s)£«6SiO2(s)£«10C(s)=6CaSiO3(s)£«P4(s)£«10CO(g)¡¡ ¦¤H3
Ôò¦¤H3£½________kJ¡¤mol£1¡£
(2)°×Á×Öж¾ºó¿ÉÓÃCuSO4ÈÜÒº½â¶¾£¬½â¶¾ÔÀí¿ÉÓÃÏÂÁл¯Ñ§·½³Ìʽ±íʾ£º
11P4£«60CuSO4£«96H2O=20Cu3P£«24H3PO4£«60H2SO4
60 mol CuSO4ÄÜÑõ»¯°×Á×µÄÎïÖʵÄÁ¿ÊÇ________¡£
(3)Á×µÄÖØÒª»¯ºÏÎïNaH2PO4¡¢Na2HPO4ºÍNa3PO4¿Éͨ¹ýH3PO4ÓëNaOHÈÜÒº·´Ó¦»ñµÃ£¬º¬Á׸÷ÎïÖֵķֲ¼·ÖÊý(ƽºâʱijÎïÖÖµÄŨ¶ÈÕ¼¸÷ÎïÖÖŨ¶ÈÖ®ºÍµÄ·ÖÊý)ÓëpHµÄ¹ØϵÈçÏÂͼËùʾ¡£
¢ÙΪ»ñµÃ¾¡¿ÉÄÜ´¿µÄNaH2PO4£¬pHÓ¦¿ØÖÆÔÚ________£»pH£½8ʱ£¬ÈÜÒºÖÐÖ÷Òªº¬Á×ÎïÖÖŨ¶È´óС¹ØϵΪ________¡£
¢ÚNa2HPO4ÈÜÒºÏÔ¼îÐÔ£¬ÈôÏòÈÜÒºÖмÓÈë×ãÁ¿µÄCaCl2ÈÜÒº£¬ÈÜÒºÔòÏÔËáÐÔ£¬ÆäÔÒòÊÇ________(ÓÃÀë×Ó·½³Ìʽ±íʾ)¡£
£¨4£©Á׵Ļ¯ºÏÎïÈýÂÈÑõÁ×£¨£©Óë¼¾ÎìËÄ´¼£¨£©ÒÔÎïÖʵÄÁ¿Ö®±È2:1·´Ó¦Ê±£¬¿É»ñµÃÒ»ÖÖÐÂÐÍ×èȼ¼ÁÖмäÌåX£¬²¢ÊͷųöÒ»ÖÖËáÐÔÆøÌå¡£¼¾ÎìËÄ´¼ÓëXµÄºË´Å¹²ÕñÇâÂùÈçÏÂͼËùʾ£º
¢ÙËáÐÔÆøÌåÊÇ______________________(Ìѧʽ)¡£
¢ÚXµÄ½á¹¹¼òʽΪ__________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
CuÓëÒ»¶¨Å¨¶ÈµÄHNO3·´Ó¦Îª:3Cu+2NO3¡ª+xH+3Cu2++2R+yH2O¡£
(1)·´Ó¦ÖеÄx=¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
(2)·´Ó¦²úÎïRµÄ»¯Ñ§Ê½Îª¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
(3)²Î¼Ó·´Ó¦µÄCuºÍÏûºÄHNO3µÄÎïÖʵÄÁ¿Ö®±ÈΪ¡¡¡¡¡¡¡¡¡¡¡£
(4)1.5 mol CuÍêÈ«·´Ó¦Ê±×ªÒƵĵç×ÓÊýΪ¡¡¡¡¡¡¡¡¡¡¡¡¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
Ϊ̽Ë÷¹¤Òµº¬ÂÁ¡¢Ìú¡¢ÍºÏ½ð·ÏÁϵÄÔÙÀûÓÃ,¼×ͬѧÉè¼ÆµÄʵÑé·½°¸ÈçÏÂ:
Çë»Ø´ð:
(1)²Ù×÷¢ÙÓõ½µÄ²£Á§ÒÇÆ÷ÓÐ ¡£
(2)д³ö·´Ó¦¢ÙµÄ»¯Ñ§·½³Ìʽ: ,·´Ó¦¢ÚµÄÀë×Ó·´Ó¦·½³Ìʽ: ¡£
(3)Éè¼ÆʵÑé·½°¸,¼ì²âÂËÒºDÖк¬ÓеĽðÊôÀë×Ó(ÊÔ¼Á×ÔÑ¡) ¡£
(4)ÔÚÂËÔüEÖмÓÈëÏ¡ÁòËáºÍÊÔ¼ÁYÖƵ¨·¯¾§ÌåÊÇÒ»ÖÖÂÌÉ«»¯Ñ§¹¤ÒÕ,ÊÔ¼ÁYΪÎÞÉ«ÒºÌå,·´Ó¦¢ÜµÄ×Ü»¯Ñ§·½³ÌʽÊÇ ¡£
(5)ÒÒͬѧÔÚ¼×ͬѧ·½°¸µÄ»ù´¡ÉÏÌá³öÓÃÂËÔüBÀ´ÖƱ¸FeCl3¡¤6H2O¾§Ìå,ÔÚÂËÔüÖеμÓÑÎËáʱ,·¢ÏÖ·´Ó¦ËÙÂʱÈͬŨ¶ÈÑÎËáÓë´¿Ìú·Û·´Ó¦Òª¿ì,ÆäÔÒòÊÇ ¡£
½«ËùµÃÂÈ»¯ÌúÈÜÒºÓüÓÈÈŨËõ¡¢½µÎ½ᾧ·¨ÖƵÃFeCl3¡¤6H2O ¾§Ìå,¶ø²»ÓÃÖ±½ÓÕô·¢½á¾§µÄ·½·¨À´ÖƵþ§ÌåµÄÀíÓÉÊÇ ¡£
(6)½«ÂËÔüBµÄ¾ùÔÈ»ìºÏÎïƽ¾ù·Ö³ÉËĵȷÝ,·Ö±ð¼ÓÈëͬŨ¶ÈµÄÏ¡ÏõËá,³ä·Ö·´Ó¦ºó,ÔÚ±ê×¼×´¿öÏÂÉú³ÉNOµÄÌå»ýÓëÊ£Óà½ðÊôµÄÖÊÁ¿¼ûϱí(ÉèÏõËáµÄ»¹Ô²úÎïÖ»ÓÐNO)¡£
ʵÑé±àºÅ | ¢Ù | ¢Ú | ¢Û | ¢Ü |
Ï¡ÏõËáÌå»ý(mL) | 100 | 200 | 300 | 400 |
Ê£Óà½ðÊôÖÊÁ¿(g) | 9.0 | 4.8 | 0 | 0 |
NOÌå»ý(L) | 1.12 | 2.24 | 3.36 | V |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÀûÓ÷ϾɶÆпÌúƤ¿ÉÖƱ¸´ÅÐÔFe3O4½ºÌåÁ£×Ó¼°¸±²úÎïZnO¡£ÖƱ¸Á÷³ÌͼÈçÏ£º
ÒÑÖª£ºZn¼°Æ仯ºÏÎïµÄÐÔÖÊÓëAl¼°Æ仯ºÏÎïµÄÐÔÖÊÏàËÆ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÃNaOHÈÜÒº´¦Àí·Ï¾É¶ÆпÌúƤµÄ×÷ÓÃÓÐ________¡£
A£®È¥³ýÓÍÎÛ | B£®Èܽâ¶Æп²ã | C£®È¥³ýÌúÐâ | D£®¶Û»¯ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
Ñо¿Ì¼¼°Æ仯ºÏÎïµÄ×ÛºÏÀûÓöԴٽøµÍ̼Éç»áµÄ¹¹½¨¾ßÓÐÖØÒªµÄÒâÒå¡£ÇëÔËÓÃÏà¹Ø֪ʶÑо¿Ì¼¼°Æ仯ºÏÎïµÄÐÔÖÊ¡£
£¨1£©½üÄêÀ´£¬ÎÒ¹úÓõ绡·¨ºÏ³ÉµÄ̼ÄÉÃ×¹ÜÖг£°éÓдóÁ¿Ì¼ÄÉÃ׿ÅÁ££¨ÔÓÖÊ£©£¬ÕâÖÖ̼ÄÉÃ׿ÅÁ£¿ÉÓÃÑõ»¯Æø»¯·¨Ìá´¿£¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
C+ K2Cr2O7+ ¡ª CO2¡ü+ K2SO4 + Cr2(SO4)3+ H2O
¢ÙÍê³É²¢ÅäƽÉÏÊö»¯Ñ§·½³Ìʽ¡£
¢ÚÔÚÉÏÊö·½³ÌʽÉÏÓõ¥ÏßÇűê³ö¸Ã·´Ó¦µç×ÓתÒƵķ½ÏòÓëÊýÄ¿¡£
£¨2£©¸ßÎÂʱ£¬ÓÃCO»¹ÔMgSO4¿ÉÖƱ¸¸ß´¿MgO¡£
¢Ù750¡æʱ£¬²âµÃÆøÌåÖꬵÈÎïÖʵÄÁ¿SO2ºÍSO3£¬´Ëʱ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ ¡£
¢ÚÓÉMgO¿ÉÖƳɡ°Ã¾£´ÎÂÈËáÑΡ±µç³Ø£¬Æä×°ÖÃʾÒâͼÈçͼ1£¬¸Ãµç³Ø·´Ó¦µÄÀë×Ó·½³ÌʽΪ ¡£
¡¡
ͼ1 ͼ2 ͼ3
£¨3£©¶þÑõ»¯Ì¼ºÏ³É¼×´¼ÊÇ̼¼õÅŵÄз½Ïò£¬½«CO2ת»¯Îª¼×´¼µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCO2(g) +3H2(g)CH3OH(g) +H2O(g) ¡÷H
¢Ù¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪK£½ ¡£
¢ÚÈ¡Îå·ÝµÈÌå»ýCO2ºÍH2µÄ»ìºÏÆøÌå(ÎïÖʵÄÁ¿Ö®±È¾ùΪ1¡Ã3)£¬·Ö±ð¼ÓÈëζȲ»Í¬¡¢ÈÝ»ýÏàͬµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÉÏÊö·´Ó¦£¬·´Ó¦Ïàͬʱ¼äºó£¬²âµÃ¼×´¼µÄÌå»ý·ÖÊý¦Õ(CH3OH)Ó뷴ӦζÈTµÄ¹ØϵÇúÏßÈçͼ2Ëùʾ£¬ÔòÉÏÊöCO2ת»¯Îª¼×´¼·´Ó¦µÄ¡÷H 0(Ìî¡°>¡±¡°<¡±»ò¡°£½¡±)¡£
¢ÛÔÚÁ½ÖÖ²»Í¬Ìõ¼þÏ·¢Éú·´Ó¦£¬²âµÃCH3OHµÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯Èçͼ3Ëùʾ£¬ÇúÏßI¡¢II¶ÔÓ¦µÄƽºâ³£Êý´óС¹ØϵΪK¢ñ KII(Ìî¡°£¾¡± ¡°£¼¡±»ò¡°£½¡±)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÒÑÖªµªÔªËØÓÐÐí¶à»¯ºÏ¼Û£¬Æ仯ºÏ¼ÛÓÐÒ»3¡¢0¡¢+1¡¢+2¡¢+3¡¢+4¡¢+5
£¨1£©Ï±íÊÇÏõËáÓëijÖÖ½ðÊôMÔÚijЩÌõ¼þÏ·´Ó¦ËùµÃ»¹Ô²úÎïµÄ¹Øϵ£º
½ðÊôпÓëijŨ¶ÈµÄÏõËᷴӦʱ£¬ÎÞÆøÌå²úÉú£¬Ôò´Ë»¯Ñ§·´Ó¦Öб»»¹ÔµÄÏõËáÓë²Î¼Ó·´Ó¦µÄÏõËáÎïÖʵÄÁ¿Ö®±ÈÊÇ____________¡£
£¨2£©ÒÔ°±×÷ΪȼÁϵĹÌÌåÑõ»¯Îº¬ÓÐ02£)ȼÁϵç³Ø£¬¾ßÓÐÈ«¹Ì̬½á¹¹¡¢ÄÜÁ¿Ð§Âʸߡ¢ÎÞÎÛȾµÈÌص㣬ÁíÍâ°±Æøº¬ÇâÁ¿¸ß£¬²»º¬Ì¼£¬Ò×Òº»¯£¬·½±ãÔËÊäºÍÖü´æ£¬ÊǺܺõÄÇâÔ´ÔØÌå¡£Æ乤×÷ÔÀíÈçͼËùʾ£¬
¢Ù¸Ãµç³Ø¹¤×÷ʱµÄ×Ü·´Ó¦Îª_______________________
¢Ú¹ÌÌåÑõ»¯Îï×÷Ϊµç³Ø¹¤×÷µÄµç½âÖÊ£¬O2-Òƶ¯·½ÏòΪ________(Ñ¡Ìî¡°Óɵ缫aÏòµç¼«b¡±»ò¡°Óɵ缫bÏòµç¼«a¡±)¡£
¢Û¸Ãµç³Ø¹¤×÷ʱ£¬ÔÚ½Ó´¥ÃæÉÏ·¢ÉúµÄµç¼«·´Ó¦Îª________¡£
£¨3£©N2O5ÊÇÒ»ÖÖÐÂÐÍÏõ»¯¼Á£¬ÆäÐÔÖʺÍÖƱ¸Êܵ½ÈËÃǵĹØ×¢¡£ÈçͼËùʾװÖÿÉÓÃÓÚÖƱ¸N2O5,д³öÔÚµç½â³ØÖÐÉú³ÉN2O5µÄµç¼«·´Ó¦Ê½________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com