ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿¡°·Ö×Óɸ¡±ÊÇÒ»ÖÖ¾ßÓжà¿×½á¹¹µÄÂÁ¹èËáÑÎ(NaAlSiO4¡¤nH2O)£¬ÆäÖÐÓÐÐí¶àÁý×´¿×ѨºÍͨµÀ£¬ÄÜÈÃÖ±¾¶±È¿×ѨСµÄ·Ö×Óͨ¹ý¶ø½«´óµÄ·Ö×ÓÁôÔÚÍâÃ棬¹Ê´ËµÃÃû¡£ÀûÓÃÂÁ»Ò(Ö÷Òª³É·ÖΪAl¡¢Al2O3¡¢AlN¡¢FeOµÈ)ÖƱ¸¡°·Ö×Óɸ¡±µÄÒ»ÖÖ¹¤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©¡°·Ö×Óɸ¡±µÄ»¯Ñ§Ê½ÓÃÑõ»¯ÎïÐÎʽ¿É±íʾΪ_______________¡£

£¨2£©ÂÁ»ÒË®½â²úÉúµÄÆøÌåΪ________(Ìѧʽ)£¬¸ÃÆøÌå·Ö×ÓÖм«ÐÔ¼üµÄÊýĿΪ___________£»¡°Ë®½â¡±ÔÚ¼ÓÈÈÌõ¼þ϶ø²»ÔÚÊÒÎÂϽøÐеÄÔ­ÒòÊÇ________________________¡£

£¨3£©¡°ËáÈÜ¡±Ê±£¬·¢ÉúÑõ»¯»¹Ô­·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________________¡£

£¨4£©¸Ã¹¤ÒÕÖÐÂËÔüµÄÑÕɫΪ________________________¡£

£¨5£©Ä³Ñ§Ï°Ð¡×éÉè¼ÆʵÑéÄ£Äâ´ÓŨËõº£Ë®(º¬Ca2+¡¢Mg2+¡¢SO42-)ÖÐÌáÈ¡ÊÔ¼Á¼¶NaCl£º

¢ÙʵÑéÖÐÈôÏòŨËõº£Ë®ÖмÓÈëµÄÊÇNa2CO3ŨÈÜÒº£¬ÔòÓÐÄÑÈܵÄMg2(OH)2CO3Éú³É£¬Í¬Ê±ÓÐÆøÌåÒݳö¡£¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________¡£

¢Ú¸ÃѧϰС×é·¢ÏÖÉÏÊöʵÑ鼴ʹBaCl2ÓÃÁ¿²»×㣬µÚ¢ó²½³ÁµíÖÐÒÀÈ»º¬ÓÐÉÙÁ¿BaCO3¡£´Óƽºâ½Ç¶È·ÖÎöÆäÔ­Òò£º_____________________________________________¡£

¡¾´ð°¸¡¿ Na2O¡¤Al2O3¡¤2SiO2¡¤2nH2O NH3 3 ¼Ó¿ìAlNË®½â·´Ó¦ËÙÂÊ£¬½µµÍNH3ÔÚË®ÖеÄÈܽâ¶È£¬´ÙʹNH3Òݳö 2Al+6H+=2Al3++3H2¡ü ºìºÖÉ« 2CO32-+2Mg2++H2O=Mg2(OH)2CO3¡ý+CO2¡ü ¼ÓÈëµÄCO32-ʹBaSO4µÄ³ÁµíÈܽâƽºâÏòÈܽⷽÏòÒƶ¯£¬´Ó¶øÐγÉBaCO3

¡¾½âÎö¡¿£¨1£©ÂÁ¹èËáÑÎ(NaAlSiO4¡¤nH2O)×é³É¸´ÔÓ£¬ÓÃÑõ»¯Îï±íʾʱ£¬½ðÊôÑõ»¯ÎïÔÚÇ°£¬·Ç½ðÊôÑõ»¯ÎïÔÚºó£¬ÀûÓÃÔ­×ÓÊغã¹æÂɿɵÃNa2O¡¤Al2O3¡¤2SiO2¡¤2nH2O£»ÕýÈ·´ð°¸£ºNa2O¡¤Al2O3¡¤2SiO2¡¤2nH2O¡£

£¨2£©ÂÁ»Ò(Ö÷Òª³É·ÖΪAl¡¢Al2O3¡¢AlN¡¢FeOµÈ)ÖÐÖ»ÓÐAlNºÍË®·¢ÉúË®½â·´Ó¦Éú³ÉÇâÑõ»¯ÂÁ³ÁµíºÍ°±Æø£»°±Æø·Ö×ÓΪ¼«ÐÔ·Ö×Ó£¬º¬Óм«ÐÔ¼üµÄÊýĿΪ3¸ö£»¼ÓÈÈ¿ÉÒÔ´Ù½øµª»¯ÂÁË®½âÉú³É°±Æø,½µµÍÔÚË®ÖеÄÈܽâ¶È,´ÙʹÒݳö£»ÕýÈ·´ð°¸£ºNH3£»3£»¼Ó¿ìAlNË®½â·´Ó¦ËÙÂÊ£¬½µµÍNH3ÔÚË®ÖеÄÈܽâ¶È£¬´ÙʹNH3Òݳö¡£

£¨3£©¡°ËáÈÜ¡±Ê±£¬½ðÊôÂÁÓëËá·´Ó¦Éú³ÉÂÁÑκÍÇâÆø£¬Àë×Ó·½³ÌʽΪ£º2Al+6H+=2Al3++3H2¡ü£»ÕýÈ·´ð°¸£º2Al+6H+=2Al3++3H2¡ü¡£

£¨4£©ÌúÀë×ÓÔÚ¼îÐÔÌõ¼þÏÂÉú³ÉÇâÑõ»¯ÌúºìºÖÉ«³Áµí£¬¸Ã¹¤ÒÕÖÐÂËÔüµÄÑÕɫΪºìºÖÉ«£»ÕýÈ·´ð°¸£ººìºÖÉ«¡£

£¨5£©¢ÙʵÑéÖÐÈôÏòŨËõº£Ë®ÖмÓÈëµÄÊÇNa2CO3ŨÈÜÒº£¬ÔòÓÐÄÑÈܵÄMg2(OH)2CO3Éú³É£¬Í¬Ê±ÓжþÑõ»¯Ì¼ÆøÌåÒݳö£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2CO32-+2Mg2++H2O=Mg2(OH)2CO3¡ý+CO2¡ü£»ÕýÈ·´ð°¸£º2CO32-+2Mg2++H2O=Mg2(OH)2CO3¡ý+CO2¡ü¡£

¢Ú¼ÓÈëµÄCO32-ʹBaSO4µÄ³ÁµíÈܽâƽºâÏòÈܽⷽÏòÒƶ¯£¬´Ó¶øÐγÉBaCO3£¬Òò´ËÉÏÊöʵÑ鼴ʹBaCl2ÓÃÁ¿²»×㣬µÚ¢ó²½³ÁµíÖÐÒÀÈ»º¬ÓÐÉÙÁ¿BaCO3£»ÕýÈ·´ð°¸£º¼ÓÈëµÄCO32-ʹBaSO4µÄ³ÁµíÈܽâƽºâÏòÈܽⷽÏòÒƶ¯£¬´Ó¶øÐγÉBaCO3¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ê¯ÓͲúÆ·Öгýº¬ÓÐH2SÍ⣬»¹º¬Óи÷ÖÖÐÎ̬µÄÓлúÁò£¬ÈçCOS¡¢CH3SHµÈ¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©CH3SH£¨¼×Áò´¼£©µÄµç×ÓʽΪ__________¡£

£¨2£©COºÍH2S·´Ó¦¿É²úÉúôÊ»ùÁò( COS)¡£ÔÚÒ»ºãÈÝÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦£ºCO(g)+H2S(g)COS(g)+H2(g)²¢´ïµ½Æ½ºâ£¬Êý¾ÝÈçϱíËùʾ£º

ʵÑé

ζÈ/¡æ

Æðʼʱ

ƽºâʱ

n(CO)/mol

n(H2S)/mol

n(COS)/mol

n(H2)/mol

n(CO)/mol

1

150

10.0

10.0

0

0

7.0

2

150

7.0

8.0

2.0

4.5

a

3

400

20.0

20.0

0

0

16.0

¢Ù¸Ã·´Ó¦ÊÇ________·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©¡£

¢ÚʵÑé1´ïƽºâʱ£¬COµÄת»¯ÂÊΪ_______¡£

¢ÛʵÑé2´ïµ½Æ½ºâʱ£¬a_______7.0£¨Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©¡£

¢ÜʵÑé3´ïƽºâºó£¬ÔÙ³äÈë1.0 molH2£¬Æ½ºâ³£ÊýÖµ____£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©¡£

£¨3£©COSÊÇ´óÆøÎÛȾÎï¡£ÔÚËáÐÔÈÜÒºÖпÉÓÃH2O2Ñõ»¯COSÉú³ÉÒ»ÖÖÇ¿ËáÍÑÁò¡£¸ÃÍѳý·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________¡£

£¨4£©ÓÐÒ»ÖÖÍÑÁò¹¤ÒÕΪ£ºÕæ¿ÕK2CO3Ò»¿ËÀÍ˹·¨¡£

¢ÙK2CO3ÈÜÒºÎüÊÕH2SµÄ·´Ó¦ÎªK2CO3 +H2S =KHS +KHCO3£¬¸Ã·´Ó¦µÄƽºâ³£ÊýµÄ¶ÔÊýֵΪlgK=_____£¨ÒÑÖª£ºH2CO3 lgK1=-6.4£¬lgK,2=- 10.3£»H2S lgKl=-7.0£¬lgK2

¢ÚÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º

a. 2H2S(g)+3O2(g)=2SO2(g)+2H2O(1) ¡÷H1=-1172kJ/mol

b. 2H2S(g)+O2(g)=2S(s)+2H2O(1) ¡÷H2 = 632 kJ/mol

¿ËÀÍ˹·¨»ØÊÕÁòµÄ·´Ó¦ÎªSO2ºÍH2SÆøÌå·´Ó¦Éú³ÉS(s)£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ_________¡£

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

¾«Ó¢¼Ò½ÌÍø