¼×¡¢ÒÒ¡¢±ûÈý¸öÈÝÆ÷ÖÐ×î³õ´æÔÚµÄÎïÖʼ°ÊýÁ¿ÈçͼËùʾ£¬Èý¸öÈÝÆ÷×î³õµÄÈÝ»ýÏàµÈ£¬Î¶ÈÏàͬ£¬·´Ó¦Öмס¢±ûµÄÈÝ»ý²»±ä£¬ÒÒÖеÄѹǿ²»±ä£¬ÔÚÒ»¶¨Î¶ÈÏ·´Ó¦´ïµ½Æ½ºâ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®Æ½ºâʱ¸÷ÈÝÆ÷ÄÚc£¨NO2£©µÄ´óС˳ÐòΪÒÒ>¼×>±û

B£®Æ½ºâʱN2O4µÄ°Ù·Öº¬Á¿£ºÒÒ>¼×£½±û

C£®Æ½ºâʱ¼×ÖÐNO2Óë±ûÖÐN2O4µÄת»¯ÂÊÏàͬ

D£®Æ½ºâʱ»ìºÏÎïµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿£º¼×>ÒÒ>±û

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016-2017ѧÄê½­Î÷Ê¡¸ß¶þÉϵڶþ´Î¿¼ÊÔ»¯Ñ§¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

( 14·Ö)

I .£¨1£©ÒÑÖªH2µÄȼÉÕÈÈ285.8KJ/mol£¬Ð´³öҺ̬ˮµç½âÉú³ÉH2ºÍO2µÄÈÈ»¯Ñ§·½³Ìʽ £»

£¨2£©ÒÑÖª2SO2(g)+O2(g) = 2SO3(g) ¦¤H£½£­197 kJ/mol£¬ÏàͬζȺÍѹǿÌõ¼þÏ£¬4molSO2ºÍ2molO2³äÈëÈÝÆ÷Öз´Ó¦ÖÁƽºâʱ·Å³öµÄÄÜÁ¿ÎªQ KJ£¬ÔòQ 394KJ(Ìî¡°>¡±¡°<¡±»ò¡°=¡±)

II.¼×´¼ÊÇÒ»ÖÖ¿ÉÔÙÉúÄÜÔ´£¬¾ßÓй㷺µÄ¿ª·¢ºÍÓ¦ÓÃÇ°¾°¡£¹¤ÒµÉÏÒ»°ã²ÉÓÃÏÂÁÐÁ½ÖÖ·´Ó¦ºÏ³É¼×´¼£º

·´Ó¦I£ºCO2(g)+3H2(g)CH3OH(g)+H2O(g) ¦¤H1

·´Ó¦II£ºCO(g)+2H2(g)CH3OH(g) ¦¤H2

ϱíËùÁÐÊý¾ÝÊÇ·´Ó¦¢òÔÚ²»Í¬Î¶ÈϵĻ¯Ñ§Æ½ºâ³£Êý(K)

ζÈ

250¡æ 

300¡æ 

350¡æ 

K

2.041

0.270

0.012

£¨1£©ÓɱíÖÐÊý¾ÝÅжϦ¤H2 0(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±)£¬

£¨2£©ÈôÈÝÆ÷ÈÝ»ý²»±ä£¬ÏÂÁдëÊ©¿ÉÔö¼Ó¼×´¼²úÂʵÄÊÇ £¬

A£®Éý¸ßζÈ

B£®½«CH3OH(g)´ÓÌåϵÖзÖÀë

C£®Ê¹ÓúÏÊʵĴ߻¯¼Á

D£®ºãκãÈݳäÈëHe£¬Ê¹Ìåϵ×ÜѹǿÔö´ó

E£®°´Ô­±ÈÀýÔÙ³äÈë COºÍ H2

£¨3£©Ä³Î¶ÈÏ£¬½«2 mol COºÍ6 mol H2³äÈë2LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬³ä·Ö·´Ó¦£¬´ïµ½Æ½ºâºó£¬²âµÃc(CO)=0.2 mol¡¤L-1£¬ÔòCOµÄת»¯ÂÊΪ £¬´ËʱµÄζÈΪ (´ÓÉϱíÖÐÑ¡Ôñ)£»

£¨4£©ºãÎÂÏ£¬1mol COºÍnmol H2ÔÚÒ»¸öÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷Öз´Ó¦´ïµ½Æ½ºâºó£¬Éú³Éa molCH3OH£®ÈôÆðʼʱ·ÅÈë3molCO+3nmolH2£¬Ôò´ïƽºâʱÉú³ÉCH3OH_______mol¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìºþÄÏʦ´ó¸½ÖиßÈýÉÏÔ¿¼ËÄ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ

ij»¯Ñ§Ð¡×éµÄͬѧģÄ⹤ҵÖÆÏõËáÉè¼ÆÁËÈçÏÂͼËùʾµÄ×°Öá£ÒÑÖª£º

CaCl2£«nH2O¡úCaCl2¡¤nH2O£»CaCl2£«8NH3¡ú[Ca(NH3)8]Cl2£¬¸ù¾ÝÌâÒâÍê³ÉÏÂÁÐÌî¿Õ£º

£¨1£©·ÖҺ©¶·Öа±Ë®µÄŨ¶ÈΪ9.0 mol/L¡£ÏÖÓÃÖÊÁ¿·ÖÊýΪ35%¡¢ÃܶÈΪ0.88 g/cm3µÄ°±Ë®ÅäÖÆ9.0 mol/LµÄ°±Ë®100 mL£¬ÐèÒªµÄ¶¨Á¿ÒÇÆ÷ÓÐ (Ñ¡Ìî±àºÅ)¡£

a£®100 mLÈÝÁ¿Æ¿ b£®10 mLÁ¿Í² c£®50 mLÁ¿Í² d£®µç×ÓÌìƽ

£¨2£©ÊÜÈÈʱ£¬ÒÒÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ ¡£

£¨3£©ÊµÑ鿪ʼÏȼÓÈÈ´ß»¯¼Á£¬µ±´ß»¯¼Á´ïºìÈÈʱÔÙ´ò¿ª·ÖҺ©¶·»îÈû²¢ÒÆ×߾ƾ«µÆ£¬¿É¹Û²ìµ½µÄÏÖÏóÓÐ ¡£

£¨4£©¸ÉÔï¹Ü¼×µÄ×÷ÓÃÊÇ £»±ûÖÐÊ¢·ÅµÄҩƷΪ (Ñ¡ÌîÏÂÁбàºÅ)£¬ÆäÄ¿µÄÊÇ ¡£

a£®Å¨H2SO4 b£®ÎÞË®CaCl2 c£®¼îʯ»Ò d£®ÎÞË®CuSO4

£¨5£©¶¡ÖгýÁËNOÖ®Í⣬»¹¿ÉÄÜ´æÔÚµÄÆøÌåÓÐ (Ìîд»¯Ñ§Ê½)¡£ÉÕ±­Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìºÚÁú½­Äµµ¤½­µÚÒ»¸ß¼¶ÖÐѧ¸ßÈý12ÔÂÔ¿¼»¯Ñ§¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ

ÏÖÓÐÒ»·Ýº¬ÓÐFeCl3ºÍFeCl2¹ÌÌå»ìºÏÎΪ²â¶¨¸÷³É·ÖµÄº¬Á¿½øÐÐÈçÏÂÁ½¸öʵÑ飺

ʵÑé1£º¢Ù ³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÑùÆ·£¬½«ÑùÆ·Èܽ⣻

¢Ú ÏòÈܽâºóµÄÈÜÒºÖмÓÈë×ãÁ¿µÄAgNO3ÈÜÒº£¬²úÉú³Áµí£»

¢Û ½«³Áµí¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïµÃµ½°×É«¹ÌÌå17.22 g¡£

ʵÑé2£º¢Ù ³ÆÈ¡ÓëʵÑé1ÖÐÏàͬÖÊÁ¿µÄÑùÆ·£¬½«ÑùÆ·Èܽ⣻

¢Ú ÏòÈܽâºóµÄÈÜÒºÖÐͨÈë×ãÁ¿µÄCl2£»

¢Û ÔÙÏò¢ÚËùµÃÈÜÒºÖмÓÈë×ãÁ¿µÄNaOHÈÜÒº£¬µÃµ½ºìºÖÉ«³Áµí£»

¢Ü ½«³Áµí¹ýÂË¡¢Ï´µÓºó£¬¼ÓÈÈ×ÆÉÕ£¬µ½ÖÊÁ¿²»ÔÙ¼õÉÙ£¬µÃµ½¹ÌÌåÎïÖÊ4g¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Èܽâ¹ý³ÌÖÐËùÓõ½µÄ²£Á§ÒÇÆ÷ÓÐ___________________¡£

£¨2£©ÊµÑéÊÒ±£´æFeCl2ÈÜҺʱͨ³£»áÏòÆäÖмÓÈëÉÙÁ¿ÊÔ¼Á_____________ºÍ_____________¡£

£¨3£©ÊµÑé2ͨÈë×ãÁ¿Cl2µÄÄ¿µÄÊÇ________ £»

Éæ¼°µÄ»¯Ñ§·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______ _¡£

£¨4£©ÈçºÎ¼ìÑéÈܽâÑùÆ·Öк¬ÓÐFe2+ ¡£

£¨5£©Í¨¹ýʵÑéËùµÃÊý¾Ý£¬¼ÆËã¹ÌÌåÑùÆ·ÖÐFeCl3ºÍFeCl2µÄÎïÖʵÄÁ¿Ö®±ÈΪ_________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìºÚÁú½­Äµµ¤½­µÚÒ»¸ß¼¶ÖÐѧ¸ßÈý12ÔÂÔ¿¼»¯Ñ§¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÔÚ2LÃܱÕÈÝÆ÷ÖнøÐз´Ó¦C(s)+H2O(g)CO(g)+H2(g)¡÷H>0£¬²âµÃc(H2O)Ë淴Ӧʱ¼ä(t)µÄ±ä»¯Èçͼ¡£ÏÂÁÐÅжÏÕýÈ·µÄÊÇ

A£®5minʱ¸Ã·´Ó¦µÄKÖµÒ»¶¨Ð¡ÓÚ12 minʱµÄKÖµ

B£®0~5minÄÚ£¬v (H2)=0.05mol/(L•min)

C£®10 minʱ£¬¸Ä±äµÄÍâ½çÌõ¼þ¿ÉÄÜÊǼõСѹǿ

D£®¸Ã·´Ó¦»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿£º5minʱСÓÚ12 min ʱµÄ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìºÚÁú½­Äµµ¤½­µÚÒ»¸ß¼¶ÖÐѧ¸ßÈý12ÔÂÔ¿¼»¯Ñ§¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ijÁòË᳧·ÏÆøÖÐSO2µÄ»ØÊÕÀûÓ÷½°¸ÈçÏÂͼËùʾ£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

A£®X¿ÉÄܺ¬ÓÐ2ÖÖÑÎ B£®aÊÇSO3

C£®YÖк¬ÓУ¨NH4£©2SO4 D£®£¨NH4£©2S2O8ÖÐSµÄ»¯ºÏ¼Û²»¿ÉÄÜΪ+7

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìºÚÁú½­Äµµ¤½­µÚÒ»¸ß¼¶ÖÐѧ¸ßÈý12ÔÂÔ¿¼»¯Ñ§¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ijѧÉúÒÔÌúË¿ºÍCl2ΪԭÁϽøÐÐÏÂÁÐÈý¸öʵÑé¡£´Ó·ÖÀà½Ç¶È·ÖÎö£¬ÏÂÁÐÑ¡ÏîÕýÈ·µÄÊÇ

A£®ÊµÑé¢Ù¡¢¢ÚËùÉæ¼°µÄÎïÖʾùΪµç½âÖÊ

B£®ÊµÑé¢Ú¡¢¢Û¾ùΪ·ÅÈÈ·´Ó¦

C£®ÊµÑé¢Ú¡¢¢Û¾ùδ·¢ÉúÑõ»¯»¹Ô­·´Ó¦

D£®ÊµÑé¢Ù¡¢¢Û·´Ó¦ÖƵõÄÎïÖʾùΪ´¿¾»Îï

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìºÚÁú½­¹þ¶û±õÈýÖиßÈýÉÏÑéÊÕ¿¼ÊÔÈý»¯Ñ§¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

25¡æʱ£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

A£®ÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄÑÎËáºÍ°±Ë®µÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºµÄpH<7

B£®ÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄÁòËáºÍ°±Ë®µÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºµÄpH<7

C£®pH=3µÄÑÎËáºÍpH=11µÄ°±Ë®µÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºµÄpH>7

D£®pH=3µÄÁòËáºÍpH=11µÄ°±Ë®µÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºµÄpH<7

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016-2017ѧÄêÕã½­Ê¡¸ß¶þÉÏÆÚÖл¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

¢ÅÊÒÎÂʱ£¬pH=aµÄ°±Ë®ÓëPH =bµÄÑÎËáµÈÌå»ý»ìºÏ£¬Ç¡ºÃÍêÈ«·´Ó¦£¬Ôò°±Ë®µçÀë¶È¿É±íʾΪ_______________ (ÓðٷÖÊýÀ´±íʾ£©

£¨2£©ÊÒÎÂÏ£¬0.1 mol/LNaClOÈÜÒºµÄpH_________0.1 mol/LNa2SO3ÈÜÒºµÄpH¡££¨Ñ¡Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±£¬»ò¡°µÈÓÚ¡±)¡£Å¨¶È¾ùΪ0.1mol/LµÄNa2SO3ºÍNa2CO3µÄ»ìºÏÈÜÒºÖУ¬SO32-¡¢CO32-¡¢HSO3-¡¢HCO3-Ũ¶È´Ó´óµ½Ð¡µÄ˳ÐòΪ ¡£

ÒÑÖª£º

H2SO3

Kal=1.54¡Ál0-2

Ka2=1.02¡Á10-7

HCIO

Ka1=2.95¡Á10-8

H2CO3

Kal=4.3¡Á10-7

Ka2=5.6¡Á10-11

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸