¡¾ÌâÄ¿¡¿ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©

A.101kPaʱ£¬2H2(g)+O2(g)=2H2O(l) ¦¤H=-572kJ¡¤mol-1£¬ÔòH2µÄȼÉÕÈȦ¤H=-572kJ¡¤mol-1

B.Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºN2(g)+3H2(g)2NH3(g) ¦¤H=-92.4kJ¡¤mol-1£¬´ËÌõ¼þϽ«1.5mol H2ºÍ¹ýÁ¿N2³ä·Ö·´Ó¦£¬·Å³öÈÈÁ¿46.2kJ

C.Èô½«µÈÖÊÁ¿µÄÁòÕôÆøºÍÁò¹ÌÌå·Ö±ðÍêȫȼÉÕ£¬ºóÕ߷ųöÈÈÁ¿¶à

D.ÒÑÖª£º2C(s)+2O2(g)=2CO2(g) ¦¤H1£¬2C(s)+O2(g)=2CO(g) ¦¤H2£¬Ôò¦¤H1£¼¦¤H2

¡¾´ð°¸¡¿D

¡¾½âÎö¡¿

A£®È¼ÉÕÈÈÊÇ1mol¿ÉȼÎïÍêȫȼÉÕÉú³ÉÎȶ¨µÄÑõ»¯Îïʱ·Å³öµÄÈÈÁ¿£¬¶ø572kJÊÇ2molÇâÆøȼÉշųöµÄÈÈÁ¿£¬A´íÎó£»

B£®N2(g)+3H2(g)2NH3(g)ÊÇ¿ÉÄæ·´Ó¦£¬²»ÄܽøÐе½µ×£¬·Å³öµÄÈÈÁ¿»áСÓÚ46.2kJ£¬B´íÎó£»

C£®Æø̬Áò±È¹Ì̬ÁòÄÜÁ¿¸ß£¬È¼ÉÕ·ÅÈȶ࣬C´íÎó£»

D£®µÈÁ¿µÄ̼ÍêȫȼÉÕÉú³É¶þÑõ»¯Ì¼·Å³öµÄÈÈÁ¿¶àÓÚ²»ÍêȫȼÉÕÉú³ÉCO·Å³öµÄÈÈÁ¿£¬¶ø¦¤HΪ¸ºÖµ£¬ËùÒÔ¦¤H1£¼¦¤H2£¬DÕýÈ·£»

¹ÊÑ¡D¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ó°Ï컯ѧ·´Ó¦ËÙÂʵÄÒòËغܶ࣬ij¿ÎÍâÐËȤС×éÓÃʵÑé·½·¨½øÐÐ̽¾¿¡£

(1)È¡µÈÎïÖʵÄÁ¿Å¨¶È¡¢µÈÌå»ýµÄH2O2ÈÜÒº·Ö±ð½øÐÐH2O2µÄ·Ö½âʵÑ飬ʵÑ鱨¸æÈçϱíËùʾ£¨ÏÖÏóºÍ½áÂÛÂÔ£©¡£

¢ÙʵÑé1¡¢2Ñо¿µÄÊÇ__________¶ÔH2O2·Ö½âËÙÂʵÄÓ°Ïì¡£

¢ÚʵÑé2¡¢3µÄÄ¿µÄÊÇ_______________¶ÔH2O2·Ö½âËÙÂʵÄÓ°Ïì¡£

(2)²éÎÄÏ׿ÉÖª£¬Cu2£«¶ÔH2O2·Ö½âÒ²Óд߻¯×÷Óã¬Îª±È½ÏFe3£«¡¢Cu2£«¶ÔH2O2·Ö½âµÄ´ß»¯Ð§¹û£¬¸ÃС×éµÄͬѧ·Ö±ðÉè¼ÆÁËÈçͼ¼×¡¢ÒÒËùʾµÄʵÑé¡£»Ø´ðÏà¹ØÎÊÌ⣺

¢Ù¶¨ÐÔÈçͼ¼×¿Éͨ¹ý¹Û²ì_______£¬¶¨ÐԱȽϵóö½áÂÛ¡£ÓÐͬѧÌá³ö½«CuSO4ÈÜÒº¸ÄΪCuCl2ÈÜÒº¸üºÏÀí£¬ÆäÀíÓÉÊÇ_________¡£

¢Ú¶¨Á¿ÈçͼÒÒËùʾ£¬ÊµÑéʱÒÔÊÕ¼¯µ½40 mLÆøÌåΪ׼£¬ºöÂÔÆäËû¿ÉÄÜÓ°ÏìʵÑéµÄÒòËØ£¬ÊµÑéÖÐÐèÒª²âÁ¿µÄÊý¾ÝÊÇ__________¡£

(3)ËáÐÔ¸ßÃÌËá¼ØÈÜÒººÍ²ÝËáÈÜÒº¿É·¢Éú·´Ó¦£º2KMnO4£«5H2C2O4£«3H2SO4=K2SO4£«2MnSO4£«8H2O£«10CO2¡ü£¬ÊµÑéʱ·¢ÏÖ¿ªÊ¼·´Ó¦ËÙÂʽÏÂý£¬ÈÜÒºÍÊÉ«²»Ã÷ÏÔ£¬µ«Ò»¶Îʱ¼äºóͻȻÍÊÉ«£¬·´Ó¦ËÙÂÊÃ÷ÏÔ¼Ó¿ì¡£¶Ô´ËÕ¹¿ªÌÖÂÛ£º

¢ÙijͬѧÈÏΪKMnO4ÓëH2C2O4µÄ·´Ó¦ÊÇ______ÈÈ·´Ó¦£¬µ¼ÖÂ_______________£»

¢Ú´ÓÓ°Ï컯ѧ·´Ó¦ËÙÂʵÄÒòËØ¿´£¬ÄãÈÏΪ»¹¿ÉÄÜÊÇ________µÄÓ°Ïì¡£ÒªÖ¤Ã÷ÄãµÄ²ÂÏ룬ʵÑé·½°¸ÊÇ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖªÖظõËá¼Ø(K2Cr2O7)¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Æ仹ԭ²úÎïCr3+ÔÚË®ÈÜÒºÖгÊÂÌÉ«»òÀ¶ÂÌÉ«¡£ÔÚK2Cr2O7ÈÜÒºÖдæÔÚÏÂÁÐƽºâ£ºCr2O72- (³ÈÉ«)+H2O2CrO42- (»ÆÉ«)+2H+¡£ÓÃK2Cr2O7ÈÜÒº½øÐÐʵÑ飬½áºÏʵÑ飬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A.¢ÙÖÐÈÜÒº³ÈÉ«¼ÓÉ¢ÛÖÐÈÜÒº±ä»Æ

B.¢ÚÖÐCr2O72-±»C2H5OHÑõ»¯

C.¶Ô±È¢ÚºÍ¢Ü¿ÉÖªK2Cr2O7¼îÐÔÈÜÒºÑõ»¯ÐÔÇ¿

D.ÈôÏò¢ÜÖмÓÈë70% H2SO4ÈÜÒºÖÁ¹ýÁ¿£¬ÈÜÒº±äΪ³ÈÉ«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÔÏ©ÌþΪԭÁÏ¿ÉÒԺϳɶàÖָ߾ÛÎïµÄºÏ³É·ÏßÈçÏ£º

ÒÑÖª£ºÏ©ÌþºÍX2ÔÚÒ»¶¨Ìõ¼þÏÂÄÜ·¢ÉúÈ¡´ú£¬ÇÒÄÜ·¢ÉúË«Ï©ºÏ³ÉÈç¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)XÖк¬ÓеĹÙÄÜÍÅΪ_____________________________________£»

(2)Y¡úZµÄ»¯Ñ§·½³ÌʽΪ_____________________________£»

(3)¸ß¾ÛÎïEµÄ½á¹¹¼òʽΪ____________________________£»¼×ÊÇAµÄÒ»ÖÖͬ·ÖÒì¹¹Ì壬ÆäÄÜʵÏÖת»¯£º£¬¼×µÄÃû³ÆΪ________£»

(4)ÓÉ¿ÉÒԺϳɡ£°´ºÏ³É·ÏßµÄ˳Ðò£¬Éæ¼°·´Ó¦µÄ·´Ó¦ÀàÐÍÓУº______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÎªÁ˲ⶨ²ÝËᾧÌåH2C2O4¡¤xH2OÖеÄxÖµ£¬Ä³ÊµÑéС×é½øÐÐʵÑ飬²½ÖèÈçÏ£º

¢Ù³ÆÈ¡1.260 g²ÝËᾧÌ壬Åä³É100 mLÈÜÒº¡£

¢ÚÈ¡25.00 mL¸ÃH2C2O4ÈÜÒº¼ÓÈë׶ÐÎÆ¿ÄÚ£¬ÔÙ¼ÓÈëÊÊÁ¿Ï¡ÁòËá¡£

¢ÛÓÃŨ¶ÈΪ0.1000 mol/LµÄKMnO4ÈÜÒºµÎ¶¨H2C2O4ÈÜÒº£¬µ±__________________ʱ£¬µÎ¶¨½áÊø¡£

¢Ü¼Ç¼Êý¾Ý£¬Öظ´ÊµÑé¡£ÕûÀíÊý¾ÝÈ磺

ʵÑéÐòºÅ

V(KMnO4ÈÜÒº)

µÎ¶¨Ç°¿Ì¶È/mL

µÎ¶¨ºó¿Ì¶È/mL

1

0.10

10.00

2

1.10

11.10

3

1.50

11.50

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)²½Öè¢ÙÐèʹÓÃÉÕ±­¡¢Á¿Í²¡¢²£Á§°ô£¬»¹È±ÉٵIJ£Á§ÒÇÆ÷Ϊ______________(ÌîÃû³Æ)£»²½Öè¢ÛµÎ¶¨¹ý³ÌÖУ¬Ê¢×°KMnO4ÈÜÒºµÄÒÇÆ÷Ϊ_______________(ÌîÃû³Æ)¡£

(2)¸Ã·´Ó¦Ô­ÀíµÄ»¯Ñ§·½³ÌʽΪ_____________________________________________¡£

(3)Ç뽫²½Öè¢Û²¹³äÍêÕû_____________________________________________________¡£

(4)¸ù¾ÝÊý¾Ý£¬¼ÆËãH2C2O4ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ_________mol/L£¬x=________¡£

(5)ÈôµÎ¶¨ÖÕµã¶ÁÊýʱ¸©ÊÓKMnO4ÈÜÒºÒºÃ棬ÔòxÖµ»á_________(Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Îª²â¶¨CO2µÄÏà¶Ô·Ö×ÓÖÊÁ¿£¬Ä³ÊµÑéС×éÈýλͬѧѡÓú¬NaHCO3µÄÑùÆ·(ÖÊÁ¿¾ùΪm1g)ºÍÆäËüºÏÀíµÄÊÔ¼Á£¬½øÐÐÁËÒÔÏÂÈý¸öʵÑé¡£Íê³ÉÏÂÁÐÌî¿Õ£º

¼×ÓÃÖØÁ¿·¨È·¶¨CO2µÄÖÊÁ¿£¬×°ÖÃÈçÏÂͼ£º

(1)BÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________________________________________________¡£

(2)ʵÑéÖгÖÐø»º»ºÍ¨Èë¿ÕÆø£¬Æä×÷ÓÃÖ®Ò»ÊÇ°ÑÉú³ÉµÄCO2È«²¿ÅÅÈëºóÐø×°ÖÃÖУ¬Ê¹Ö®ÍêÈ«±»ÎüÊÕ£»ÁíÓÐ×÷ÓÃΪ___________________________________________________________¡£

(3)²»ÄÜÌá¸ß²â¶¨¾«È·¶ÈµÄ´ëÊ©ÊÇ___________¡£

a£®ÏòBÄÚ¼ÓÈëËá֮ǰ£¬Åž¡×°ÖÃÄÚµÄCO2ÆøÌå

b£®ÏòBÄڵμÓËáʱ²»Ò˹ý¿ì

c£®ÔÚB¡¢CÖ®¼äÔöÌíÊ¢Óб¥ºÍNaHCO3ÈÜÒºµÄÏ´Æø×°ÖÃ

d£®ÔÚDºóÔöÌíÊ¢Óмîʯ»ÒµÄ¸ÉÔï¹Ü

ÒÒÓõζ¨·¨È·¶¨CO2µÄÎïÖʵÄÁ¿£¬½«ÑùÆ·ÅäÖƳÉ100mLÈÜÒº£¬´ÓÖÐÈ¡³ö20.00 mL£¬ÓÃc mol¡¤L£­1µÄÑÎËáµÎ¶¨(¼×»ù³È×÷ָʾ¼Á)¡£µ±______________________________________________________ʱ£¬Í£Ö¹µÎ¶¨¡£Æ½ÐвⶨÈý´Î£¬ÓйØʵÑéÊý¾Ý¼Ç¼ÈçÏÂ±í¡£m1 gÑùÆ·²úÉúCO2µÄÎïÖʵÄÁ¿Îª_____________¡£

ʵÑé±àºÅ

´ý²âÒºÌå»ý

(mL)

ÏûºÄÑÎËáÌå»ý(mL)

³õ¶ÁÊý

Ä©¶ÁÊý

1

20.00

0.00

25.02

2

20.00

0.20

28.80

3

20.00

1.30

26.28

±ûÓÃÆøÌåÌå»ý·¨È·¶¨CO2µÄÌå»ý£¬×°ÖÃÈçͼËùʾ¡£

(4)ΪÁ˼õСʵÑéÎó²î£¬Á¿Æø¹ÜÖмÓÈëµÄÒºÌåXΪ___________________ÈÜÒº£»

(5)Èô¸Ã×°ÖÃÆøÃÜÐÔÁ¼ºÃ£¬¶ÁÊýƽÊÓ£¬µ«²âµÃµÄ¡°CO2Ìå»ý¡±Êý¾ÝÈÔȻƫС£¬ÆäÔ­Òò¿ÉÄÜÊÇ____________________________________________________________________________¡£

(6)È·¶¨CO2µÄÏà¶Ô·Ö×ÓÖÊÁ¿£¬Ñ¡ÓÃ___________________(ÏÞÓ᰼ס±¡¢¡°ÒÒ¡±¡¢¡°±û¡±½øÐÐÌîд)µÄʵÑéÊý¾ÝΪ×î¼Ñ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³Î¶ÈÏÂ2LÃܱÕÈÝÆ÷ÖУ¬3ÖÖÆøÌåÆðʼ״̬ºÍƽºâ״̬ʱµÄÎïÖʵÄÁ¿£¨n£©ÈçϱíËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©

X

Y

W

n£¨Æðʼ״̬£©/mol

2

1

0

n£¨Æ½ºâ״̬£©/mol

1

0.5

1.5

A.¸ÃζÈÏ´ïƽºâºó£¬Ôö´óѹǿƽºâ²»Òƶ¯

B.¸Ã·´Ó¦·½³Ìʽ¿É±íʾΪ£ºX+2Y=3W

C.Éý¸ßζȣ¬ÈôWµÄÌå»ý·ÖÊý¼õС£¬Ôò´Ë·´Ó¦¦¤H£¾0

D.ºãκãÈÝʱ£¬Ôö¼ÓXµÄÎïÖʵÄÁ¿£¬Æ½ºâÏòÕýÏòÒƶ¯£¬XµÄת»¯ÂÊÌá¸ß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ

A.ÒÑÖªt1¡æʱ£¬·´Ó¦C+CO22CO ¡÷H>0µÄËÙÂÊΪ¦Ô£¬ÈôÉý¸ßζȣ¬Äæ·´Ó¦ËÙÂʼõС

B.ºãѹÈÝÆ÷Öз¢Éú·´Ó¦N2+O22NO£¬ÈôÔÚÈÝÆ÷ÖгäÈëHe£¬ÕýÄæ·´Ó¦ËÙÂʾù²»±ä

C.µ±Ò»¶¨Á¿µÄп·ÛºÍ¹ýÁ¿µÄ6molL1ÑÎËᷴӦʱ£¬ÎªÁ˼õÂý·´Ó¦ËÙÂÊ,ÓÖ²»Ó°Ïì²úÉúH2µÄ×ÜÁ¿,¿ÉÏò·´Ó¦Æ÷ÖмÓÈëÉÙÁ¿µÄCuSO4ÈÜÒº

D.´ý·´Ó¦PCl5(g)PCl3(g)£«Cl2(g) ´ïµ½Æ½ºâºó£¬±£³ÖζȺÍÌå»ý²»±ä£¬ÔÙ³äPCl5(g)´ïµ½ÐµÄƽºâ£¬ÐÂƽºâºÍԭƽºâÏà±ÈPCl5(g)µÄת»¯ÂʼõÉÙ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑ֪ij¿ÉÄæ·´Ó¦mA(g)+nB(g)pC(g)ÔÚÃܱÕÈÝÆ÷ÖнøÐУ¬Èçͼ±íʾÔÚ²»Í¬·´Ó¦Ê±¼ätʱ£¬Î¶ÈTºÍѹǿpÓë·´Ó¦ÎïBÔÚ»ìºÏÆøÌåÖеÄÌå»ý·ÖÊý[¦Õ(B)]µÄ¹ØϵÇúÏߣ¬ÓÉÇúÏß·ÖÎö£¬ÏÂÁÐÅжÏÕýÈ·µÄÊÇ(¡¡ ¡¡)

A.T1<T2£¬p1>p2£¬m+n>p£¬·ÅÈÈ·´Ó¦

B.T1<T2£¬p1>p2£¬m+n<p£¬ÎüÈÈ·´Ó¦

C.T1>T2£¬p1<p2£¬m+n<p£¬ÎüÈÈ·´Ó¦

D.T1>T2£¬p1<p2£¬m+n>p£¬·ÅÈÈ·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸