¡¾ÌâÄ¿¡¿NH3¡¢N2H4ÔÚ¹¤ÒµÉú²úºÍ¹ú·À½¨ÉèÖж¼Óй㷺ӦÓ᣻شðÏÂÁÐÎÊÌ⣺

£¨1£©¢ÙN2H4 (g) N2(g)+2H2(g) ¡÷H1

¢ÚN2(g)+3H2(g) 2NH3(g) ¡÷H2

¢Û7N2H4(g) 8NH3(g)+3N2(g)+2H2(g) ¡÷H3

¡÷H3=___________£¨Óú¬¡÷H1ºÍ¡÷H2µÄ´úÊýʽ±íʾ£©¡£

£¨2£©ÄÉÃ×îܵĴ߻¯×÷ÓÃÏ£¬N2H4¿É·Ö½âÉú³ÉÁ½ÖÖÆøÌ壬ÆäÖÐÒ»ÖÖÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶¡£µ±·´Ó¦ÔÚ²»Í¬Î¶ÈÏ´ﵽƽºâʱ£¬»ìºÏÆøÌåÖи÷×é·ÖµÄÌå»ý·ÖÊýÈçͼËùʾ¡£

¸Ã·´Ó¦µÄ¡÷H________£¨Ìî¡°>¡±»ò¡°<¡±£©0£¬N2H4·¢Éú·Ö½â·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________¡£

£¨3£©T¡æʱ£¬ÏòÒ»Ìå»ýΪ5L µÄºãÈÝÃܱÕÈÝÆ÷ÖгäÈë×ÜÎïÖʵÄÁ¿Îª2 molµÄCO2ºÍNH3£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º2NH3(g)+CO2(g£©====CO(NH2)2(s)+H2O(g)¡£»ìºÏÆøÌåÖа±ÆøµÄÌå»ý·ÖÊýË淴Ӧʱ¼äµÄ±ä»¯ÈçͼËùʾ¡£

¢Ù 0¡«60sÄÚ£¬·´Ó¦ËÙÂÊv(CO2)=________mol/(L¡¤s)

¢Ú T¡æʱ£¬¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýK=________¡£

¡¾´ð°¸¡¿ 7¡÷H1+4¡÷H2 > 3N2H44NH3+N2 1.25¡Á10-3 240

¡¾½âÎö¡¿£¨1£©ÒÑÖª£º

¢ÙN2H4(g)N2(g)+2H2(g) ¡÷H1

¢ÚN2(g)+3H2(g)2NH3(g) ¡÷H2

¢Û7N2H4(g)8NH3(g)+3N2(g)+2H2(g) ¡÷H3

Ôò¸ù¾Ý¸Ç˹¶¨ÂÉ¿ÉÖª¢Ù¡Á7+¢Ú¡Á4¼´µÃµ½·´Ó¦¢ÛµÄ¡÷H3=7¡÷H1+4¡÷H2¡£

£¨2£©ÆäÖÐÒ»ÖÖÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬¸ÃÆøÌåÊÇ°±Æø¡£¸ù¾ÝͼÏñ¿ÉÖªËæζÈÉý¸ß£¬Éú³ÉÎïÌå»ý·ÖÊýÔö¼Ó£¬Õâ˵Ã÷Éý¸ßζÈƽºâÏòÕý·´Ó¦·½Ïò½øÐУ¬Ôò¸Ã·´Ó¦µÄ¡÷H£¾0¡£¸ù¾ÝͼÏñ¿ÉÖªÁ½ÖÖÉú³ÉÎïµÄÌå»ýÖ®±È½üËÆΪ4:1£¬ËùÒÔ¸ù¾ÝÔ­×ÓÊغã¿ÉÖªN2H4·¢Éú·Ö½â·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ3N2H44NH3+N2¡£

£¨3£©¢Ù¸ù¾ÝͼÏñ¿ÉÖªÆðʼʱ¶þÕßµÄÌå»ýÏàµÈ£¬Òò´Ë¾ùÊÇ1mol£¬Å¨¶È¾ùÊÇ0.2mol/L£¬Æ½ºâʱ°±ÆøÌå»ý·ÖÊýÊÇ0.2£¬Ôò¸ù¾Ý·½³Ìʽ¿ÉÖª

2NH3(g)+CO2(g£©CO(NH2)2(s)+H2O(g)

ÆðʼŨ¶È£¨mol/L£© 0.2 0.2 0

ת»¯Å¨¶È£¨mol/L£© 2x x x

ƽºâŨ¶È£¨mol/L£© 0.2-2x 0.2-x x

Ôò£¬½âµÃx£½0.075

Ôò0¡«60sÄÚ£¬·´Ó¦ËÙÂÊv(CO2)=0.075mol/L¡Â60s£½1.25¡Á10-3mol/(L¡¤s)

¢Ú T¡æʱ£¬¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýK=¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¶ÌÖÜÆÚÔªËØW¡¢X¡¢Y¡¢ZµÄÔ­×ÓÐòÊýÒÀ´ÎÔö¼Ó¡£m¡¢n¡¢pÊÇÓÉÕâЩԪËØ×é³ÉµÄ¶þÔª»¯ºÏÎW2¡¢X2¡¢Z2·Ö±ðÊÇÔªËØW¡¢X¡¢ZµÄµ¥ÖÊ¡£ÒÑÖªI.Ò»¶¨Ìõ¼þÏÂijÃܱÕÈÝÆ÷Öпɷ¢Éú·´Ó¦aX2+bW2cm£¬·´Ó¦¹ý³ÌÖÐÎïÖʵÄŨ¶È±ä»¯ÈçÏ£º

X2

W2

m

ÆðʼŨ¶È/mol/L

0.4

0.4

0

ƽºâŨ¶È/mol/L

0.3

0.1

0.2

¢ò£®ËüÃÇ¿É·¢ÉúÈçÏ·´Ó¦£º2m(g)+3Z2(g)=6n(g)£«X2(g)£»4n(g)+Y2(g)2p(l)+2Z2(g)¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. m¡¢n¡¢pÈýÖÖÎïÖʾùΪ¹²¼Û»¯ºÏÎï

B. a:b:c=3:1:2

C. XµÄÑõ»¯ÎïÒ»¶¨ÊÇÎÞÉ«ÆøÌå

D. Ô­×Ӱ뾶£ºW<X<Y

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ð´ÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ¡£

£¨1£©Na2CO3ÈÜÒºÓëÑÎËáµÄ·´Ó¦£º___________________¡£

£¨2£©Ba(OH)2ÈÜÒºÓëCuSO4µÄ·´Ó¦£º_________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁйØÓÚÓлú»¯ºÏÎïµÄ˵·¨ÕýÈ·µÄÊÇ

A. ÒÒ´¼µÄË®ÈÜÒºË׳ƾƾ«

B. Óɱ½ÓëÂÈÆøÔÚÒ»¶¨Ìõ¼þÏÂÉú³ÉC6H6Cl6µÄ·´Ó¦ÊôÓÚÈ¡´ú·´Ó¦

C. »¯Ñ§Ê½ÎªC4H10OµÄ±¥ºÍÒ»Ôª´¼ÓÐ4ÖÖ

D. ÌÇÀà·¢ÉúË®½â·´Ó¦µÄ×îÖÕ²úÎﶼÊÇÆÏÌÑÌÇ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÈçͼËùʾ£¬Á½µç¼«Ò»ÎªÌ¼°ô£¬Ò»ÎªÌúƬ£¬ÈôµçÁ÷±íµÄÖ¸Õë·¢Éúƫת£¬ÇÒa¼«ÉÏÓдóÁ¿ÆøÅÝÉú³É£¬ÔòÒÔÏÂÐðÊöÕýÈ·µÄÊÇ

A. aΪ¸º¼«£¬ÊÇÌúƬ£¬ÉÕ±­ÖеÄÈÜҺΪÁòËá

B. bΪ¸º¼«£¬ÊÇÌúƬ£¬ÉÕ±­ÖеÄÈÜҺΪÁòËáÍ­ÈÜÒº

C. aΪÕý¼«£¬ÊÇ̼°ô£¬ÉÕ±­ÖеÄÈÜҺΪÁòËá

D. bΪÕý¼«£¬ÊÇ̼°ô£¬ÉÕ±­ÖеÄÈÜҺΪÁòËáÍ­

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐʵÑé²Ù×÷ÕýÈ·µÄÊÇ£¨ £©
A. ³ÆÁ¿ÇâÑõ»¯ÄƹÌÌå
B. ·ÖÀë²ñÓͺÍË®
C. °Ñ±¥ºÍʳÑÎË®ÖеÄʳÑÎÌáÈ¡³öÀ´
D. ·ÖÀëÁ½ÖÖ»¥Èܵ«·ÐµãÏà²î½Ï´óµÄÒºÌå»ìºÏÎï

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨1£©·ÀÖλ·¾³ÎÛȾ£¬¸ÄÉÆÉú̬»·¾³ÒѳÉΪȫÇòµÄ¹²Ê¶¡£

¢Ù¿ÕÆøÖÊÁ¿±¨¸æµÄ¸÷ÏîÖ¸±êÄÜÌåÏÖ¿ÕÆøµÄÖÊÁ¿¡£ÏÂÁи÷ÏîÖ¸±ê²»ÐèÒª¼à²âµÄÊÇ___________¡£

A£®¿ÉÎüÈë¿ÅÁ£ÎPM2.5£© B£®NO2Ũ¶È C£®SO2Ũ¶È D£®CO2Ũ¶È

¢ÚÏÂÁдëÊ©ÓÐÀûÓÚ¸ÄÉÆ»·¾³ÖÊÁ¿µÄÊÇ___________¡£

A£®½«·Ïµç³ØÉîÂñ£¬ÒÔ¼õÉÙÖؽðÊôµÄÎÛȾ

B£®½«µØ¹µÓÍ»ØÊÕÔÙ¼Ó¹¤ÎªÊ³ÓÃÓÍ£¬ÒÔ¼õ»ºË®Ì帻ӪÑø»¯

C£®ÀûÓöþÑõ»¯Ì¼µÈÔ­ÁϺϳɾÛ̼Ëáõ¥Àà¿É½µ½âËÜÁÏ´úÌæ¾ÛÒÒÏ©

£¨2£©²ÉÈ¡Êʵ±µÄ´ëÊ©ÄܼõÉÙ¶Ô»·¾³µÄÎÛȾ¡£

¢ÙËáÓêÊÇÖ¸pH___________µÄ½µË®£¬´óÁ¿È¼ÉÕº¬Áòú²úÉúÆøÌåËæÓêË®½µÂäµ½µØÃ棬pHËæʱ¼ä±ä³¤»áÓÐËù¼õС£¬ÊÔÓû¯Ñ§·½³Ìʽ½âÊÍÆäÔ­Òò___________¡£

¢ÚÒ»Ñõ»¯Ì¼Ò²Êdz£¼ûµÄ´óÆøÎÛȾÎï¡£ÆûÓÍÔÚÆû³µ·¢¶¯»úÖеIJ»ÍêȫȼÉտɱíʾΪ2C8H18+23O212CO2+4CO+18H2O

ij»·¾³¼à²âÐËȤС×éµÄͬѧÃè»æµÄÊÐÖÐÐĵØÇø¿ÕÆøÖÐCOº¬Á¿±ä»¯ÇúÏß(ºá×ø±ê±íʾ±±¾©Ê±¼ä0µ½24Сʱ£¬×Ý×ø±ê±íʾCOº¬Á¿)£¬ÄãÈÏΪ±È½Ï·ûºÏʵ¼ÊµÄÊÇ____________¡£

A B C D

¢ÛÌìȻˮÖк¬ÓеÄϸСÐü¸¡¿ÅÁ£¿ÉÒÔÓÃ___________×ö»ìÄý¼Á£¨Ð´Ãû³Æ£©½øÐо»»¯´¦Àí¡£¹¤Òµ·ÏË®Öк¬ÓеÄCr3+Àë×Ó£¬¿ÉÓÃÊìʯ»Ò×÷³Áµí¼Á£¬ÔÚpHΪ8ÖÁ9ʱÉú³É³Áµí¶ø³ýÈ¥£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_____________¡£

£¨3£©Ê³Æ·ºÍÒ©Æ·¹ØϵÈ˵ÄÉú´æºÍ½¡¿µ¡£

¢ÙÊг¡ÉÏÏúÊÛµÄʳÑÎÆ·ÖֺܶࡣÏÂÁÐʳÑÎÖУ¬ËùÌí¼ÓµÄÔªËز»ÊôÓÚÈËÌå±ØÐè΢Á¿ÔªËصÄÊÇ___________£¨Ìî×Öĸ£©¡£

A£®¼ÓпÑÎ B£®¼ÓµâÑÎ C£®¼Ó¸ÆÑÎ

¢ÚÏËάËر»³ÆΪ¡°µÚÆßÓªÑøËØ¡±¡£Ê³ÎïÖеÄÏËάËØËäÈ»²»ÄÜΪÈËÌåÌṩÄÜÁ¿£¬µ«ÄÜ´Ù½ø³¦µÀÈ䶯¡¢Îü¸½ÅųöÓк¦ÎïÖÊ¡£´Ó»¯Ñ§³É·Ö¿´£¬ÏËάËØÊÇÒ»ÖÖ___________£¨Ìî×Öĸ£©¡£

A£®¶àÌÇ B£®µ°°×ÖÊ C£®Ö¬·¾

¢Ûijͬѧ¸Ðð·¢ÉÕ£¬Ëû¿É·þÓÃÏÂÁÐÄÄÖÖÒ©Æ·½øÐÐÖÎÁÆ____________£¨Ìî×Öĸ£©¡£

A£®Âé»Æ¼î¡¡ ¡¡ B£®°¢Ë¾Æ¥ÁÖ ¡¡C£®¿¹ËáÒ©

£¨4£©»¯Ñ§Óë²ÄÁÏÃÜÇÐÏà¹Ø¡£

¢ÙÈ˹¤ºÏ³ÉµÄËÜÁÏÓжàÖÖ£¬ÆäÖоÛÒÒÏ©Êdz£¼ûµÄËÜÁÏÖÆÆ·£¬½á¹¹¼òʽÊÇ___________¡£

¢Ú¸ÖÌúµÄ¸¯Ê´Ö÷Òª·¢ÉúÎüÑõ¸¯Ê´£¬ÆäÕý¼«µÄµç¼«·´Ó¦Îª___________¡£Îª·ÀÖ¹ÂÖ´¬µÄ´¬ÌåÔÚº£Ë®Öб»¸¯Ê´£¬Ò»°ãÔÚ´¬ÉíÁ¬½Ó_____________£¨Ìп¿é¡±»ò¡°Í­¿é¡±£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Î¶ÈΪTʱ.Ïò2.0 LºãÈÝÃܱÕÈÝÆ÷ÖгäÈë1.0 mol PCl5£¬·¢Éú·´Ó¦PCl5 (g)PCl3(g)+Cl2(g)£¬¾­¹ýÒ»¶Îʱ¼äºó´ïµ½Æ½ºâ£¬·´Ó¦¹ý³ÌÖвⶨµÄ²¿·ÖÊý¾ÝÁÐÓÚÏÂ±í¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

t/s

0

50

150

250

350

n(PCl3)/mol

0

0.16

0.19

0.20

0.20

A. ÔÚÇ°50s£¬PCl3µÄƽ¾ù·´Ó¦ËÙÂÊ v(PCl3)=0.0032mol/(L¡¤s)

B. ´ïµ½Æ½ºâʱ£¬ÈÝÆ÷ÖеÄѹǿÊÇÆðʼʱµÄ1.2±¶

C. ÏàͬζÈÏ£¬ÆðʼʱÏòÈÝÆ÷ÖгäÈë1.0mo PCl5¡¢0.20 mo1 PCl3ºÍ0.20 mo1 Cl2£¬·´Ó¦´ïµ½Æ½ºâÇ°v(Õý)>v(Äæ)

D. ÏàͬζÈÏ£¬ÆðʼʱÏòÈÝÆ÷ÖгäÈë2.0 mol PCl3ºÍ2.0 mol Cl2£¬´ïµ½Æ½ºâʱ£¬PCl3µÄת»¯ÂÊСÓÚ80%

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿°´ÒªÇóÌî¿Õ
£¨1£©1.5mol HNO3Ô¼º¬ÓиöÑõÔ­×Ó£¬º¬0.6mol H µÄC6H12O6µÄÎïÖʵÄÁ¿ÊÇmol£®
£¨2£©22¿ËCO2µÄÎïÖʵÄÁ¿Îªmol£¬¸ÃÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪL£®
£¨3£©4gNaOH¹ÌÌåµÄÎïÖʵÄÁ¿Îªmol£¬½«ÆäÈÜÓÚË®Åä³É500mLÈÜÒº£¬´ËÈÜÒºÖÐNaOHµÄÎïÖʵÄÁ¿Å¨¶ÈΪ £®
£¨4£©½«5mol/LÑÎËá10mLÏ¡Ê͵½200mL£¬Ï¡ÊͺóÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ £®
£¨5£©ÊµÑéÊÒÖƱ¸Fe £¨OH£©3½ºÌåµÄ»¯Ñ§·½³Ìʽ £®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸