Ç⻯¸Æ£¨CaH2£©¹ÌÌåÊǵÇɽÔ˶¯Ô±³£ÓõÄÄÜÔ´Ìṩ¼Á¡£Ç⻯¸ÆÒªÃÜ·â±£´æ£¬Ò»µ©½Ó´¥µ½Ë®¾Í·¢Éú·´Ó¦Éú³ÉÇâÑõ»¯¸ÆºÍÇâÆø¡£Ç⻯¸ÆÍ¨³£ÓÃÇâÆøÓë½ðÊô¸Æ¼ÓÈÈÖÆÈ¡£¬Í¼1ÊÇÄ£ÄâÖÆÈ¡×°Öãº

£¨1£©×°ÖÃBµÄ×÷ÓÃÊÇ                      £»×°ÖÃDµÄ×÷ÓÃÊÇ                      £»

£¨2£©ÀûÓÃͼ1ʵÑé×°ÖýøÐÐʵÑ飬ʵÑé²½ÖèÈçÏ£º¼ì²é×°ÖÃÆøÃÜÐÔºó×°ÈëÒ©Æ·£»´ò¿ª·ÖҺ©¶·»îÈû£¬                             £¨Çë°´ÕýÈ·µÄ˳ÐòÌîÈëÏÂÁв½ÖèµÄÐòºÅ£©¡£

¢Ù¼ÓÈÈ·´Ó¦Ò»¶Îʱ¼ä       ¢ÚÊÕ¼¯ÆøÌå²¢¼ìÑéÆä´¿¶È

    ¢Û¹Ø±Õ·ÖҺ©¶·»îÈû       ¢ÜÍ£Ö¹¼ÓÈÈ£¬³ä·ÖÀäÈ´

£¨3£©ÎªÁËÈ·ÈϽøÈë×°ÖÃCµÄÇâÆøÒѾ­¸ÉÔӦÔÚB¡¢CÖ®¼äÔÙ½ÓÒ»×°Ö㬸Ã×°ÖÃÖмÓÈëµÄÊÔ¼ÁÊÇ£º                ¡£

£¨4£©¼×ͬѧÉè¼ÆÒ»¸öʵÑ飬²â¶¨ÉÏÊöʵÑéÖеõ½µÄÇ⻯¸ÆµÄ´¿¶È£¨ÔÓÖÊÖв»º¬¸ÆÔªËØ¡£ÇëÍêÉÆÏÂÁÐʵÑé²½Ö裺

¢ÙÑùÆ·³ÆÁ¿£»¢Ú¼ÓÈë________ÈÜÒº£¨Ìѧʽ£©£¬½Á°è¡¢¹ýÂË£»¢Û________£¨Ìî²Ù×÷Ãû³Æ£©£»¢Ü_______   £¨Ìî²Ù×÷Ãû³Æ£©£» ¢Ý³ÆÁ¿Ì¼Ëá¸Æ¡£

£¨5£©ÒÒͬѧÀûÓÃ×¢ÉäÆ÷²âÁ¿Ç⻯¸ÆºÍË®·´Ó¦ÇâÆøÌå»ýµÄ·½·¨£¬²â¶¨ÉÏÊöʵÑéÖеõ½µÄÇ⻯¸ÆµÄ´¿¶È¡£Ëû³ÆÈ¡46 mg ËùÖÆµÃµÄÇ⻯¸ÆÑùÆ·£¬¼Ç¼¿ªÊ¼Ê±×¢ÉäÆ÷»î˨ͣÁôÔÚ10.00mL¿Ì¶È´¦£¬·´Ó¦½áÊøºó³ä·ÖÀäÈ´£¬»î˨×îÖÕÍ£ÁôÔÚ57.04mL¿Ì¶È´¦£¨ÉÏÊöÆøÌåÌå»ý¾ùÔÚ±ê×¼×´¿öϲⶨ£©¡£ÊÔͨ¹ý¼ÆËãÇóÑùÆ·ÖÐÇ⻯¸ÆµÄ´¿¶È£º                           ¡£

£¨6£©ÇëÄãÔÙÉè¼ÆÒ»ÖÖÇ⻯¸Æ´¿¶ÈµÄ²â¶¨·½·¨£º                                       

                                                                               

                                                                              ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ijѧÉú¿ÎÍâ»î¶¯Ð¡×éÀûÓÃÓÒͼËùʾװÖ÷ֱð×öÈçÏÂʵÑ飺

£¨1£©ÔÚÊÔ¹ÜÖÐ×¢ÈëijºìÉ«ÈÜÒº£¬¼ÓÈÈÊԹܣ¬ÈÜÒºÑÕÉ«Öð½¥±ädz£¬ÀäÈ´ºó»Ö¸´ºìÉ«£¬ÔòÔ­ÈÜÒº¿ÉÄÜÊÇ________________ÈÜÒº£»¼ÓÈÈʱÈÜÒºÓɺìÉ«Öð½¥±ädzµÄÔ­ÒòÊÇ£º_________    ____                                                                                                              __                                             ¡£

£¨2£©ÔÚÊÔ¹ÜÖÐ×¢ÈëijÎÞÉ«ÈÜÒº£¬¼ÓÈÈÊԹܣ¬ÈÜÒº±äΪºìÉ«£¬ÀäÈ´ºó»Ö¸´ÎÞÉ«£¬Ôò´ËÈÜÒº¿ÉÄÜÊÇ________________ÈÜÒº£»¼ÓÈÈʱÈÜÒºÓÉÎÞÉ«±äΪºìÉ«µÄÔ­ÒòÊÇ _________                                                                 ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔÚ2A + B£½3C + 4DµÄ·´Ó¦ÖУ¬ÏÂÁбíʾ¸Ã·´Ó¦µÄ»¯Ñ§·´Ó¦ËÙ¶È×î¿ìµÄÊÇ

    A£®v (A)£½0.5 mol/(L¡¤s)                   B£®v (B)£½ 0.3 mol/(L¡¤s)

C£®v (C)£½0.8 mol/(L¡¤s)                   D£®v (D)£½1.0 mol/(L¡¤s)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÓÃÖк͵ζ¨·¨²â¶¨Ä³ÉÕ¼îÑùÆ·µÄ´¿¶È¡£ÓÐÒÔϲ½Ö裺

(1) ÅäÖÆ´ý²âÒº£ºÓÃÒѳƺõÄ5.0gº¬ÓÐÉÙÁ¿ÔÓÖÊ(ÔÓÖʲ»ÓëÑÎËá·´Ó¦)µÄ¹ÌÌåÉÕ¼îÑùÆ·ÅäÖÆ1000mLÈÜÒº¡£³ýÉÕ±­ºÍ²£°ôÍ⣬»¹ÐèÒªÓõ½µÄÖ÷ÒªÒÇÆ÷ÓÐ__________£¬__________£»

(2) µÎ¶¨¹ý³Ì£º

¢ÙÊ¢×°0.10 mol/LµÄÑÎËá±ê×¼ÒºÓ¦¸ÃʹÓÃ_______µÎ¶¨¹Ü£»

¢ÚµÎ¶¨Ê±Ë«ÑÛӦעÒâ¹Û²ì___________________________________£»

(3) Îó²îÌÖÂÛ£º(Ñ¡Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족 )

¢Ù ÓÃÕôÁóË®³åÏ´×¶ÐÎÆ¿£¬²â¶¨½á¹û__________£»

¢Ú Ôڵζ¨¹ý³ÌÖв»É÷½«ÊýµÎËáÒºµÎÔÚ×¶ÐÎÆ¿Í⣬²â¶¨½á¹û__________£»

¢Û ¶ÁÊýʱ£¬µÎ¶¨Ç°ÑöÊÓ£¬µÎ¶¨ºó¸©ÊÓ£¬²â¶¨½á¹û____________£»

¢Ü ×°±ê׼Һ֮ǰ£¬Ã»ÓÐÓñê×¼ÒºÈóÏ´µÎ¶¨¹Ü£¬²â¶¨½á¹û____________£»

(4) ÓйØÊý¾Ý¼Ç¼ÈçÏ£º

²â¶¨ÐòºÅ

´ý²âÈÜÒºµÄÌå»ý(mL)

ËùºÄÑÎËá±ê×¼ÒºµÄÌå»ý(mL)

µÎ¶¨Ç°¶ÁÊý

µÎ¶¨ºó¶ÁÊý

1

20.00

0.50

20.78

2

20.00

1.20

21.32

¼ÆËã´¿¶È£ºÉÕ¼îÑùÆ·µÄ´¿¶ÈÊÇ____________£¨È¡Á½´ÎʵÑéËùºÄÑÎËáµÄƽ¾ùÖµ½øÐмÆË㣬²»Ð´¼ÆËã¹ý³Ì£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


D£®ÈôA¡¢B¡¢C¡¢D¾ùΪ10µç×Ó΢Á££¬ÇÒCÊÇ¿ÉʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壬ÔòD³£ÎÂÏÂÒ»¶¨³ÊҺ̬

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÓйؾ§ÌåµÄ˵·¨ÖУ¬ÕýÈ·µÄÊÇ

A£®·²ÊǾßÓйæÔòÍâÐεĹÌÌå¶¼ÊǾ§Ìå

B£®Âêè§ÊôÓÚ²£Á§ÌåÓй̶¨ÈÛµã

C£®¹Ì̬Ñõ£¨O2£©ºÍ¹Ì̬Áò£¨S8£©¶¼ÊÇ·Ö×Ó¾§Ìå

D£®ÂÈ»¯ÄƺÍÂÈ»¯ï¤¶¼ÊÇÀë×Ó¾§Ì壬¾§Ìå½á¹¹Ïàͬ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁйØÓÚÓлúÎïµÄ˵·¨ÕýÈ·µÄÊÇ

A£®±ûÍéµÄÒ»ÂÈ´úÎïÖ»ÓÐÒ»ÖÖ

B£®Âȷ¼ÈÄÜ·¢ÉúË®½â·´Ó¦£¬ÓÖÄÜ·¢ÉúÏûÈ¥·´Ó¦

C£®±ûÏ©£¨CH3—CH = CH2£©·Ö×Ó¾ßÓÐ˳·´Òì¹¹

D£®ÓÃÒÒÏ©ºÍä廯Çâ·´Ó¦ÖÆµÃµÄäåÒÒÍé±ÈÓÃÒÒÍéºÍäåÕôÆø·´Ó¦ËùµÃµÄ²úÎï¸ü´¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¼×ÍéÖеÄ̼ԭ×ÓÊÇsp3ÔÓ»¯£¬ÏÂÁÐÓÃ*±íʾ̼ԭ×ÓµÄÔÓ»¯ºÍ¼×ÍéÖеÄ̼ԭ×ÓÔÓ»¯×´Ì¬Ò»ÖµÄÊÇ                                                               £¨    £©

    A£®CH3C*H2CH3                        B£®C*H2£½CHCH3    

C£®CH2£½C*HCH3     D£®CH2£½CHC*H3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ñõ»¯»¹Ô­·´Ó¦ÓëËÄÖÖ»ù±¾·´Ó¦ÀàÐ͵ĹØÏµÈçͼËùʾ¡£ÏÂÁл¯Ñ§·´Ó¦ÊôÓÚÒõÓ°²¿·ÖµÄÊÇ

A£®4NH3 +5O2 £½4NO+6 H2O

B£®4Fe(OH)2+O2+2H2O=4Fe(OH)3

C£®2NaHCO3=Na2CO3+H2O+CO2¡ü

D£®Cl2+2NaBr=2NaCl +Br2

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸