ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ºÏ³ÉÕý¶¡È©µÄʵÑé×°ÖÃÈçͼËùʾ¡£

·¢ÉúµÄ·´Ó¦ÈçÏÂ:CH3CH2CH2CH2OHCH3CH2CH2CHO

£¨Õý¶¡´¼£© £¨Õý¶¡È©£©

·´Ó¦ÎïºÍ²úÎïµÄÏà¹ØÊý¾ÝÁбíÈçÏÂ:

·Ðµã/¡æ

ÃܶÈ(g¡¤cm-3)

Ë®ÖÐÈܽâÐÔ

Õý¶¡´¼

11.72

0.8109

΢ÈÜ

Õý¶¡È©

75.7

0.8017

΢ÈÜ

ʵÑé²½ÖèÈçÏ£º

½«6.0gNa2Cr2O7·ÅÈë100mLÉÕ±­ÖУ¬¼Ó30mLË®Èܽ⣬ÔÙ»ºÂý¼ÓÈë5mLŨÁòËᣬ½«ËùµÃÈÜҺСÐÄתÒÆÖÁBÖÐÔÚAÖмÓÈë4.0g Õý¶¡´¼ºÍ¼¸Á£·Ðʯ£¬¼ÓÈÈ¡£µ±ÓÐÕôÆû³öÏÖʱ£¬¿ªÊ¼µÎ¼ÓBÖÐÈÜÒº¡£µÎ¼Ó¹ý³ÌÖб£³Ö·´Ó¦Î¶ÈΪ90¡«95¡æÔÚFÖÐÊÕ¼¯90¡æÒÔϵÄÁó·Ö¡£½«Áó³öÎïµ¹Èë·ÖҺ©¶·ÖУ¬·Öȥˮ²ã£¬Óлú²ã¸ÉÔïºóÕôÁó£¬ÊÕ¼¯75¡«77¡æÁó·Ö£¬²úÁ¿2.0g¡£»Ø´ðÏÂÁÐÎÊÌâ:

£¨1£©ÊµÑéÖУ¬ÄÜ·ñ½«Na2Cr2O7ÈÜÒº¼Óµ½Å¨ÁòËáÖУ¬²¢ËµÃ÷ÀíÓÉ________¡£

£¨2£©¼ÓÈë·ÐʯµÄ×÷ÓÃÊÇ________¡£Èô¼ÓÈȺó·¢ÏÖδ¼Ó·Ðʯ£¬Ó¦²ÉÈ¡µÄÕýÈ··½·¨ÊÇ________¡£

£¨3£©ÉÏÊö×°ÖÃͼÖУ¬EÒÇÆ÷µÄÃû³ÆÊÇ________£¬DÒÇÆ÷µÄÃû³ÆÊÇ________¡£

£¨4£©·ÖҺ©¶·Ê¹ÓÃÇ°±ØÐë½øÐеIJÙ×÷ÊÇ________(ÌîÕýÈ·´ð°¸±êºÅ)¡£

a.Èóʪ b.¸ÉÔï c.¼ì© d.±ê¶¨

£¨5£©½«Õý¶¡È©´Ö²úÆ·ÖÃÓÚ·ÖҺ©¶·ÖзÖˮʱ£¬Ë®ÔÚ________²ã(Ìî¡°ÉÏ¡±»ò¡°Ï¡±)¡£

¡¾´ð°¸¡¿ ²»ÄÜ ÒױŽ¦ ·ÀÖ¹±©·Ð ÀäÈ´ºó²¹¼Ó Å£½Ç¹Ü Ö±ÐÎÀäÄý¹Ü c ÏÂ

¡¾½âÎö¡¿ÊÔÌâ·ÖÎö£º±¾Ì⿼²éŨÁòËáµÄÏ¡ÊÍ£¬ÒÇÆ÷µÄʶ±ð£¬»ù±¾ÊµÑé²Ù×÷£¬ÎïÖʵķÖÀëÌá´¿¡£

£¨1£©Å¨ÁòËáµÄÃܶȱÈË®´ó£¬Å¨ÁòËáÈÜÓÚË®·Å³ö´óÁ¿ÈÈ£¬Ó¦½«Å¨ÁòËá¼Óµ½Na2Cr2O7ÈÜÒºÖв¢¼°Ê±É¢ÈÈ¡£²»Äܽ«Na2Cr2O7ÈÜÒº¼Óµ½Å¨ÁòËáÖУ¬ÀíÓÉÊÇ£ºÈܽâʱ·Å³öµÄÈÈÁ¿Ê¹ÒºµÎ±Å½¦¡£

£¨2£©¼ÓÈë·ÐʯµÄ×÷ÓÃÊÇ£º·ÀÖ¹±©·Ð¡£Èô¼ÓÈȺó·¢ÏÖδ¼Ó·Ðʯ£¬Ó¦Á¢¼´Í£Ö¹¼ÓÈÈ£¬ÀäÈ´ºó²¹¼Ó·Ðʯ¡£

£¨3£©¸ù¾ÝÒÇÆ÷µÄ¹¹ÔìÌص㣬EÒÇÆ÷Ϊţ½Ç¹Ü£¬DÒÇÆ÷ΪֱÐÎÀäÄý¹Ü¡£

£¨4£©·ÖҺ©¶·ÔÚʹÓÃÇ°±ØÐë¼ì©£¬´ð°¸Ñ¡c¡£

£¨5£©ÓÉÓÚÕý¶¡È©µÄÃܶÈСÓÚË®µÄÃܶȣ¬ÇÒÕý¶¡È©Î¢ÈÜÓÚË®£¬ËùÒÔÕý¶¡È©´Ö²úÆ·ÖÃÓÚ·ÖҺ©¶·ÖзÖˮʱ£¬Ë®ÔÚϲ㡣

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿150¡æʱ£¬ÏòÈçͼËùʾµÄÈÝÆ÷(ÃÜ·âµÄ¸ô°å¿É×ÔÓÉ»¬¶¯£©ÖмÓÈë4LN2ºÍH2µÄ»ìºÏÆøÌ壬 ÔÚ´ß»¯¼Á×÷ÓÃϳä·Ö·´Ó¦(´ß»¯¼ÁÌå»ýºöÂÔ²»¼Æ£©£¬·´Ó¦ºó»Ö¸´µ½Ô­Î¶ȡ£Æ½ºâºóÈÝÆ÷Ìå»ý±äΪ3.4L£¬ÈÝÆ÷ÄÚÆøÌå¶ÔÏàͬÌõ¼þµÄÇâÆøµÄÏà¶ÔÃܶÈΪ5¡£

£¨1£©·´Ó¦Ç°»ìºÏÆøÌåÖÐV(N2)£ºV(H2)= _______, ·´Ó¦´ïµ½Æ½ºâºóV(NH3)= _______L£¬¸Ã·´Ó¦ÖÐN2ת»¯ÂÊΪ______¡£

£¨2£©ÏòƽºâºóµÄÈÝÆ÷ÖгäÈë0. 2molµÄNH3,Ò»¶Î¶Ô¼äºó·´Ó¦Ôٴδﵽƽºâ£¬»Ö¸´µ½150¡æʱ²âµÃ´Ë¹ý³ÌÖдÓÍâ½çÎüÊÕÁË 6.44kJµÄÈÈÁ¿£»

¢Ù³äÈëNH3ʱ£¬»ìºÏÆøÌåµÄÃܶȽ«_________£¬ÔڴﵽƽºâµÄ¹ý³ÌÖУ¬»ìºÏÆøÌåµÄÃܶȽ«_______(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©;·´Ó¦ÖØдïƽºâµÄ»ìºÏÆøÌå¶ÔÇâÆøµÄÏà¶ÔÃܶȽ«_____5(Ìî¡° >¡±¡¢¡° <¡±»ò¡°=¡±£©¡£

¢ÚÏÂÁÐÄÄЩʵÑéÏÖÏó»òÊý¾ÝÄÜ˵Ã÷·´Ó¦ÖØе½´ïÁËƽºâ _______£»

A.ÈÝÆ÷²»ÔÙÓëÍâ½ç·¢ÉúÈȽ»»»

B.»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿±£³Ö²»±ä

C.µ±K( N2):V( H2):F( NH3) = 1:3:2ʱ

D.µ±ÈÝÆ÷µÄÌå»ý²»ÔÙ·¢Éú±ä»¯Ê±

¢Ûд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ: __________¡£

¡¾ÌâÄ¿¡¿ÀûÓû¯Ñ§Ô­Àí¶Ô¹¤³§ÅŷŵķÏË®¡¢·ÏÔüµÈ½øÐÐÓÐЧ¼ì²âÓëºÏÀí´¦Àí¡£

(Ò»)ȾÁϹ¤ÒµÅŷŵķÏË®Öк¬ÓдóÁ¿Óж¾µÄNO£¬¿ÉÒÔÔÚ¼îÐÔÌõ¼þϼÓÈëÂÁ·Û³ýÈ¥(¼ÓÈÈ´¦ÀíºóµÄ·ÏË®»á²úÉúÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌå)¡£³ýÈ¥NOµÄÀë×Ó·½³ÌʽΪ____________________¡£

(¶þ)ij¹¤³§¶ÔÖƸ﹤ҵÎÛÄàÖÐCrÔªËصĻØÊÕÓëÔÙÀûÓù¤ÒÕÈçÏÂ(ÁòËá½þÈ¡ÒºÖнðÊôÀë×ÓÖ÷ÒªÊÇCr3£«£¬Æä´ÎÊÇFe3£«¡¢Fe2£«¡¢Al3£«¡¢Ca2£«¡¢Mg2£«)£º

³£ÎÂϲ¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ³ÁµíʱÈÜÒºµÄpH¼ûÏÂ±í£º

ÑôÀë×Ó

Fe3£«

Fe2£«

Mg2£«

Al3£«

Cu2£«

Cr3£«

¿ªÊ¼³ÁµíʱµÄpH

1.9

7.0

£­

£­

4.7

£­

³ÁµíÍêȫʱµÄpH

3.2

9.0

11.1

8

6.7

9(£¾9Èܽâ)

(1)Ëá½þʱ£¬ÎªÁËÌá¸ß½þÈ¡ÂʿɲÉÈ¡µÄ´ëÊ©ÊÇ________(ÖÁÉÙдһÌõ)¡£

(2)H2O2µÄ×÷ÓÃÊÇ______________¡£

(3)µ÷pH£½8ÊÇΪÁ˳ýÈ¥__________(ÌîFe3£«¡¢Fe2£«¡¢Al3£«¡¢Ca2£«¡¢Mg2£«)¡£

(4)ÄÆÀë×Ó½»»»Ê÷Ö¬µÄÔ­ÀíΪMn£«£«nNaR¨D¡úMRn£«nNa£«£¬±»½»»»µÄÔÓÖÊÀë×ÓÊÇ____________(ÌîFe3£«¡¢Fe2£«¡¢Al3£«¡¢Ca2£«¡¢Mg2£«)¡£

(5)ÊÔÅäƽÑõ»¯»¹Ô­·´Ó¦·½³Ìʽ£º______

Na2Cr2O7£«SO2£«H2O===Cr(OH)(H2O)5SO4£«Na2SO4£»

ÿÉú³É1 mol Cr(OH)(H2O)5SO4תÒƵç×ÓµÄÊýĿΪ____________¡£

(Èý)Ó¡Ë¢µç·ͭ°å¸¯Ê´¼Á³£ÓÃFeCl3¡£¸¯Ê´Í­°åºóµÄ»ìºÏÈÜÒºÖУ¬ÈôCu2£«¡¢Fe3£«ºÍFe2£«µÄŨ¶È¾ùΪ0.10 mol¡¤L£­1£¬Çë²ÎÕÕÉϱí¸ø³öµÄÊý¾ÝºÍÌṩµÄÒ©Æ·£¬¼òÊö³ýÈ¥CuCl2ÈÜÒºÖÐFe3£« ºÍFe2£«µÄʵÑé²½Ö裺

¢Ù________________________________________________________________________£»

¢Ú________________________________________________________________________£»

¢Û¹ýÂË¡£(ÌṩµÄÒ©Æ·£ºCl2¡¢Å¨H2SO4¡¢NaOHÈÜÒº¡¢CuO¡¢Cu)¡£

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

¾«Ó¢¼Ò½ÌÍø