ÌâÄ¿ÄÚÈÝ

ʵÑéÊÒÖÆÈ¡ÆøÌåËùÐè×°ÖÃÈçÏÂͼËùʾ¡£

 

A           B         C         D         E

Çë»Ø´ðÒÔÏÂÎÊÌ⣺

£¨1£©×°ÖÃÖбꡰa¡±¡°b¡±µÄÒÇÆ÷Ãû³Æ·Ö±ðÊÇ_____       ¡¢            ¡£

£¨2£©·¢Éú×°ÖÃAÓëCÁ¬½ÓÖÆÈ¡ÑõÆø£¬µ±ÊÕ¼¯ÍêÑõÆøºó£¬Ó¦ÏÈ                       £¬ÔÙ                               ¡£

£¨3£©ÓÃ×°ÖÃBÓë     Á¬½Ó¿ÉÒÔÖÆÈ¡¶þÑõ»¯Ì¼£¬¼ì²é×°ÖÃBÆøÃÜÐԵķ½·¨ÊÇ____              .   

                                                  

£¨4£©ÊµÑéÊÒ¿ÉÒÔÓÃÎÞË®´×ËáÄƺͼîʯ»ÒÁ½ÖÖ¹Ì̬ҩƷ£¬¾­ÑÐÄ¥¾ùÔÈ»ìºÏºó×°È뷴ӦװÖÃÖУ¬¼ÓÈȲúÉú¼×ÍéÆøÌå¡£±¾ÊµÑé¿ÉÑ¡ÓõÄÆøÌå·¢Éú×°ÖÃΪ       £¨Ìî×ÖĸÐòºÅ£©£¬ÒÀ¾ÝÊÇ                            

£¨5£©°Ñ25gʯ»ÒʯÑùÆ··ÅÈë×ãÁ¿µÄÏ¡ÑÎËáÖУ¬Ê¯»ÒʯÖеÄ̼Ëá¸ÆÓëÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¨ÔÓÖʲ»·´Ó¦£¬Ò²²»Èܽ⣩£¬Éú³ÉÆøÌåµÄÖÊÁ¿Îª8.8g¡£Ôòʯ»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ¶àÉÙ£¿

 

(1) ׶ÐÎÆ¿ ¼¯ÆøÆ¿  £¨2·Ö£©

(2) °Ñµ¼Æø¹Ü´ÓË®²ÛÖÐÄóö   ϨÃð¾Æ¾«µÆ £¨2·Ö£©

(3)  D£¨1·Ö£©   Óõ¯»É¼Ð¼ÐסÏ𽺹ܣ¬Ïò³¤¾±Â©¶·ÖÐ×¢ÈëË®£¬Èô³¤¾±Â©¶·ÖÐÐγÉÒ»¶ÎÎȶ¨µÄË®Öù£¬ËµÃ÷×°Öò»Â©Æø£¨2·Ö£©  

(4) A  £¨1·Ö£©  ·´Ó¦ÎïÊǹÌÌ壬·´Ó¦Ðè¼ÓÈÈ£¨2·Ö£©

(5) £¨6·Ö£©

½â£ºÉè²Î¼Ó·´Ó¦Ì¼Ëá¸ÆµÄÖÊÁ¿Îªx

CaCO3£«2HCl£½CaCl2£«H2O£«CO2¡ü                        

100                    44

x                     8.8 g

100 :44=x: 8.8 g               

x= 20g                  

       ʯ»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ£º

           20g/25g¡Á100£¥ =80£¥      

   ´ð£ºÊ¯»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ80£¥ 

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ʵÑéÊÒÖÆÈ¡ÆøÌåËùÐè×°ÖÃÈçͼËùʾ£¬Çë»Ø´ðÒÔÏÂÎÊÌ⣺

£¨1£©ÒÇÆ÷aµÄÃû³ÆÊÇ
Ë®²Û
Ë®²Û
£®
£¨2£©ÓøßÃÌËá¼ØÖÆÈ¡ÑõÆøʱ£¬ËùÑ¡Óõķ¢Éú×°ÖÃÊÇ
A
A
£¨Ìî×ÖĸÐòºÅ£¬ÏÂͬ£©£¬Æ仯ѧ·´Ó¦·½³ÌʽΪ
2KMnO4
  ¡÷  
.
 
K2MnO4+MnO2+O2¡ü
2KMnO4
  ¡÷  
.
 
K2MnO4+MnO2+O2¡ü
£®Èç¹ûÒªÖƵýϴ¿¾»µÄÑõÆøÓ¦¸ÃÑ¡Ôñ×°ÖÃ
C
C
ÊÕ¼¯£®
 £¨3£©Áò»¯Ç⣨H2S£©ÊÇÒ»ÖÖÓж¾ÆøÌ壬ÆäÃܶȱȿÕÆøÃܶȴó£¬ÇÒÄÜÈÜÓÚË®ÐγÉÇâÁòËᣮʵÑéÊÒͨ³£Óÿé×´¹ÌÌåÁò»¯ÑÇÌú£¨FeS£©ºÍÏ¡ÁòËá»ìºÏ£¬ÔÚ³£ÎÂÏÂͨ¹ý¸´·Ö½â·´Ó¦ÖÆÈ¡Áò»¯ÇâÆøÌ壮ÊԻشð£º
¢ÙʵÑéÊÒÖÆÈ¡Áò»¯ÇâÆøÌåµÄ»¯Ñ§·½³Ìʽ
H2SO4+FeS¨TH2S¡ü+FeSO4
H2SO4+FeS¨TH2S¡ü+FeSO4

ÖÆÈ¡Áò»¯ÇâÆøÌåµÄ·¢Éú×°ÖÿÉÑ¡Ôñ
B
B
ÈôÓÃͼ1×°ÖÃÀ´ÊÕ¼¯Áò»¯ÇâÆøÌ壬½øÆø¿ÚӦΪ
a
a
£¬Ô­ÒòÊÇ
H2SµÄÃܶȱȿÕÆø´ó
H2SµÄÃܶȱȿÕÆø´ó
£®
¢ÚΪÁËÑéÖ¤ÇâÁòËáË®ÈÜÒº³ÊËáÐÔ£¬Í¼1×°ÖÃÖÐӦװµÄÊÔ¼ÁÊÇ
×ÏɫʯÈïÊÔÒº
×ÏɫʯÈïÊÔÒº
 ÏÖÏóÊÇ
×ÏɫʯÈïÊÔÒº±äºì
×ÏɫʯÈïÊÔÒº±äºì
£®
¢ÛΪÁË·ÀÖ¹¶àÓàÁò»¯ÇâÒݳöÎÛȾ¿ÕÆø£¬¿ÉÓÃ
ÇâÑõ»¯ÄÆÈÜÒº
ÇâÑõ»¯ÄÆÈÜÒº
 À´ÎüÊÕ£®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2NaOH+H2S¨TNa2S+2H2O
2NaOH+H2S¨TNa2S+2H2O
£®
ʵÑéÊÒÖÆÈ¡ÆøÌåËùÐè×°ÖÃÈçÏÂͼËùʾ£¬Çë»Ø´ðÒÔÏÂÎÊÌ⣺

£¨1£©ÒÇÆ÷aµÄÃû³ÆÊÇ
Ë®²Û
Ë®²Û
£®
£¨2£©ÓøßÃÌËá¼ØÖÆÈ¡ÑõÆøʱ£¬ËùÑ¡Óõķ¢Éú×°ÖÃÊÇ
A
A
£¨Ìî×ÖĸÐòºÅ£¬ÏÂͬ£©£¬Æ仯ѧ·´Ó¦·½³ÌʽΪ
2KMnO4
  ¡÷  
.
 
K2MnO4+MnO2+O2¡ü
2KMnO4
  ¡÷  
.
 
K2MnO4+MnO2+O2¡ü
£®
£¨3£©Áò»¯Ç⣨H2S£©ÊÇÒ»ÖÖÓж¾ÆøÌ壬ÆäÃܶȱȿÕÆøÃܶȴó£¬ÇÒÄÜÈÜÓÚË®ÐγÉÇâÁòËᣮʵÑéÊÒͨ³£Óÿé×´¹ÌÌåÁò»¯ÑÇÌú£¨FeS£©ºÍÏ¡ÁòËá»ìºÏ£¬ÔÚ³£ÎÂÏÂͨ¹ý¸´·Ö½â·´Ó¦ÖÆÈ¡Áò»¯ÇâÆøÌ壮ÊԻشð
¢ÙʵÑéÊÒÖÆÈ¡Áò»¯ÇâÆøÌåµÄ»¯Ñ§·½³Ìʽ
FeS+H2SO4=FeSO4+H2S¡ü
FeS+H2SO4=FeSO4+H2S¡ü

ÖÆÈ¡Áò»¯ÇâÆøÌåµÄ·¢Éú×°ÖÿÉÑ¡Ôñ
B
B
£¨Ìî×ÖĸÐòºÅ£©ÈôÓÃÓÒͼװÖÃÀ´ÊÕ¼¯Áò»¯ÇâÆøÌ壬´ÓaͨÈ룬ԭÒòÊÇ
Áò»¯ÇâÆøÌåÃܶȴóÓÚ¿ÕÆøÃܶÈ
Áò»¯ÇâÆøÌåÃܶȴóÓÚ¿ÕÆøÃܶÈ
£®

¢ÚΪÁËÑéÖ¤ÇâÁòËáË®ÈÜÒº³ÊËáÐÔ£¬ÓÒͼװÖÃÖÐӦװµÄÊÔ¼ÁÊÇ
×ÏɫʯÈïÊÔÒº
×ÏɫʯÈïÊÔÒº

¢ÛΪÁË·ÀÖ¹¶àÓàÁò»¯ÇâÒݳöÎÛȾ¿ÕÆø£¬¿ÉÓÃ
ÇâÑõ»¯ÄÆÈÜÒº
ÇâÑõ»¯ÄÆÈÜÒº
 À´ÎüÊÕ£®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2NaOH+H2S¨TNa2S+2H2O
2NaOH+H2S¨TNa2S+2H2O
£®

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

¾«Ó¢¼Ò½ÌÍø