äå±½ÊÇÒ»ÖÖ³£ÓõĻ¯¹¤Ô­ÁÏ¡£ÊµÑéÊÒÖƱ¸äå±½

µÄʵÑé²½ÖèÈçÏ£º

²½Öè1£ºÔÚaÖмÓÈë15 mL±½ºÍÉÙÁ¿Ìúм£¬ÔÙ½«bÖÐ4.0 mL

ÒºäåÂýÂý¼ÓÈëµ½aÖУ¬³ä·Ö·´Ó¦¡£

²½Öè2£ºÏòaÖмÓÈë10 mLË®£¬È»ºó¹ýÂ˳ýȥδ·´Ó¦µÄÌúм¡£¸ß¿¼×ÊÔ´Íø

²½Öè3£ºÂËÒºÒÀ´ÎÓÃ10 mLË®¡¢8 mL 10%µÄNaOHÈÜÒº¡¢10 mL ˮϴµÓ£¬·ÖÒºµÃ´Öäå±½¡£

²½Öè4£ºÏò·Ö³öµÄ´Öäå±½ÖмÓÈëÉÙÁ¿µÄÎÞË®ÂÈ»¯¸Æ£¬¾²ÖᢹýÂ˼´µÃ´Ö²úÆ·¡£

±½

äå

äå±½

ÃܶÈ/g¡¤cm£­3

0.88

3.10

1.50

·Ðµã/¡æ

80

59

156

ÔÚË®ÖеÄÈܽâ¶È

΢ÈÜ

΢ÈÜ

΢ÈÜ

(1)ÒÇÆ÷dµÄ×÷ÓÃÊÇ                ¡£

(2)½«bÖеÄÒºäåÂýÂý¼ÓÈëµ½aÖУ¬¶ø²»ÄÜ¿ìËÙ¼ÓÈëµÄÔ­ÒòÊÇ                         ¡£

(3)ÒÇÆ÷cµÄ×÷ÓÃÊÇÀäÄý»ØÁ÷£¬»ØÁ÷µÄÖ÷ÒªÎïÖÊÓР                    (Ìѧʽ)¡£

(4)²½Öè4µÃµ½µÄ´Ö²úÆ·Öл¹º¬ÓÐÔÓÖʱ½¡£ÒÑÖª±½¡¢äå±½µÄÓйØÎïÀíÐÔÖÊÈçÉÏ±í£¬ÔòÒª½øÒ»²½Ìá´¿´Ö²úÆ·£¬»¹±ØÐë½øÐеÄʵÑé²Ù×÷Ãû³ÆÊÇ                  ¡£


  (1)ÎüÊÕHBr·ÀÎÛȾ£¨2·Ö£©

(2)·ÀÖ¹·´Ó¦·Å³öµÄÈÈʹC6H6¡¢Br2»Ó·¢¶øÓ°Ïì²úÂÊ£¨2·Ö£©

(3)C6H6¡¢Br2¡¡£¨2·Ö£©

(4)ÕôÁó¡¡£¨2·Ö£©


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁйØÓÚÓлú»¯ºÏÎïµÄ˵·¨ÕýÈ·µÄÊÇ £¨  £©

A£®ÒÒËáºÍÒÒËáÒÒõ¥¿ÉÓÃNa2CO3ÈÜÒº¼ÓÒÔÇø±ð  

B£®ÎìÍ飨C5H12£©ÓÐÁ½ÖÖͬ·ÖÒì¹¹Ìå 

C£®ÒÒÏ©¡¢¾ÛÂÈÒÒÏ©ºÍ±½·Ö×ÓÖоùº¬ÓÐ̼̼˫¼ü   

D£®ÌÇÀà¡¢ÓÍÖ¬ºÍµ°°×Öʾù¿É·¢ÉúË®½â·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


äåË®ÔÚ¿ÆѧʵÑéÖоßÓÐÖØÒªµÄ×÷Óã¬Èç¿ÉÓÃÓÚÎïÖʵļø±ðºÍ·ÖÀëÌá´¿¡£½«6.72L£¨±ê×¼×´¿ö£©ÒÒÏ©ºÍ¼×ÍéµÄ»ìºÏÆøͨÈë×ãÁ¿µÄäåË®ÖУ¬³ä·Ö·´Ó¦ºó£¬äåË®µÄÖÊÁ¿Ôö¼ÓÁË1.4g£¬ÇëÁÐʽ¼ÆËãÔ­»ìºÏÆøÌåÖÐÒÒÏ©ºÍ¼×ÍéµÄÎïÖʵÄÁ¿Ö®±ÈÊǶàÉÙ£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¶Ô·úÀï°º£¨½á¹¹ÈçÓÒͼ£©µÄÓйØÐðÊöÕýÈ·µÄÊÇ       £¨      £©

A£®ÓÐÁ½ÖÖͬ·ÖÒì¹¹Ìå                            B£®ÊÇƽÃæÐÍ·Ö×Ó

C£®ÊÇÒ»ÖÖÎÂÊÒÆøÌå                       D£®Ôڸ߿տÉÆÆ»µ³ôÑõ²ã

 


²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁй²ÓÐÊ®¶þÖÖ»¯Ñ§·ûºÅ£º

¢ÙO2 ¢Ú614C  ¢Û238U  ¢Ü1123Na  ¢ÝO¢Þ714N  ¢ß1224Mg  ¢à    

¢á235U   ¢â£¨11£©  £¨12£©

ÆäÖУº

£¨1£©±íʾºËËصķûºÅ¹²ÓУߣߣßÖÖ.

£¨2£©»¥ÎªÍ¬Î»ËصÄÊǣߣߣߣ¨ÌîÐòºÅ£¬ÒÔÏÂÏàͬ£©.

£¨3£©ÖÊÁ¿ÊýÏàµÈ£¬µ«²»ÄÜ»¥³ÆͬλËصÄÊǣߣߣß.
£¨4£©ÖÐ×ÓÊýÏàµÈ£¬µ«ÖÊ×ÓÊý²»ÏàµÈµÄÊǣߣߣß.
£¨5£©»¥ÎªÍ¬ËØÒìÐÎÌåµÄÊǣߣߣß.
£¨6£©»¥ÎªÍ¬·ÖÒì¹¹ÌåµÄÊǣߣߣߣ®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÓйØÎïÖÊÐÔÖʵÄ˵·¨´íÎóµÄÊÇ

A£®ÈÈÎȶ¨ÐÔ£ºHCl£¾HI             B£®Ô­×Ӱ뾶£ºNa£¾Mg

C£®ËáÐÔ£ºH2SO3£¾H2SO4                 D£®»¹Ô­ÐÔ£ºS2£­£¾Cl£­

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔÚŨÑÎËáÖÐH3AsO3ÓëSnCl2µÄÀë×Ó·½³ÌʽΪ£º3SnCl2£«12Cl£­£«6 H£«£«2H3AsO3=== 2As £«3SnCl62£­£«6M£¬¹ØÓڸ÷´Ó¦µÄ˵·¨ÖÐÕýÈ·µÄ×éºÏÊÇ

¢ÙÑõ»¯¼ÁÊÇH3AsO3          ¢Ú»¹Ô­ÐÔ£ºCl£­£¾As     ¢Û ÿÉú³É1mol As£¬·´Ó¦ÖÐתÒƵç×ÓµÄÎïÖʵÄÁ¿Îª3 mol       ¢Ü MΪOH£­         ¢Ý SnCl62£­ÊÇÑõ»¯²úÎï

A£®¢Ù¢Û¢Ý        B£®¢Ù¢Ú¢Ü¢Ý       C£®¢Ù¢Ú¢Û¢Ü         D£®Ö»ÓТ٢Û

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ijÆø̬ÌþAÓëH2µÄÏà¶ÔÃܶÈΪ14£¬Æä²úÁ¿¿ÉÒÔÓÃÀ´ºâÁ¿Ò»¸ö¹ú¼ÒµÄʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½¡£ÒԸû¯ºÏÎïΪԭÁϺϳɻ¯ºÏÎïG¡¢EºÍIµÄÁ÷³ÌÈçÏ£º

    ÒÑÖª£º¢ñ£®·¼Ïã×廯ºÏÎïFΪC¡¢H¡¢O»¯ºÏÎÆäÏà¶Ô·Ö×ÓÖÊÁ¿Îª166£¬»·ÉϵÄÒ»ÂÈ´úÎïÓÐÒ»ÖÖ£¬1 mol FÓë×ãÁ¿NaHCO3ÈÜÒº·´Ó¦ÄÜÉú³É2 mol CO2£¬FÓë×ãÁ¿B·´Ó¦Éú³ÉG¡£

¢ò£®HΪ¶þÔª´¼£¬ÆäÕôÆøÃܶÈÕÛËã³É±ê×¼×´¿öΪ2.77 g/L£¬HÓë×ãÁ¿D·´Ó¦Éú³ÉI¡£

(1)AÖйÙÄÜÍŵÄÃû³ÆΪ            £¬EµÄ½á¹¹¼òʽ                         ¡£

(2)GµÄ·Ö×ÓʽΪ              £¬·´Ó¦¢ÝµÄ·´Ó¦ÀàÐÍΪ                ¡£

(3)д³öÏÂÁл¯Ñ§·½³Ìʽ£º

¢Ú                                                                  £»

¢Þ                                                               £»

(4)FÓëH¿ÉÉú³É¸ß·Ö×Ó»¯ºÏÎïJ£¬Ð´³öÉú³ÉJµÄ»¯Ñ§·´Ó¦·½³Ìʽ£º

                                                                   ¡£

(5)IÓжàÖÖͬ·ÖÒì¹¹Ì壬ÆäÖÐÒ»Ààͬ·ÖÒì¹¹ÌåÓÐÈçÏÂÌØÕ÷£º

¢Ù·Ö×ÓÖк¬ÓÐÎåÔª»·½á¹¹£»¢Ú1 mol¸ÃÓлúÎïÓë×ãÁ¿NaHCO3ÈÜÒº·´Ó¦£¬ÄÜÉú³É1 mol CO2£»¢Û1 mol¸ÃÓлúÎïÓë×ãÁ¿Na·´Ó¦£¬ÄÜÉú³É1.5 mol H2£»¢Ü»·ÉϵÄÒ»ÂÈ´úÎïÖ»ÓÐÈýÖÖ¡£ÔòÂú×ãÒÔÉÏÌõ¼þµÄÓлúÎïµÄËùÓпÉÄܵĽṹ¼òʽΪ£º

 

                                                                                   

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÎåÖÖÓÐÉ«ÈÜÒºÓëSO2×÷Ó㬾ùÄÜÍÊÉ«£¬Æ䷴ӦʵÖÊÏàͬµÄÊÇ£¨      £©

 ¢ÙÆ·ºìÈÜÒº¢ÚËáÐÔKMnO4ÈÜÒº¢ÛäåË®¢ÜµÎÈë·Ó̪µÄÉÕ¼îÈÜÒº¢Ýµí·Û¡ªµâÈÜÒº

A£®¢Ù¢Ú¢Û          B£®¢Ú¢Û¢Ü        C£®¢Û¢Ü¢Ý        D£®¢Ú¢Û¢Ý

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸