3£®·ÏÆúµç·°åÖк¬ÓдóÁ¿¿ÉÀûÓõÄËÜÁÏ¡¢²£Á§¡¢½ðÊôµÈÎïÖÊ£¬ÀûÓó¬ÁÙ½çË®¼¼Êõ´¦Àí·ÏÆúµç·°å£¬Äܵõ½ÓÉCuOºÍCu2O×é³ÉµÄ¹ÌÌå²ÐÔü£¬½«²ÐÔü½øÐнøÒ»²½´¦Àí¿ÉµÃÁòËáÍ­¾§ÌåµÈÎïÖÊ£®´¦ÀíÁ÷³ÌÈçÏÂͼËùʾ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ÒÑÖª£ºCu2O+H2SO4¨TCuSO4+Cu+H2O£©£¨¡¡¡¡£©
A£®IIÖк¬¹ýÂ˲Ù×÷£¬IÖв»º¬
B£®ÈÜÒº¼×ºÍÈÜÒºÒÒÖоùº¬Í­ÔªËØ
C£®²ÐÔüµÄ´¦ÀíÁ÷³ÌÖУ¬H2SO4ºÍCuSO4ʵÏÖÁËÑ­»·ÀûÓÃ
D£®·ÏÆúµç·°åµÄ¿É»ØÊÕÀûÓÃÎïÖÊÖУ¬ÊôÓÚ¸´ºÏ²ÄÁϵÄÊÇËÜÁÏ

·ÖÎö A¡¢¸ù¾Ý¾­¹ýI¡¢¢ò²Ù×÷£¬µÃµ½Á˹ÌÌåºÍÒºÌå½øÐзÖÎö£»
B¡¢¸ù¾ÝÈÜÒº¼×ºÍÈÜÒºÒÒÖоùº¬ÁòËáÍ­½øÐзÖÎö£»
C¡¢¸ù¾ÝÏ¡ÁòËá¿ÉÒÔÓÃÓÚÈܽâÂËÔü£¬ÁòËáÍ­ÈÜÒº¿ÉÓÃÓÚÔٴνᾧ½øÐзÖÎö£»
D¡¢¸ù¾ÝËÜÁÏÊôÓÚÈý´óºÏ³É²ÄÁÏÖ®Ò»½øÐзÖÎö£®

½â´ð ½â£ºA¡¢¾­¹ýI¡¢¢ò²Ù×÷£¬µÃµ½Á˹ÌÌåºÍÒºÌ壬ËùÒÔI¡¢¢ò²Ù×÷Öж¼ÓйýÂË£¬¹ÊA´íÎó£»
B¡¢ÈÜÒº¼×ºÍÈÜÒºÒÒÖоùº¬ÁòËáÍ­£¬ËùÒÔÈÜÒº¼×ºÍÈÜÒºÒÒÖоùº¬Í­ÔªËØ£¬¹ÊBÕýÈ·£»
C¡¢Ï¡ÁòËá¿ÉÒÔÓÃÓÚÈܽâÂËÔü£¬ÁòËáÍ­ÈÜÒº¿ÉÓÃÓÚÔٴνᾧ£¬¹ÊCÕýÈ·£»
D¡¢ËÜÁÏÊôÓÚÈý´óºÏ³É²ÄÁÏÖ®Ò»£¬²»ÊôÓÚ¸´ºÏ²ÄÁÏ£¬¹ÊD´íÎó£®
¹ÊÑ¡£ºBC£®

µãÆÀ ÔÚ½â´ËÀàÌâʱ£¬Ê×ÏÈ·ÖÎöÌâÖп¼²éµÄÎÊÌ⣬Ȼºó½áºÏѧ¹ýµÄ֪ʶºÍÌâÖÐËù¸øµÄÌáʾ½øÐнâ´ð£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ

12£®Ï±íÁгöµÄ³ýÈ¥ÎïÖÊÖÐËùº¬ÉÙÁ¿ÔÓÖʵķ½·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
ÎïÖÊÔÓÖʳýÔÓÖʵÄÊÔ¼ÁºÍ·½·¨
ACO2COÓÃ×ãÁ¿NaOHÈÜÒºÎüÊÕ£¬¸ÉÔï
BNaOHNa2CO3¼ÓÏ¡ÑÎËᣬÕô·¢
CN2O2ͨ¹ýׯÈȵÄÍ­Íø
DFeCl2ÈÜÒºCuCl2¹ýÁ¿Ìú·Û£¬¹ýÂË
A£®A¡¢B£®B¡¢C£®C¡¢D£®D¡¢

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¼ÆËãÌâ

13£®Ä³¸¯Ê´Ó¡Ë¢µç·°åµÄ·ÏÒº¾­´¦ÀíºóÖ»º¬ÓÐCuCl2ºÍFeCl2Á½ÖÖÈÜÖÊ£¬ÎªÁË·ÖÎö´¦Àíºó·ÏÒºµÄ×é³É£¬È¡200g¸Ã·ÏÒº¼ÓÈë40g·ÏÌúм£¨ÔÓÖʲ»ÈÜÓÚË®£¬Ò²²»²ÎÓë·´Ó¦£©£¬Ç¡ºÃÍêÈ«·´Ó¦£¬¹ýÂ˾­´¦ÀíµÃµ½16gÍ­£¬ÍùÂËÒºÖмÓÈë×ãÁ¿µÄÏõËáÒøÈÜÒº£¬¾­¹ýÂË¡¢¸ÉÔï¡¢³ÆÁ¿£¬×îÖյõ½114.8gAgCl¹ÌÌ壮»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Íù·ÏÒºÖмÓÈë·ÏÌúм·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪFe+CuCl2=FeCl2+Cu£¬
ÍùÂËÒºÖмÓÈëÏõËáÒøÈÜÒº·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪFeCl2+2AgNO3=2AgCl¡ý+Fe£¨NO3£©2£®
£¨2£©ÇóÂËÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£¨¾«È·µ½1%£©£®
£¨3£©Çó200g·ÏÒºÖÐFeCl2µÄÖÊÁ¿·ÖÊý£¨¾«È·µ½1%£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

10£®ÏÂÁÐÎïÖʵķÖÀàÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®»îÆÃ½ðÊô£ºÃ¾¡¢ÂÁ¡¢½ðB£®Ñõ»¯Îˮ¡¢Ñõ»¯Í­¡¢Éúʯ»Ò
C£®»ìºÏÎ¿ÕÆø¡¢Ê¯ÓÍ¡¢¼×ÍéD£®ÑΣºÉռ´¿¼î¡¢Ì¼Ëá¸Æ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

17£®Éú²ú»òÉú»îÖÐÔ̺¬×ÅÐí¶àµÄ»¯Ñ§Ô­Àí£¬ÇëÓû¯Ñ§·½³Ìʽ½âÊÍÏÂÁÐÔ­Àí£º
£¨1£©¹¤ÒµÉÏÀûÓÃÒ»Ñõ»¯Ì¼»¹Ô­Ñõ»¯ÌúÁ¶ÌúFe2O3+3CO$\frac{\underline{\;¸ßÎÂ\;}}{\;}$2Fe+3CO2£®
£¨2£©Éúʯ»Ò×ö¸ÉÔï¼ÁCaO+H2O=Ca£¨OH£©2£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ

8£®Å¯±¦±¦Ìù£¨Ö÷Òª³É·ÖΪÌú·Û¡¢Ä¾Ì¿¡¢Ê³ÑΣ©µÄÈÈÁ¿À´Ô´ÓÚÌú·ÛµÄÑõ»¯Ê±·´Ó¦µôÑõÆøÉú³ÉÌúÐ⣮СÌÎͬѧÉè¼ÆÊ¹ÓÃů±¦±¦ÌùÀ´²â¶¨¿ÕÆøÖÐÑõÆøµÄº¬Á¿£¬ÊµÑ鿪ʼǰµÄ×°ÖÃÈçͼËùʾ£¬ÊµÑéºó´ÓÁ¿Í²ÖÐÁ÷Èë²£Á§Æ¿£¨ÈÝ»ýΪ250mL£©ÖеÄË®µÄÌå»ýΪ45mL£¨Ìú·ÛÉúÐâÏûºÄµÄË®ºöÂÔ²»¼Æ£©£®ÏÂÁÐ˵·¨´íÎóµÄÊÇ£¨¡¡¡¡£©
A£®ÊµÑéǰ±ØÐë¼ì²é×°ÖÃµÄÆøÃÜÐÔ
B£®±¾´ÎʵÑéÊý¾Ý²âµÃ¿ÕÆøÖÐÑõÆøµÄÌå»ý·ÖÊýΪ18%
C£®ÈôʵÑé²âµÃ¿ÕÆøÖÐÑõÆøÌå»ý·ÖÊýÆ«¸ß£¬¿ÉÄÜÊÇů±¦±¦ÌùµÄʹÓÃÊýÁ¿²»×ã
D£®±ØÐëµÈζȼƵĶÁÊý»Ö¸´ÖÁʵÑéǰµÄζȺó²ÅÄܼǼÁ¿Í²ÄÚÊ£ÓàË®µÄÌå»ý

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

15£®ÈÕ³£Éú»îÖУ¬ÈËÃÇÀûÓÃ̼ËáÄÆÈÜÒº¾ßÓмîÐÔ£¬ÇåÏ´²Í¾ßÉϵÄÓÍÎÛ£¬¼îÐÔԽǿ£¬È¥ÓÍÎÛµÄЧ¹ûÔ½ºÃ£®ÏÂÃæÊǶÔÓ°Ïì̼ËáÄÆÈÜÒº¼îÐÔµÄÒòËØÕ¹¿ªÌ½¾¿£®
ÓÃ̼ËáÄÆ¹ÌÌåºÍ²»Í¬Î¶ȵÄË®£¬ÅäÖÆÈÜÖÊÖÊÁ¿·ÖÊý·Ö±ðΪ2%¡¢6%ºÍ10%µÄ̼ËáÄÆÈÜÒº£¬Á¢¼´²âÁ¿ÈÜÒºµÄpH£¬¼Ç¼Êý¾ÝÈçÏÂ±í£º
ʵÑé±àºÅ¢Ù¢Ú¢Û¢Ü¢Ý¢Þ¢ß¢à¢á
ÈÜÖÊÖÊÁ¿·ÖÊý2%2%2%6%6%6%]10%10%10%
Ë®µÄζȣ¨¡æ£©204060205060 204060
ÈÜÒºpH10.9011.1811.2611.0811.2711.3011.2211.4611.50
ÇëÄã·ÖÎö±íÖÐÊý¾Ý»Ø´ð£º
£¨1£©È¥ÓÍÎÛµÄЧ¹û×îºÃµÄÊǢᣨÌîʵÑé±àºÅ£©£®
£¨2£©ÔÚÒ»¶¨Î¶ȷ¶Î§ÄÚ£¬Î¶ȶÔ̼ËáÄÆÈÜÒºpHµÄÓ°ÏìÊÇ£ºµ±Ì¼ËáÄÆÈÜÒºµÄÖÊÁ¿·ÖÊýÏàͬʱ£¬Î¶ÈÔ½¸ß£¬pHÖµÔ½´ó£®
£¨3£©Òª»­³ö̼ËáÄÆÈÜÒºµÄpHËæÈÜÒºÖÐÈÜÖÊÖÊÁ¿·ÖÊýµÄ±ä»¯¹ØÏµÇúÏߣ¬¿ÉÑ¡ÔñµÄÒ»×éʵÑéÊÇ
¢Ù¢Ü¢ß£¬»ò¢Û¢Þ¢á£¨ÌîʵÑé±àºÅ£©£¬½áÂÛÊÇÔÚÒ»¶¨µÄÖÊÁ¿·ÖÊý·¶Î§ÄÚ£¬Î¶ÈÒ»¶¨Ê±£¬ÈÜÖÊÖÊÁ¿·ÖÊýÔ½´ó£¬ÈÜÒºpHÔ½´ó£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

12£®40¡æÊ±£¬ÉÕ±­ÖÐÊ¢ÓÐ100gÏõËá¼ØµÄ±¥ºÍÈÜÒº£¬ºãÎÂÏÂÕô·¢²¿·ÖÈܼÁ£¬È»ºóÔÙ½«ÈÜÒºÖð½¥Éýε½60¡æ£®Äܱíʾ´Ë¹ý³ÌÈÜÖÊÖÊÁ¿·ÖÊýa%Óëʱ¼ät¹ØÏµµÄʾÒâͼµÄÊÇ£¨¡¡¡¡£©
A£®B£®C£®D£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑé̽¾¿Ìâ

13£®ÓÐÒ»°×É«¹ÌÌ壬ËüÊÇÓÉ Ba£¨NO3£©2¡¢MgCl2¡¢Na2CO3¡¢Na2SO4ºÍ MgSO4µÈÖеÄÈýÖÖÎïÖÊ»ìºÏ¶ø³É µÄ£®ÏֶԸùÌÌå½øÐÐÁËÈçÏÂʵÑ飺
¢ÙÈ¡Ñù£¬¼Ó×ãÁ¿µÄË®£¬³ä·Ö½Á°èºó¹ýÂË£¬µÃµ½°×É«³Áµí A ºÍÎÞÉ«ÈÜÒº B£®
¢ÚÔÚ°×É«³Áµí A ÖмÓÈë×ãÁ¿µÄÏ¡ÏõËᣬ³Áµí²¿·ÖÈܽ⣬²¢·Å³öÄÜʹʯ»ÒË®±ä»ë×ǵÄÎÞÉ«ÆøÌ壮
¢ÛÔÚÎÞÉ«ÈÜÒº B ÖмÓÈë×ãÁ¿µÄ Ba£¨OH£©2ÈÜÒº£¬Óֵõ½°×É«³Áµí£»ÔÙ¼ÓÈë×ãÁ¿µÄÏ¡ÏõËᣬ½ö¹Û²ìµ½³Áµí²¿ ·ÖÈܽâµÄÏÖÏó£®
Çë»Ø´ð£º
£¨1£©Ô­°×É«¹ÌÌåÓÉÏõËá±µ¡¢Ì¼ËáÄÆ¡¢ÁòËáþ×é³É£®
£¨2£©ÊµÑé¢ÛÖгÁµíÈܽâµÄ»¯Ñ§·½³ÌʽΪ£ºMg£¨OH£©2+2HNO3=Mg£¨NO3£©2+2H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸