ÎÒÃÇÔÚʵÑéÊÒ×öÍê¡°¶þÑõ»¯Ì¼µÄÖÆÈ¡¡±ÊµÑéºó£¬½«·ÏÒºµ¹ÈëÍ°ÖÐÊÕ¼¯ÁË´óÁ¿µÄ»ìºÏÈÜÒº£¨¿¼ÂǹÌÌåÔÓÖÊ£©£®Îª±ÜÃâÎÛȾ»·¾³£¬»¯Ñ§ÐËȤС×é×öÁËÈçÏÂʵÑ飺
È¡·ÏҺͰÉϲãÇåÒº¹²59.4kg£¬ÏòÆäÖмÓÈëÈÜÖÊÖÊÁ¿·ÖÊýΪ21.2%µÄ̼ËáÄÆÈÜÒº£®ËùµÃÈÜÒºpHÓë¼ÓÈëµÄ̼ËáÄÆÈÜÒºµÄÖÊÁ¿¹ØϵÈçͼËùʾ£º
С×ÊÁÏ£º
ÂÈ»¯¸ÆÈÜÒºµÄPHΪ7£¬
NaCO3+CaCl2=2NaCl+CaCO3¡ý¾«Ó¢¼Ò½ÌÍø
£¨1£©Í¨¹ýͼ·ÖÎö£¬Í°ÖеķÏÒºÀﺬÓеÄÈÜÖÊÊÇ
 
£®
£¨2£©Í¨¹ýͼ¿ÉÖª£¬µ±Ì¼ËáÄÆÈÜÒºÖÊÁ¿¼Óµ½
 
kgʱ£¬·ÏҺǡºÃ´¦ÀíÍ꣮
£¨3£©Í¼ÖбêʾaµãµÄº¬Òå
 
£®
£¨4£©×îÖÕËùµÃÈÜÒºÖÐÂÈ»¯ÄƵÄÖÊÁ¿Îª
 
£®
·ÖÎö£ºÌ¼Ëá¸ÆÓëÏ¡ÑÎËá·´Ó¦Éú³ÉÂÈ»¯¸ÆÈÜÓÚË®£¬¸ù¾ÝͼÏóÖеÄPHÅжÏÈÜÒºÖеÄÈÜÖÊ£¬¼ÓÈë̼ËáÄÆ·ÖÎö·´Ó¦µÄÏȺó˳Ðò£¬ÔÙ½áºÏͼÏó½øÐмÆË㣮
½â´ð£º½â£º£¨1£©ÓÉͼÏó¿É¿´³ö£¬·´Ó¦ºóµÄ·ÏÒºÖÐpHСÓÚ7£¬¹ÊÑÎËáÓÐÊ£Ó࣬Òò´ËÍ°ÖзÏÒºµÄÈÜÖÊΪHClºÍÉú³ÉµÄCaCl2£¬
£¨2£©ÔÚ·ÏÒºÖмÓÈë̼ËáÄÆ£¬ÏÈÓëÑÎËá·´Ó¦Éú³ÉÂÈ»¯ÄÆ¡¢Ë®ºÍ¶þÑõ»¯Ì¼£¬È»ºó̼ËáÄÆÓëÂÈ»¯¸Æ·´Ó¦Éú³É̼Ëá¸Æ³Áµí£¬ÒòΪ¼ÈʹÏÈÓëÂÈ»¯¸Æ·´Ó¦Éú³É̼Ëá¸Æ£¬½Ó×Å̼Ëá¸Æ¾Í±»ÑÎËáÈܽâÁË£»Ëæ×Å̼ËáÄƵIJ»¶ÏµÎ¼Ó£¬PH²»¶ÏÉý¸ß£¬µ±ÑÎËáÇ¡ºÃÍêÈ«·´Ó¦ÍêʱpHµÈÓÚ7£¬Ö®ºó̼ËáÄÆÓëÂÈ»¯¸Æ·´Ó¦£¬pH²»±ä£¬µ±ÂÈ»¯¸Æ·´Ó¦Í̼꣬ËáÄÆÓÐÊ£ÓàʱpH´óÓÚ7£¬¹ÊͼÏóµÄÈý¶Î·Ö±ð±íʾ̼ËáÄÆÓëÑÎËá·´Ó¦£¬Ì¼ËáÄÆÓëÂÈ»¯¸Æ·´Ó¦£¬Ì¼ËáÄÆÊ£Ó࣮ÓÉͼÖÐÊý¾Ý¿ÉÖªµ±Ì¼ËáÄÆÈÜÒºÖÊÁ¿¼Óµ½75kgʱ£¬·ÏҺǡºÃ´¦ÀíÍ꣮
£¨3£©Í¼ÖбêʾaµãµÄº¬ÒåaµãÊÇ̼ËáÄÆÓëÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¬pHµÈÓÚ7£¬ÈÜÒºÏÔÖÐÐÔ£®
£¨4£©¸ù¾ÝÌâÒâд³ö»¯Ñ§·½³ÌʽNa2CO3+2HCl¨T2NaCl+H2O+CO2¡ü£»Na2CO3+CaCl2=CaCO3¡ý+2NaCl¸ù¾Ý·´Ó¦Ç°ºóÔªËØÖÊÁ¿²»±ä¿É¿´³ö²Î¼Ó·´Ó¦µÄ̼ËáÄÆÖеÄÄÆÔªËØÈ«²¿Éú³ÉÁËÂÈ»¯ÄÆ£¬¿ÉÕÒ³ö̼ËáÄÆÓëÂÈ»¯ÄƵĹØϵʽ£¬¸ù¾Ý¹Øϵʽ¿ÉÇó³öÂÈ»¯ÄƵÄÖÊÁ¿£®
ÉèÂÈ»¯ÄƵÄÖÊÁ¿Îªx
ÓÉ»¯Ñ§·½³Ìʽ
Na2CO3+2HCl¨T2NaCl+H2O+CO2¡ü£»Na2CO3+CaCl2=CaCO3¡ý+2NaCl£®
¿ÉµÃ¹Øϵʽ£ºNa2CO3¡«2NaCl
             106         117
             75kg¡Á21.2%   x
           
106
75kg¡Á21.2%
=
117
x

              
            x=17.55kg
¹Ê´ð°¸Îª£º
£¨1£©HClºÍCaCl2 £®
£¨2£©75
£¨3£©aµãÊÇ̼ËáÄÆÓëÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¬pHµÈÓÚ7£¬ÈÜÒºÏÔÖÐÐÔ£®
£¨4£©11.7 kg+5.85kg=17.6 kg£®
µãÆÀ£º±¾Ìâ¹Ø½¡ÊǽáºÏÌâÒâ·ÖÎöͼÏóµÄÈý¶Î±íʾµÄÒâÒ壮ÀûÓÃÔªËØÖÊÁ¿Êغ㷨½âÌâ±È¸ù¾Ý»¯Ñ§·½³Ìʽ¸ü¼òµ¥£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2009?¾£ÃÅ£©Í¨³£ÔÚʵÑéÊÒ×öÍê¡°ÇâÆøµÄÖÆÈ¡¡±ÊµÑéºó£¬Òª½«·ÏÒºµ¹Èë·ÏҺͰÖУ¬ÕâÑù¾Í¿ÉÊÕ¼¯Ò»Ð©ÁòËáºÍÁòËáпµÄ»ìºÏÈÜÒº£¨²»¿¼ÂÇÆäËûÔÓÖÊ£©£®Îª±ÜÃâÎÛȾ»·¾³£¬Ä³Ð£»¯Ñ§ÐËȤС×éµÄͬѧ×öÁËÈçÏÂʵÑ飺
È¡·ÏҺͰÉϲãÇåÒº¹²12.2kg£¬ÏòÆäÖмÓÈëÈÜÖÊÖÊÁ¿·ÖÊýΪ8%µÄÇâÑõ»¯ÄÆÈÜÒº£®ËùµÃÈÜÒºpHÓë¼ÓÈëµÄÇâÑõ»¯ÄÆÈÜÒºµÄÖÊÁ¿¹ØϵÈçͼËùʾ£®£¨ÒÑÖªZnSO4+2NaOH¨TZn£¨OH£©2¡ý+Na2SO4£¬ZnSO4ÈÜÒºµÄpH½Ó½ü7£©
£¨1£©Í¼ÖбêʾµÄaµãµÄº¬ÒåÊÇ£º
±íʾÇâÑõ»¯ÄÆÓëÁòËáÇ¡ºÃÍêÈ«·´Ó¦
±íʾÇâÑõ»¯ÄÆÓëÁòËáÇ¡ºÃÍêÈ«·´Ó¦
£®
£¨2£©Í¨¹ýͼ¿ÉÖª£¬µ±ÇâÑõ»¯ÄÆÈÜÒºµÄÖÊÁ¿¼Óµ½
15
15
kgʱ£¬·ÏҺǡºÃ´¦ÀíÍ꣮

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

ÎÒÃÇÔÚʵÑéÊÒ×öÍê¡°¶þÑõ»¯Ì¼µÄÖÆÈ¡¡±ÊµÑéºó£¬½«·ÏÒºµ¹ÈëÍ°ÖÐÊÕ¼¯ÁË´óÁ¿µÄ»ìºÏÈÜÒº£¨¿¼ÂǹÌÌåÔÓÖÊ£©£®Îª±ÜÃâÎÛȾ»·¾³£¬»¯Ñ§ÐËȤС×é×öÁËÈçÏÂʵÑ飺
È¡·ÏҺͰÉϲãÇåÒº¹²59.4kg£¬ÏòÆäÖмÓÈëÈÜÖÊÖÊÁ¿·ÖÊýΪ21.2%µÄ̼ËáÄÆÈÜÒº£®ËùµÃÈÜÒºpHÓë¼ÓÈëµÄ̼ËáÄÆÈÜÒºµÄÖÊÁ¿¹ØϵÈçͼËùʾ£º

С×ÊÁÏ£º
ÂÈ»¯¸ÆÈÜÒºµÄPHΪ7£¬
NaCO3+CaCl2=2NaCl+CaCO3¡ý


£¨1£©Í¨¹ýͼ·ÖÎö£¬Í°ÖеķÏÒºÀﺬÓеÄÈÜÖÊÊÇ______£®
£¨2£©Í¨¹ýͼ¿ÉÖª£¬µ±Ì¼ËáÄÆÈÜÒºÖÊÁ¿¼Óµ½______kgʱ£¬·ÏҺǡºÃ´¦ÀíÍ꣮
£¨3£©Í¼ÖбêʾaµãµÄº¬Òå______£®
£¨4£©×îÖÕËùµÃÈÜÒºÖÐÂÈ»¯ÄƵÄÖÊÁ¿Îª______£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º²ýƽÇøһģ ÌâÐÍ£ºÌî¿ÕÌâ

ÎÒÃÇÔÚʵÑéÊÒ×öÍê¡°¶þÑõ»¯Ì¼µÄÖÆÈ¡¡±ÊµÑéºó£¬½«·ÏÒºµ¹ÈëÍ°ÖÐÊÕ¼¯ÁË´óÁ¿µÄ»ìºÏÈÜÒº£¨¿¼ÂǹÌÌåÔÓÖÊ£©£®Îª±ÜÃâÎÛȾ»·¾³£¬»¯Ñ§ÐËȤС×é×öÁËÈçÏÂʵÑ飺
È¡·ÏҺͰÉϲãÇåÒº¹²59.4kg£¬ÏòÆäÖмÓÈëÈÜÖÊÖÊÁ¿·ÖÊýΪ21.2%µÄ̼ËáÄÆÈÜÒº£®ËùµÃÈÜÒºpHÓë¼ÓÈëµÄ̼ËáÄÆÈÜÒºµÄÖÊÁ¿¹ØϵÈçͼËùʾ£º
С×ÊÁÏ£º
ÂÈ»¯¸ÆÈÜÒºµÄPHΪ7£¬
NaCO3+CaCl2=2NaCl+CaCO3¡ý

¾«Ó¢¼Ò½ÌÍø

£¨1£©Í¨¹ýͼ·ÖÎö£¬Í°ÖеķÏÒºÀﺬÓеÄÈÜÖÊÊÇ______£®
£¨2£©Í¨¹ýͼ¿ÉÖª£¬µ±Ì¼ËáÄÆÈÜÒºÖÊÁ¿¼Óµ½______kgʱ£¬·ÏҺǡºÃ´¦ÀíÍ꣮
£¨3£©Í¼ÖбêʾaµãµÄº¬Òå______£®
£¨4£©×îÖÕËùµÃÈÜÒºÖÐÂÈ»¯ÄƵÄÖÊÁ¿Îª______£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2009Äê±±¾©ÊвýƽÇøÖп¼»¯Ñ§Ò»Ä£ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

£¨2009?²ýƽÇøһģ£©ÎÒÃÇÔÚʵÑéÊÒ×öÍê¡°¶þÑõ»¯Ì¼µÄÖÆÈ¡¡±ÊµÑéºó£¬½«·ÏÒºµ¹ÈëÍ°ÖÐÊÕ¼¯ÁË´óÁ¿µÄ»ìºÏÈÜÒº£¨¿¼ÂǹÌÌåÔÓÖÊ£©£®Îª±ÜÃâÎÛȾ»·¾³£¬»¯Ñ§ÐËȤС×é×öÁËÈçÏÂʵÑ飺
È¡·ÏҺͰÉϲãÇåÒº¹²59.4kg£¬ÏòÆäÖмÓÈëÈÜÖÊÖÊÁ¿·ÖÊýΪ21.2%µÄ̼ËáÄÆÈÜÒº£®ËùµÃÈÜÒºpHÓë¼ÓÈëµÄ̼ËáÄÆÈÜÒºµÄÖÊÁ¿¹ØϵÈçͼËùʾ£º
С×ÊÁÏ£º
ÂÈ»¯¸ÆÈÜÒºµÄPHΪ7£¬
NaCO3+CaCl2=2NaCl+CaCO3¡ý

£¨1£©Í¨¹ýͼ·ÖÎö£¬Í°ÖеķÏÒºÀﺬÓеÄÈÜÖÊÊÇ£®
£¨2£©Í¨¹ýͼ¿ÉÖª£¬µ±Ì¼ËáÄÆÈÜÒºÖÊÁ¿¼Óµ½kgʱ£¬·ÏҺǡºÃ´¦ÀíÍ꣮
£¨3£©Í¼ÖбêʾaµãµÄº¬Ò壮
£¨4£©×îÖÕËùµÃÈÜÒºÖÐÂÈ»¯ÄƵÄÖÊÁ¿Îª£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸