ÌâÄ¿ÄÚÈÝ
19£®º£Ë®ÊÇÒ»ÖַḻµÄ×ÊÔ´£¬º¬ÓдóÁ¿µÄ»¯Ñ§ÎïÖÊ£®£¨1£©Îª½µµÍÄܺģ¬º£Ë®µ»¯³§»á²ÉÈ¡Ò»¶¨µÄ·½·¨£¬Ê¹Ë®ÔÚµÍÓڷеãµÄÇé¿öÏ·ÐÌÚ£¬½«ÕôÆøÀäÄý¶øµÃµ½µË®£¬ÕâÀïËùÖ¸µÄÒ»¶¨·½·¨ÊÇÕôÁó£®
£¨2£©³£ÎÂÏ£¬ÊµÑéÊÒÓÃÇéÐÔµçÊõµç½â200 mLº¬×ãÁ¿NaC1ÈÜÒº£¬ÈôÒõÑôÁ½¼«¹²µÃµ½44.8mLÆøÌ壨±ê×¼×´¿ö£©£¬ÔòËùµÃÈÜÒºµÄpHΪ13£¨ºöÂÔ·´Ó¦Ç°ºóÈÜÒºµÄÌå»ý±ä»¯£©£®ÀûÓÃÂȼҵ²úÎïÖеÄÁ½ÖÖÎïÖÊ·´Ó¦ÖƱ¸Æ¯°×Òº£¬´ËƯ°×ÒºÓÐЧ³É·ÖµÄ»¯Ñ§Ê½ÊÇNaClO£®
£¨3£©´µ³ö·¨Êdz£ÓõÄÒ»ÖÖº£Ë®Ìáäå¼¼Êõ£¬Ò»°ã²½ÖèÊÇ£º
¢ÙÏÈÓÃÂÈÆø½«º£Ë®ÖеÄBrÑõ»¯ÎªBr2£¬¸Ã²½Öè³ÆΪÑõ»¯£»
¢ÚÔÙÓÿÕÆø´µ³öäåÕôÆøÓÃÑÇÁòËáÄÆÈÜÒºÎüÊÕ£º
¢Û×îºó¾½øÒ»²½×ª»¯µÃµ½äåµ¥ÖÊ£®
д³ö²½Öè¢Ù¡¢¢ÚµÄÀë×Ó·½³Ìʽ¢ÙCl2+2Br-=Br2+2Cl- ¢ÚBr2+2H2O+SO2¨T4H++2Br-+SO42-£®
£¨4£©º£Ë®ÌáþµÄÒ»·½·¨ÖУ¬ÊÇÍùº£Ë®Àï¼ÓÈëʯ»ÒÈé»ñÈ¡Mg£¨OH£©2¶ø²»ÓÃNaOHÈÜÒºµÄÔÒòÊÇʯ»ÒÈéÔÁϷḻ£¬³É±¾µÍ£®
£¨5£©º£Ë®ÌáþµÄºæ¸É²½ÖèÈôÖ±½Ó¼ÓÈÈMgC12•6H2O£¬ÍùÍùµÃ²»MgC12£®ÀýÈçÔÚ235¡æʱMgC12•6H2O·Ö½âÉú³ÉMg£¨OH£©C1¡¢ÂÈ»¯ÇâºÍË®ÕôÆø£¬µÃµ½µÄÑÎËáÖÊÁ¿·ÖÊý¿É´ï28.9%£®£¨ÒªÇóд³ö¼ÆËã¹ý³Ì£®½á¹û¾«È·µ½0.1£©
·ÖÎö £¨1£©Ê¹Ë®ÔÚµÍÓڷеãµÄÇé¿öÏ·ÐÌÚ£¬½«ÕôÆøÀäÄý¶øµÃµ½µË®£¬ÎªÕôÁó²Ù×÷£»
£¨2£©ÓÃÇéÐÔµçÊõµç½â200 mLº¬×ãÁ¿NaC1ÈÜÒº£¬ÈôÒõÑôÁ½¼«¹²µÃµ½44.8mLÆøÌ壨±ê×¼×´¿ö£©£¬·Ö±ðΪÂÈÆø¡¢ÇâÆø£¬¿É¼ÆËãÉú³ÉNaOHµÄÎïÖʵÄÁ¿£¬½ø¶ø¼ÆËãpH£¬ÂÈÆøºÍÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉÂÈ»¯ÄƺʹÎÂÈËáÄÆ£»
£¨3£©´µ³ö·¨Êdz£ÓõÄÒ»ÖÖº£Ë®Ìáäå¼¼Êõ£¬Ò»°ã²½ÖèÊÇŨËõ¡¢Ñõ»¯¡¢¸»¼¯¡¢ÌáÈ¡µÈ²Ù×÷£¬ÂÈÆøÑõ»¯ä廯ÄÆÉú³Éä壬äå¿ÉÓëÑÇÁòËáÄÆ·¢ÉúÑõ»¯»¹Ô·´Ó¦£»
£¨4£©ÇâÑõ»¯¸Æ³É±¾½ÏµÍ£»
£¨5£©ÔÚ235¡æʱMgC12•6H2O·Ö½âÉú³ÉMg£¨OH£©C1¡¢ÂÈ»¯ÇâºÍË®ÕôÆø£¬·´Ó¦µÄ·½³ÌʽΪMgC12•6H2O$\frac{\underline{\;\;¡÷\;\;}}{\;}$Mg£¨OH£©C1+HCl+5H2O£¬½áºÏ·½³Ìʽ¼ÆË㣮
½â´ð ½â£º£¨1£©Ê¹Ë®ÔÚµÍÓڷеãµÄÇé¿öÏ·ÐÌÚ£¬½«ÕôÆøÀäÄý¶øµÃµ½µË®£¬ÎªÕôÁó²Ù×÷£¬¹Ê´ð°¸Îª£ºÕôÁó£»
£¨2£©µç½â±¥ºÍʳÑÎË®ÔÚÒõ¼«µÃµ½ÇâÆø£¬Ñô¼«µÃµ½ÂÈÆø£¬ÈôÒõÑôÁ½¼«¹²µÃµ½44.8mLÆøÌ壨±ê×¼×´¿ö£©£¬ÔòÂÈÆøµÄÎïÖʵÄÁ¿Îª£º$\frac{0.112L}{22.4L/mol}$=0.005mol£¬ÓÉ2NaCl+2H2O$\frac{\underline{\;ͨµç\;}}{\;}$Cl2¡ü+H2¡ü+2NaOH ¿ÉÖª£¬Cl2¡ü¡«H2¡ü¡«2NaOH£¬Ôòn£¨NaOH£©=2n£¨Cl2£©=2¡Á0.005mol=0.01mol£¬c£¨NaOH£©=$\frac{0.01mol}{0.1L}$=0.1mol/L£¬ËùÒÔpH=13£¬ÂÈÆøºÍÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉÂÈ»¯ÄƺʹÎÂÈËáÄÆ£¬ÓÐЧ³É·ÖΪNaClO£¬¹Ê´ð°¸Îª£º13£»NaClO£»
£¨3£©´µ³ö·¨Êdz£ÓõÄÒ»ÖÖº£Ë®Ìáäå¼¼Êõ£¬Ò»°ã²½ÖèÊÇŨËõ¡¢Ñõ»¯¡¢¸»¼¯¡¢ÌáÈ¡µÈ²Ù×÷£¬ÂÈÆøÑõ»¯ä廯ÄÆÉú³Éä壬·´Ó¦µÄÀë×Ó·½³ÌʽΪCl2+2Br-=Br2+2Cl-£¬äå¿ÉÓëÑÇÁòËáÄÆ·¢ÉúÑõ»¯»¹Ô·´Ó¦£¬Àë×Ó·½³ÌʽΪBr2+2H2O+SO2¨T4H++2Br-+SO42-£¬
¹Ê´ð°¸Îª£ºÑõ»¯£»Cl2+2Br-=Br2+2Cl-£»Br2+2H2O+SO2¨T4H++2Br-+SO42-£»
£¨4£©Ê¯»ÒÈéÖƱ¸¼òµ¥£¬³É±¾½ÏµÍ£¬ÔÁϷḻ£¬²»ÄÜÓÃÇâÑõ»¯ÄÆ´úÌ棬¹Ê´ð°¸Îª£ºÊ¯»ÒÈéÔÁϷḻ£¬³É±¾µÍ£»
£¨5£©ÔÚ235¡æʱMgC12•6H2O·Ö½âÉú³ÉMg£¨OH£©C1¡¢ÂÈ»¯ÇâºÍË®ÕôÆø£¬·´Ó¦µÄ·½³ÌʽΪMgC12•6H2O$\frac{\underline{\;\;¡÷\;\;}}{\;}$Mg£¨OH£©C1+HCl+5H2O£¬ÓÉ·½³Ìʽ¿ÉÖªÉú³É1molHCl£¬ÔòÉú³É5molË®£¬ÔòµÃµ½µÄÑÎËáÖÊÁ¿·ÖÊý$\frac{1mol¡Á36.5g£®mol}{1mol¡Á36.5g/mol+5mol¡Á18g/mol}¡Á100%$=28.9%£¬
¹Ê´ð°¸Îª£º28.9£®
µãÆÀ ±¾Ì⿼²é½Ï×ۺϣ¬Éæ¼°Àë×Ó·´Ó¦¡¢´ÖÑÎÌá´¿µÈ£¬Îª¸ß¿¼³£¼ûÌâÐÍ£¬²àÖØÓÚѧÉúµÄ·ÖÎö¡¢¼ÆËãÄÜÁ¦µÄ¿¼²é£¬°ÑÎÕ·¢ÉúµÄ»¯Ñ§·´Ó¦Îª½â´ðµÄ¹Ø¼ü£¬×¢ÒâÀë×Ó·´Ó¦ÓëÑõ»¯»¹Ô·´Ó¦µÄ·ÖÎö£¬ÌâÄ¿ÄѶȲ»´ó£®
£¨1£©ÔÚõ¥»¯·´Ó¦µÄʵÑéÖУ¬ÒÒËá¡¢ÒÒ´¼µÄÓÃÁ¿ºÍƽºâʱÒÒËáÒÒõ¥µÄÉú³ÉÁ¿ÈçÏÂ±í£ºÓɱíÖÐÊý¾ÝÍƲ⣬ÊýÖµxµÄ·¶Î§ÊÇ1.57£¼x£¼1.76£®
·´Ó¦ | ÒÒ´¼£¨mol£© | ÒÒËᣨmol£© | ÒÒËáÒÒõ¥£¨mol£© |
1 | 2 | 2 | 1.33 |
2 | 3 | 2 | 1.57 |
3 | 4 | 2 | x |
4 | 5 | 2 | 1.76 |
¢ÙÈôÏò´×ËáÖмÓÈëÉÙÁ¿´×ËáÄƹÌÌ壬´ËʱÈÜÒºÖÐ$\frac{c£¨{H}^{+}£©}{c£¨C{H}_{3}COOH£©}$½«¼õС£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©£»
¢ÚÈôÏò´×ËáÖмÓÈëÏ¡NaOHÈÜÒº£¬Ê¹ÆäÇ¡ºÃÍêÈ«·´Ó¦£¬ËùµÃÈÜÒºµÄpH£¾7£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©£¬ÓÃÀë×Ó·½³Ìʽ½âÊÍÆäÔÒòCH3COO-+H2O?CH3COOH+OH-£®
¢ÛÈôÏò´×ËáÖмÓÈëpH=11µÄNaOHÈÜÒº£¬ÇÒ¶þÕßµÄÌå»ý±ÈΪ1£º1£¬ÔòËùµÃÈÜÒºÖи÷Àë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇc£¨CH3COO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©£®
¢Üµ±´×ËáµÄŨ¶ÈΪ0.1mol/Lʱ£¬´Ëʱ´×ËáµÄµçÀë³£ÊýԼΪ1¡Á10-5£®
A£® | ΢Á£°ë¾¶£ºO2-£¼F-£¼Na+£¼Li+ | B£® | µÚÒ»µçÀëÄÜ£ºHe£¼Ne£¼Ar | ||
C£® | ·Ö×ÓÖеļü½Ç£ºCH4£¾H2O£¾CO2 | D£® | ¹²¼Û¼üµÄ¼üÄÜ£ºC-C£¼C=C£¼C¡ÔC |
A£® | ¿ÉÓëÄÆ·´Ó¦ | |
B£® | ÆäͬÀàͬ·ÖÒì¹¹Ìå¹²4ÖÖ | |
C£® | ÆäͬÀàͬ·ÖÒì¹¹Ìå¹²6ÖÖ£¨°üÀ¨Æä±¾Éí£© | |
D£® | ÍêȫȼÉÕʱºÄÑõµÄÖÊÁ¿±È±¾ÉíµÄÖÊÁ¿Ð¡ |
ÑôÀë×Ó | K+¡¢Na+¡¢Ba2+¡¢NH4+ |
ÒõÀë×Ó | CH3COO-¡¢Cl-¡¢OH-¡¢SO42- |
¢ÙA¡¢CÈÜÒºµÄpH¾ù´óÓÚ7£¬BÈÜÒºµÄpHСÓÚ7£¬A¡¢BµÄÈÜÒºÖÐË®µÄµçÀë³Ì¶ÈÏàͬ£»DÈÜÒºÑæÉ«·´Ó¦£¨Í¸¹ýÀ¶É«îܲ£Á§£©ÏÔ×ÏÉ«£®
¢ÚCÈÜÒººÍDÈÜÒºÏàÓöʱֻÉú³É°×É«³Áµí£¬BÈÜÒººÍCÈÜÒºÏàÓöʱֻÉú³É´Ì¼¤ÐÔÆøζµÄÆøÌ壬AÈÜÒººÍDÈÜÒº»ìºÏʱÎÞÃ÷ÏÔÏÖÏó£®
£¨1£©AµÄ»¯Ñ§Ê½£ºCH3COONa£®
£¨2£©ÓÃÀë×Ó·½³Ìʽ±íʾBµÄË®ÈÜÒºpHСÓÚ7ÔÒòNH4++H2O?NH3•H2O+H+£®
£¨3£©Ð´³öCÈÜÒººÍDÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽBa2++SO42-=BaSO4¡ý£®
£¨4£©pH=10µÄAÈÜÒººÍpH=10µÄCÈÜÒºÖÐË®µÄµçÀë³Ì¶È´óµÄÊÇCH3COONa£¨ÌîдA»òCµÄ»¯Ñ§Ê½£©£®
£¨5£©½«µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄBÈÜÒººÍCÈÜÒº»ìºÏ£¬·´Ó¦ºóÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨OH-£©£¾c£¨Ba2+£©=c£¨Cl-£©£¾c£¨NH4+£©£¾c£¨H+£©£®
A£® | ÂÈÆø | B£® | äåË® | ||
C£® | ÇâÑõ»¯ÄÆÈÜÒº | D£® | ËáÐÔ¸ßÃÌËá¼ØÈÜÒº |