ÌâÄ¿ÄÚÈÝ
14£®X¡¢Y¡¢Z¡¢W¡¢QÊÇÔ×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬ËüÃÇÔÚÖÜÆÚ±íÖеÄÏà¶ÔλÖÃÈç±í£»XÔªËØ×îµÍ¸º»¯ºÏ¼ÛµÄ¾ø¶ÔÖµÓëÆäÔ×Ó×îÍâ²ãµç×ÓÊýÏàµÈ£»ZÊǵؿÇÖк¬Á¿×î¶àµÄ½ðÊôÔªËØX | Y | ||
Z | W | Q |
£¨2£©WµÄ×î¸ß¼ÛÑõ»¯ÎïµÄ»¯Ñ§Ê½ÎªP2O5£¬YÔ×ÓÓëÇâÔ×Ó¹¹³ÉµÄÎåºË10µç×Ó΢Á£µÄµç×ÓʽΪ£®
£¨3£©XºÍÇâ¿ÉÐγɶàÖÖ»¯ºÏÎï·Ö×Ó£¬ÆäÖÐÒ»ÖÖ·Ö×Óº¬18¸öµç×Ó£¬Æä·Ö×ÓʽΪC2H6£®
£¨4£©¹ØÓÚYµÄÆø̬Ç⻯Îï¼×£º
¢Ùд³ö¼×ÓëQµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯Îï·´Ó¦µÄ»¯Ñ§·½³Ìʽ2NH3+H2SO4=£¨NH4£©2SO4£®
¢ÚÔÚ΢µç×Ó¹¤ÒµÖУ¬¼×µÄË®ÈÜÒº¿É×÷¿ÌÊ´¼ÁH2O2 µÄÇå³ý¼Á£¬Ëù·¢Éú·´Ó¦µÄ²úÎï²»ÎÛȾ»·¾³£¬Æ仯ѧ·½³ÌʽΪ2NH3+3H2O2=N2+6H2O£®¸Ã·´Ó¦µÄÑõ»¯¼ÁÊÇH2O2£¬µ±ÓÐ0.1mol¼×²Î¼Ó·´Ó¦Ê±£¬µç×ÓתÒÆ0.3 mol£®
£¨5£©QºÍYÐγɵÄÒ»ÖÖ¶þÔª»¯ºÏÎï¾ßÓÐÉ«ÎÂЧӦ£¬ÆäÏà¶Ô·Ö×ÓÖÊÁ¿ÔÚ170¡«190Ö®¼ä£¬ÇÒQµÄÖÊÁ¿·ÖÊýԼΪ69.6%£¬Ôò¸Ã»¯ºÏÎïµÄ·Ö×ÓʽΪS4N4£®
·ÖÎö X¡¢Y¡¢Z¡¢W¡¢QÊÇÔ×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬ÓÉËüÃÇÔÚÖÜÆÚ±íÖеÄÏà¶ÔλÖ㬿ÉÖªX¡¢Y´¦ÓÚµÚ¶þÖÜÆÚ£¬Z¡¢W¡¢Q´¦ÓÚµÚÈýÖÜÆÚ£¬XÔªËØ×îµÍ¸º»¯ºÏ¼ÛµÄ¾ø¶ÔÖµÓëÆäÔ×Ó×îÍâ²ãµç×ÓÊýÏàµÈ£¬ÔòXΪCÔªËØ£¬ZÊǵؿÇÖк¬Á¿×î¶àµÄ½ðÊôÔªËØ£¬ÔòZΪAl£¬¿ÉÍÆÖªYΪNÔªËØ¡¢WΪPÔªËØ¡¢QΪSÔªËØ£®
£¨1£©Ö÷×åÔªËØÖÜÆÚÊý=µç×Ó²ãÊý¡¢Ö÷×å×åÐòÊý=×îÍâ²ãµç×ÓÊý£»ZµÄÀë×ÓºËÍâÓÐ10¸öµç×Ó£¬ÓÐ2¸öµç×Ӳ㣬¸÷²ãµç×ÓÊýΪ2¡¢8£»
£¨2£©WµÄ×î¸ß¼ÛÑõ»¯ÎïΪÎåÑõ»¯¶þÁ×£¬YÔ×ÓÓëÇâÔ×Ó¹¹³ÉµÄÎåºË10µç×Ó΢Á£ÎªNH4+£»
£¨3£©XºÍÇâ¿ÉÐγɶàÖÖ»¯ºÏÎï·Ö×Ó£¬ÆäÖÐÒ»ÖÖ·Ö×Óº¬18¸öµç×Ó£¬Ö»Äܺ¬ÓÐ2¸öCÔ×Ó£¬HÔ×ÓÊýĿΪ6£»
£¨4£©YµÄÆø̬Ç⻯Îï¼×ΪNH3£¬
¢ÙQµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïΪH2SO4£¬Óë°±Æø·´Ó¦Éú³ÉÁòËá泥»
¢Ú°±ÆøµÄË®ÈÜÒº¿É×÷¿ÌÊ´¼ÁH2O2 µÄÇå³ý¼Á£¬Ëù·¢Éú·´Ó¦µÄ²úÎï²»ÎÛȾ»·¾³£¬Ó¦Éú³ÉµªÆøÓëË®£¬Ëùº¬ÔªËØ»¯ºÏ¼Û½µµÍµÄ·´Ó¦ÎïΪÑõ»¯¼Á£¬¸ù¾ÝNÔªËØ»¯ºÏ¼Û±ä»¯¼ÆËãתÒƵç×Ó£»
£¨5£©¸ù¾ÝÖÊÁ¿·ÖÊý¼ÆË㻯ºÏÎïÖÐN¡¢SÔ×ÓÊýÄ¿Ö®±È£¬½áºÏÏà¶Ô·Ö×ÓÖÊÁ¿È·¶¨»¯Ñ§Ê½£®
½â´ð ½â£ºX¡¢Y¡¢Z¡¢W¡¢QÊÇÔ×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬ÓÉËüÃÇÔÚÖÜÆÚ±íÖеÄÏà¶ÔλÖ㬿ÉÖªX¡¢Y´¦ÓÚµÚ¶þÖÜÆÚ£¬Z¡¢W¡¢Q´¦ÓÚµÚÈýÖÜÆÚ£¬XÔªËØ×îµÍ¸º»¯ºÏ¼ÛµÄ¾ø¶ÔÖµÓëÆäÔ×Ó×îÍâ²ãµç×ÓÊýÏàµÈ£¬ÔòXΪCÔªËØ£¬ZÊǵؿÇÖк¬Á¿×î¶àµÄ½ðÊôÔªËØ£¬ÔòZΪAl£¬¿ÉÍÆÖªYΪNÔªËØ¡¢WΪPÔªËØ¡¢QΪSÔªËØ£®
£¨1£©QΪSÔªËØ£¬´¦ÓÚÖÜÆÚ±íÖеÚÈýÖÜÆÚ¢öA×壬Al3+Àë×ӽṹʾÒâͼΪ£¬
¹Ê´ð°¸Îª£ºµÚÈýÖÜÆÚ¢öA×壻£»
£¨2£©WµÄ×î¸ß¼ÛÑõ»¯ÎïΪP2O5£¬YÔ×ÓÓëÇâÔ×Ó¹¹³ÉµÄÎåºË10µç×Ó΢Á£ÎªNH4+£¬µç×ÓʽΪ£¬¹Ê´ð°¸Îª£ºP2O5£»£»
£¨3£©XºÍÇâ¿ÉÐγɶàÖÖ»¯ºÏÎï·Ö×Ó£¬ÆäÖÐÒ»ÖÖ·Ö×Óº¬18¸öµç×Ó£¬Ö»Äܺ¬ÓÐ2¸öCÔ×Ó£¬HÔ×ÓÊýĿΪ6£¬·Ö×ÓʽΪC2H6£¬¹Ê´ð°¸Îª£ºC2H6£»
£¨4£©YµÄÆø̬Ç⻯Îï¼×ΪNH3£¬
¢ÙQµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïΪH2SO4£¬Óë°±Æø·´Ó¦Éú³ÉÁòËá泥¬·´Ó¦·½³ÌʽΪ£º2NH3+H2SO4=£¨NH4£©2SO4£¬¹Ê´ð°¸Îª£º2NH3+H2SO4=£¨NH4£©2SO4£»
¢Ú°±ÆøµÄË®ÈÜÒº¿É×÷¿ÌÊ´¼ÁH2O2 µÄÇå³ý¼Á£¬Ëù·¢Éú·´Ó¦µÄ²úÎï²»ÎÛȾ»·¾³£¬Ó¦Éú³ÉµªÆøÓëË®£¬·´Ó¦·½³ÌʽΪ£º2NH3+3H2O2=N2+6H2O£¬H2O2ÖÐOÔªËØ»¯ºÏ¼Û½µµÍ£¬H2O2ΪÑõ»¯¼Á£¬·´Ó¦ÖÐNÔªËØ»¯ºÏ¼ÛÓÉ-3¼ÛÉý¸ßΪ0¼Û£¬0.1mol¼×²Î¼Ó·´Ó¦Ê±£¬×ªÒƵç×ÓΪ0.3mol£¬
¹Ê´ð°¸Îª£º2NH3+3H2O2=N2+6H2O£»H2O2£»0.3£»
£¨5£©YΪNÔªËØ£¬QΪSÔªËØ£¬ÐγɵĻ¯ºÏÎïÖÐSµÄÖÊÁ¿·ÖÊýΪ69.6%£¬ÔòSºÍNÔ×Ó¸öÊý±ÈΪ$\frac{69.6%}{32}$£º$\frac{1-69.6%}{14}$=1£º1£¬ÆäÏà¶Ô·Ö×ÓÖÊÁ¿ÔÚ170¡«190Ö®¼ä£¬É軯ѧʽΪ£¨SN£©x£¬Ôò170£¼46x£¼190£¬½âµÃx=4ʱ£¬¹Ê¶þÕßÐγɵķÖ×ÓʽΪS4N4£¬¹Ê´ð°¸Îª£ºS4N4£®
µãÆÀ ±¾Ì⿼²éÔªËØÖÜÆÚ±íÓëÔªËØÖÜÆÚÂÉ¡¢³£Óû¯Ñ§ÓÃÓïµÈ£¬ÄѶÈÖеȣ¬×¢Òâ¶ÔÔªËØÖÜÆÚÂɵÄÀí½âÕÆÎÕ£®
A£® | µÈÎïÖʵÄÁ¿Å¨¶ÈµÄFeBr2ºÍCuCl2»ìºÏÈÜÒºÓöèÐԵ缫µç½â×î³õ·¢Éú£ºCu 2++2Br- $\frac{\underline{\;µç½â\;}}{\;}$Cu+Br2 | |
B£® | Na2SÈÜҺˮ½â£ºS2-+2H2O?H2S+2OH- | |
C£® | Fe£¨NO3£©3µÄËáÐÔÈÜÒºÖÐͨÈë×ãÁ¿Áò»¯Ç⣺2Fe3++H2S=2Fe2++S¡ý+2H+ | |
D£® | H218OÖÐͶÈëNa2O2£º2H218O+2Na2O2=4Na++4OH-+18O2¡ü |
¹ØÓÚÕâÁ½ÖÖ¾§ÌåµÄ˵·¨£¬ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£® | Á¢·½Ï൪»¯Åðº¬ÓЦҼüºÍ¦Ð¼ü£¬ËùÒÔÓ²¶È´ó | |
B£® | Áù·½Ï൪»¯Åð²ã¼ä×÷ÓÃÁ¦Ð¡£¬ËùÒÔÖʵØÈí | |
C£® | Á½ÖÖ¾§ÌåÖеļÈÓм«ÐÔ¼üÓÖÓзǼ«ÐÔ¼ü | |
D£® | Á½ÖÖ¾§Ìå¾ùΪÔ×Ó¾§Ìå |
A£® | ½«Ò»Ð¡¿éÄÆͶÈ뵽ʢÓÐÁòËáÍÈÜÒºµÄÉÕ±ÖУ¬·¢ÏÖûÓÐ×ϺìÉ«µÄÍÎö³ö£¬²»ÄÜ˵Ã÷ÄƵĽðÊôÐÔ±ÈÍÈõ | |
B£® | ÓÃʪÈóµÄµí·Ûµâ»¯¼ØÊÔÖ½·ÅÖÃÔÚÓÐÂÈÆø²úÉúµÄµ¼¹ÜÉ϶ˣ¬·¢ÏÖÊÔÖ½±äÀ¶£¬ËµÃ÷ÂÈÔªËصķǽðÊôÐԱȵâÔªËØÇ¿ | |
C£® | °ÑH2SÆøÌåºÍCl2ÔÚ¼¯ÆøÆ¿ÖлìºÏ£¬Ò»¶Îʱ¼äºó·¢ÏÖÆ¿±ÚÓе»ÆÉ«¹ÌÌåÎö³ö£¬ËµÃ÷ClµÄ·Ç½ðÊôÐÔ±ÈSÇ¿ | |
D£® | ½«0.1 mol Na¡¢Mg¡¢Al·Ö±ðÓë×ãÁ¿Ï¡ÑÎËá·´Ó¦£¬½ðÊôʧȥµÄµç×ÓÊý·Ö±ðΪ0.1 mol¡¢0.2 mol¡¢0.3 mol£¬ËµÃ÷ÕâÈýÖÖ½ðÊôµÄ»îÆÃÐÔΪ£ºAl£¾Mg£¾Na |
A£® | ÈôÐγɻ¯Ñ§¼üÊͷŵÄÄÜÁ¿´óÓÚ¶ÏÁÑ»¯Ñ§¼üËùÎüÊÕµÄÄÜÁ¿£¬ÔòÊÇ·ÅÈÈ·´Ó¦ | |
B£® | ÔÚÏ¡ÈÜÒºÖУºH+£¨aq£©+OH-£¨aq£©¨TH2O£¨l£©¡÷H=-57.3 kJ/mol£¬Èô½«º¬0.5 mol H2SO4µÄŨÁòËáÓ뺬1 mol NaOHµÄÈÜÒº»ìºÏ£¬·Å³öµÄÈÈÁ¿´óÓÚ57.3 kJ | |
C£® | ÓÉ4P£¨s£¬ºìÁ×£©¨TP4£¨s£¬°×Á×£©¡÷H=+139.2 kJ/mol£¬¿ÉÖªºìÁױȰ×Á×Îȶ¨ | |
D£® | ÔÚ101 kPaʱ£¬2 g H2ÍêȫȼÉÕÉú³ÉҺ̬ˮ£¬·Å³ö285.8 kJ ÈÈÁ¿£¬ÇâÆøȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ¿É±íʾΪ2H2£¨g£©+O2£¨g£©¨T2H2O£¨l£©¡÷H=-285.8 kJ/mol |
£¨1£©Ï±íÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬¸ù¾Ý±íÖиø³öµÄ8ÖÖÔªËØ£¬ÓÃÔªËØ·ûºÅ»ò»¯Ñ§Ê½»Ø´ðÏÂÁÐÎÊÌ⣺
ÖÜÆÚ×å | ¢ñA | ¢òA | ¢óA | ¢ôA | ¢õA | ¢öA | ¢÷A | 0 |
2 | A | B | C | D | ||||
3 | E | F | G | H |
¢ÚÔªËØ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïÖУ¬ËáÐÔ×îÇ¿µÄÊÇHClO4£®
¢ÛBÓëCÁ½ÔªËصÄÆø̬Ç⻯ÎÈÈÎȶ¨ÐÔ½ÏÇ¿µÄÊǵÚÈýÖÜÆÚVA×壮
£¨2£©ËÄÖÖ¶ÌÖÜÆÚÔªËØÔÚÖÜÆÚ±íÖеÄÏà¶ÔλÖÃÈçͼËùʾ£¬ÆäÖÐZÔªËØÔ×ÓºËÍâµç×Ó×ÜÊýÊÇÆä×îÍâ²ãµç×ÓÊýµÄ3±¶£®
X | ||
Y | Z | W |
¢ÙÔªËØZλÓÚÖÜÆÚ±íÖеÚÈýÖÜÆÚVA ×壮
¢ÚYÔªËصÄÔ×ӽṹʾÒâͼΪ£®
¢ÛÔªËØXµÄÆø̬Ç⻯ÎïÓëÔªËØWµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯Îï·´Ó¦µÄÉú³ÉÎïÖл¯Ñ§¼üµÄÀàÐÍÊÇÀë×Ó¼ü¡¢¹²¼Û¼ü£®
A£® | 0.1 mol•L-1ij¶þÔªÈõËáµÄËáʽÑÎNaHAÈÜÒºÖУºc£¨Na+£©=2c£¨A2-£©+c£¨HA-£©+c£¨H2A£© | |
B£® | ijËáʽÑÎNaHYµÄË®ÈÜÒºÏÔ¼îÐÔ£¬¸ÃËáʽÑÎÈÜÒºÖÐÀë×ÓŨ¶È¹ØϵΪ£ºc£¨Na+£©£¾c£¨HY-£©£¾c£¨OH-£©£¾c£¨H+£© | |
C£® | ÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄCH3COOHºÍCH3COONaÈÜÒºµÈÌå»ý»ìºÏ£ºc£¨CH3COO-£©+c£¨OH-£©=c£¨H+£©+c£¨CH3COOH£© | |
D£® | 25¡æ£¬pH=12µÄ°±Ë®ºÍpH=2µÄÑÎËáµÈÌå»ý»ìºÏ£ºc£¨Cl-£©£¾c£¨NH4+£©£¾c£¨H+£©£¾c£¨OH-£© |
A£® | Ôµç³ØÊǰѵçÄÜת»¯Îª»¯Ñ§ÄܵÄ×°Öà | |
B£® | Ôµç³ØÖеç×ÓÁ÷³öµÄÒ»¼«ÊǸº¼«£¬·¢Éú»¹Ô·´Ó¦ | |
C£® | ¹¹³ÉÔµç³ØµÄÁ½¼«±ØÐëÊÇÁ½ÖÖ²»Í¬µÄ½ðÊô | |
D£® | Cu-FeÏ¡ÁòËáÔµç³Ø¹¤×÷ʱ£¬ÍÊÇÕý¼« |
A£® | Cl2+H2O¨THCl+HClO | B£® | 3Fe+4H2O£¨g£©$\frac{\underline{\;\;¡÷\;\;}}{\;}$Fe3O4+4H2 | ||
C£® | 2F2+2H2O¨T4HF+O2 | D£® | 3NO2+H2O¨T2HNO3+NO |