¡¾ÌâÄ¿¡¿ÔڲⶨÁòËá;§ÌåÖнᾧˮº¬Á¿µÄʵÑé²Ù×÷ÖУº
£¨1£©¼ÓÈÈÇ°Ó¦½«¾§Ìå·ÅÔÚ__________ÖÐÑÐË飬¼ÓÈÈÊÇ·ÅÔÚ__________ÖнøÐУ¬¼ÓÈÈʧˮºó£¬Ó¦·ÅÔÚ__________ÖÐÀäÈ´¡£
£¨2£©ÅжÏÊÇ·ñÍêȫʧˮµÄ·½·¨ÊÇ______________________________________________¡£
£¨3£©ÏÂÃæÊÇijѧÉúÒ»´ÎʵÑéµÄÊý¾Ý£¬ÇëÍê³É¼ÆË㣬ÌîÈëÏÂÃæµÄ±í¸ñÖС£
ÛáÛöÖÊÁ¿ | ÛáÛöÓ뾧Ìå×ÜÖÊÁ¿ | ¼ÓÈȺóÛáÛöÓë¹ÌÌå×ÜÖÊÁ¿ | ²âµÃ¾§ÌåÖнᾧˮ¸öÊý |
11.7 g | 22.7 g | 17. 6 g | _________ |
£¨4£©Õâ´ÎʵÑéÖвúÉúÎó²îµÄÔÒò¿ÉÄÜÊÇ__________£¨Ìîд×Öĸ£©ËùÔì³É¡£
A£®ÁòËá;§ÌåÖк¬ÓÐÒ×»Ó·¢ÐÔÔÓÖ¾ B£®ÊµÑéÇ°¾§ÌåÒѲ¿·Ö±ä°×
C£®¼ÓÈÈʱ¹ÌÌ岿·Ö±äºÚ D£®¼ÓÈÈʧˮºó©ÖÃÔÚ¿ÕÆøÖÐÀäÈ´
£¨5£©ÒÑÖªÔÚÛáÛöÖмÓÈÈÁòËá;§Ì壬ÊÜÈÈ·Ö½â¹ý³ÌÈçÏ£º
CuSO4¡¤5H2O CuSO4¡¤3H2O CuSO4¡¤H2O CuSO4
ÓÐÈ˽èÖúÈçͼ·â±Õ×°ÖýøÐÐÁòËá;§ÌåÍÑˮʵÑ飬»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù±¾ÊµÑé¿ÉÓÃÓÚÑéÖ¤µÄ»¯Ñ§¶¨ÂÉÊÇ_____________________________¡£
¢Úa´¦¼ÓÈÈƬ¿ÌºóÏÖÏó£º______________________________________¡£
¢ÛÄãÈÏΪ´Ë×°ÖÃÉè¼ÆÊÇ·ñºÏÀí£®¿Æѧ£¿Èç²»ºÏÀí£¬Çëд³öÀíÓÉ£º___________________________________________________________________________________________________¡£
¡¾´ð°¸¡¿Ñв§ ÛáÛö ¸ÉÔïÆ÷ ÈôÁ¬ÐøÁ½´Î¼ÓÈȵÄÖÊÁ¿²î²»³¬¹ý0.1 g£¬Ôò˵Ã÷Íêȫʧˮ 7.7 ACD ÖÊÁ¿Êغ㶨ÂÉ À¶É«¾§Ìå±äΪ°×É«·Û ²»ºÏÀí£»·â±ÕÒÇÆ÷²»Ò˼ÓÈÈ¡£
¡¾½âÎö¡¿
ÁòËá;§ÌåÖнᾧˮµÄÖÊÁ¿·ÖÊý£½(½á¾§Ë®µÄÖÊÁ¿=ÁòËá;§ÌåºÍ´ÉÛáÛöµÄÖÊÁ¿¡ªÎÞË®ÁòËáͺʹÉÛáÛöµÄÖÊÁ¿ )¡£²½ÖèΪ¢ÙÑÐÄ¥£ºÔÚÑв§Öн«ÁòËá;§ÌåÑÐËé¡£¢Ú³ÆÁ¿£»×¼È·³ÆÁ¿¸ÉÔïµÄ´ÉÛáÛöµÄÖÊÁ¿£¬²¢ÓôËÛáÛö׼ȷ³ÆÈ¡Ò»¶¨ÖÊÁ¿ÒÑÑÐËéµÄÁòËá;§Ìå¡£¢Û¼ÓÈÈ£º¼ÓÈȾ§Ì壬ʹÆäʧȥȫ²¿½á¾§Ë®(ÓÉÀ¶É«ÍêÈ«±äΪ°×É«)¡£¢Ü³ÆÁ¿£ºÔÚ¸ÉÔïÆ÷ÄÚÀäÈ´ºó³ÆÁ¿£¬²¢¼ÇÏ´ÉÛáÛöºÍÎÞË®ÁòËá͵ÄÖÊÁ¿¡£¢ÝÔÙ¼ÓÈÈ¡¢ÔÙ³ÆÁ¿ÖÁºãÖØ£º°ÑÊ¢ÓÐÎÞË®ÁòËá͵ĴÉÛáÛöÔÙ¼ÓÈÈ£¬ÔÙ·ÅÈë¸ÉÔïÆ÷ÀïÀäÈ´ºóÔÙ³ÆÁ¿£¬¼ÇÏÂÖÊÁ¿¡£µ½Á¬ÐøÁ½´Î³ÆÁ¿µÄÖÊÁ¿Ïà²î²»³¬¹ý0.1gΪֹ¡£¢Þ¼ÆË㣺¸ù¾ÝʵÑé²âµÃµÄ½á¹ûÇóÁòËá;§ÌåÖнᾧˮµÄÖÊÁ¿·ÖÊý¡£
£¨1£©¼ÓÈÈÇ°Ó¦½«¾§Ìå·ÅÔÚÑв§ÖÐÑÐË飬¼ÓÈÈÊÇ·ÅÔÚÛáÛöÖнøÐУ¬¼ÓÈÈʧˮºó£¬Ó¦·ÅÔÚ¸ÉÔïÆ÷ÖÐÀäÈ´¡£
£¨2£©ÅжÏÊÇ·ñÍêȫʧˮµÄ·½·¨ÊÇÈôÁ¬ÐøÁ½´Î¼ÓÈȵÄÖÊÁ¿²î²»³¬¹ý0.1 g£¬Ôò˵Ã÷Íêȫʧˮ¡£
£¨3£©½á¾§Ë®µÄÖÊÁ¿=ÁòËá;§ÌåºÍ´ÉÛáÛöµÄÖÊÁ¿¡ªÎÞË®ÁòËáͺʹÉÛáÛöµÄÖÊÁ¿=22.7g-17.6g=5.1g£¬½á¾§Ë®µÄ¸öÊý= £»
£¨4£©A£®ÁòËá;§ÌåÖк¬ÓÐÒ×»Ó·¢ÐÔÔÓÖ¾£¬µ¼Ö¼ÓÈȹý³ÌÖйÌÌå¼õÉÙÁ¿Ôö¶à£¬²â¶¨½á¹ûÆ«´ó£¬AÕýÈ·£»
B£®ÊµÑéÇ°¾§ÌåÒѲ¿·Ö±ä°×¶ÔʵÑé²»Ó°Ï죬B´íÎó£»
C£®¼ÓÈÈʱ¹ÌÌ岿·Ö±äºÚ£¬ËµÃ÷CuSO4ÒÑ·¢Éú·Ö½âCuSO4CuO+SO3¡ü£¬Ê¹²â¶¨µÄ½á¹ûÆ«´ó£¬CÕýÈ·£»
D£®¼ÓÈÈʧˮºó©ÖÃÔÚ¿ÕÆøÖÐÀäÈ´£¬¹ÌÌåÎüÊÕ¿ÕÆøÖеÄË®£¬ÊǹÌÌå¼õÉٵIJîÖµ½µµÍ£¬²â¶¨½á¹ûƫС£¬DÕýÈ·¡£
´ð°¸ÎªACD¡£
£¨5£©·â±Õ×°ÖýøÐÐÁòËá;§ÌåÍÑˮʵÑ飬¢ÙʵÑéÖÐÎïÖʲ»»áɢʧ£¬±¾ÊµÑé¿ÉÓÃÓÚÑéÖ¤µÄ»¯Ñ§¶¨ÂÉÊÇÖÊÁ¿Êغ㶨ÂÉ¡£
¢Úa´¦¼ÓÈÈƬ¿Ìºó£¬½á¾§Ë®·Ö½â£¬ÏÖÏóΪÀ¶É«¾§Ìå±äΪ°×É«·Û£»
¢Û´Ë×°Öò»ºÏÀí£¬·â±ÕÒÇÆ÷²»Ò˼ÓÈÈ¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÒÈ©ÊÇÖƱ¸ÒÒËá¡¢ÒÒËáÑÜÉúÎïµÈ»¯¹¤²úÆ·µÄÔÁÏ¡£Íê³ÉÏÂÁÐÌî¿Õ£º
(1)ÒÒÈ©·Ö×ÓÖеĹÙÄÜÍÅΪ______¡£
(2)½«ÍË¿ÔÚ¿ÕÆøÖÐ×ÆÉÕ±äºÚºó£¬Ñ¸ËÙÉìÈëÒÒ´¼ÖУ¬¹Û²ìµ½ÍË¿±íÃæ______£»·´¸´ÉÏÊö¶à´Î²Ù×÷ºó£¬Îŵ½´Ì¼¤ÐÔÆøζ£¬ËµÃ÷ÓÐ______Éú³É¡£
(3)д³ö¼ìÑéÒÒÈ©µÄ»¯Ñ§·´Ó¦·½³Ìʽ¡£______ÉÏÊö·´Ó¦ÏÔʾÒÒÈ©¾ßÓÐ______ÐÔ¡£
(4)ÒÑÖª¼×ËáÒ²ÄÜ·¢ÉúÒø¾µ·´Ó¦£¬Èôij¼×ËáÈÜÒºÖпÉÄÜ»ìÓÐÒÒÈ©£¬ÈçºÎͨ¹ýʵÑéÖ¤Ã÷ÊÇ·ñº¬ÓÐÒÒÈ©²¢Ð´³ö¼òÒª²Ù×÷²½Öè________________
(5)ÒÑÖª£ºÓлú»¯Ñ§Öн«½ôÁÚ¹ÙÄÜÍŵĵÚÒ»¸ö̼Ô×Ó³ÉΪ¦Á¡ªC£¬¦Á¡ªCÉϵÄH¾Í³ÆΪ¦Á¡ªH£¬È©µÄ¦Á¡ªH½Ï»îÆ㬿ÉÒÔºÍÁíÒ»¸öÈ©µÄôÊ»ù½øÐмӳɣ¬Éú³ÉôÇ»ùÈ©£¬È磺
Éè¼ÆÒ»ÌõÒÔÒÒϩΪÔÁÏÖƱ¸Õý¶¡´¼CH3CH2CH2CH2OHµÄºÏ³É·Ïß(ÎÞ»úÊÔ¼ÁÈÎÑ¡)£º_________(ºÏ³É·Ïß³£Óõıíʾ·½Ê½Îª£ºA B¡¡Ä¿±ê²úÎï)
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÀûÓõç½âÔÀí½«SO2¡¢NOxת»¯Îª£¨NH4£©2SO4µÄ×°ÖÃÈçͼËùʾ¡£ÏÂÁÐ˵·¨´íÎóµÄÊÇ£¨ £©
A.Òõ¼«µÄµç¼«·´Ó¦Ê½£ºNOx+£¨2x+3£©e-+£¨2x+4£©H+=NH4++xH2O
B.ÈÜÒºCµÄËáÐÔ±ÈÁòËáï§Ï¡ÈÜҺǿ
C.µç¼«AÓëµçÔ´µÄ¸º¼«ÏàÁ¬£¬·¢ÉúÑõ»¯·´Ó¦
D.תÒÆ0.2molµç×ÓʱÏûºÄ0.1molSO2
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÂñÔÚµØϵĸֹܵÀ¿ÉÒÔÓÃÈçͼËùʾ·½·¨½øÐе绯ѧ±£»¤£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©
A.¸Ö¹ÜµÀ±íÃæ·¢ÉúÁË»¹Ô·´Ó¦
B.¸Ã·½·¨½«µçÄÜת»¯ÎªÁË»¯Ñ§ÄÜ
C.¸Ã·½·¨³ÆΪÍâ¼ÓµçÁ÷Òõ¼«±£»¤·¨
D.þ¿éÉÏ·¢ÉúµÄµç¼«·´Ó¦£ºO2+2H2O+4e¡ú4OH-
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ä³»¯Ñ§Ð¡×éÔÚ³£ÎÂϲⶨһ¶¨ÖÊÁ¿µÄijÍÂÁ»ìºÏÎïÖÐ͵ÄÖÊÁ¿·ÖÊý£¬Éè¼ÆÁËÈçÏÂʵÑé·½°¸£º
·½°¸I£ºÍÂÁ»ìºÏÎï²â¶¨Éú³ÉÆøÌåµÄÌå»ý
·½°¸II£ºÍÂÁ»ìºÏÎï²â¶¨Ê£Óà¹ÌÌåµÄÖÊÁ¿
ÏÂÁÐÓйØÅжÏÖв»ÕýÈ·µÄÊÇ£¨ £©
A.ÈÜÒºAºÍB¾ù¿ÉÒÔÊÇÑÎËá
B.ÈÜÒºAºÍB¾ù¿ÉÒÔÊÇNaOHÈÜÒº
C.ÈÜÒºAºÍB¾ùÖ»ÄÜÊÇÑÎËᣬ²»¿ÉÒÔÊÇNaOHÈÜÒº
D.ʵÑéÊÒ·½°¸II¸ü±ãÓÚʵʩ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿
(1)MµÄ»¯Ñ§Ê½ÊÇAlCl3£¬ÓÃÀë×Ó·½³Ìʽ±íʾÆäË®ÈÜÒºµÄËá¼îÐÔÇé¿ö______________¡£
(2)ÔªËØCÔÚÖÜÆÚ±íÖеÄλÖÃÊǵÚÈýÖÜÆÚµÚ¢óA×壬Æä×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïµÄµçÀë·½³ÌʽÊÇ____________________¡£
(3)д³öEÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ________________________¡£
(4)µ¥ÖÊCÓöÀäŨÏõËáÎÞÃ÷ÏÔÏÖÏóµÄÔÒòÊÇ______________________¡£
(5)µç½âÈÛÈÚ̬Ñõ»¯ÎïAʱ£¬Ñô¼«·¢ÉúµÄµç¼«·´Ó¦ÊÇ___________________£»Ã»Óз¢ÏÖÈÛ¼Á֮ǰ£¬ÓÉÓÚÑõ»¯ÎïAÈÛµã¸ß£¬ÄÑÒÔʵÏÖ¹¤Òµ»¯¹ý³Ì£¬´ÓÎïÖʽṹ½Ç¶È˵Ã÷A¾ßÓиßÈÛµãµÄÔÒò_____________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿½«0.20 mol/LµÄÑÎËáºÍÎïÖʵÄÁ¿Å¨¶ÈΪc mol/LµÄNaOHÈÜÒº°´²»Í¬Ìå»ý±ÈÅäÖƳÉÁ½ÖÖÈÜÒº¡£Ï±íÊÇÅäÖÆʱËùÈ¡ÑÎËáÓë NaOH ÈÜÒºÌå»ýÓë»ìºÏºóÈÜÒºÖÐNa+ÓëCl-µÄÎïÖʵÄÁ¿Å¨¶ÈÊý¾Ý£¨ºöÂÔÈÜÒºÌå»ý±ä»¯£©£º
ÈÜÒº | »ìºÏÇ°ËùÈ¡ÈÜÒºÌå»ý£¨mL£© | »ìºÏºóÀë×ÓŨ¶È£¨mol/L£© | ||
HCl | NaOH | Na+ | Cl- | |
¢Ù | 30 | x | 1.5z | z |
¢Ú | 10 | y | z | 2z |
ÏÂÁÐ˵·¨ÕýÈ·£¨ £©
A.x=90B.y=30C.z=0.10D.c=0.10
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿½ðÊô¼°Æ仯ºÏÎïÔÚÉú²úÉú»îÖÐÕ¼Óм«ÆäÖØÒªµÄµØ룬Çë½áºÏ½ðÊô¼°Æ仯ºÏÎïµÄÏà¹Ø֪ʶ»Ø´ðÏÂÁÐÎÊÌâ¡£
£¨1£©º¬Ì¼Á¿ÔÚ0.03%2%Ö®¼äijºÏ½ð£¬ÊÇĿǰʹÓÃÁ¿×î´óµÄºÏ½ð£¬ÕâÖֺϽðÊÇ___¡£
A.ÂÁºÏ½ð B.ÇàÍ C.þºÏ½ð D.¸Ö
£¨2£©FeCl3ÈÜÒºÓÃÓÚ¸¯Ê´Í²Ó¡Ë¢Ïß·°å£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ___¡£
£¨3£©Ä³ÈÜÒºÖÐÓÐMg2+¡¢Fe2+¡¢Al3+¡¢Cu2+µÈÀë×Ó£¬ÏòÆäÖмÓÈë¹ýÁ¿µÄNa2O2ºó£¬¹ýÂË£¬½«ÂËÔüͶÈë×ãÁ¿µÄÑÎËáÖУ¬ËùµÃÈÜÒºÓëÔÈÜÒºÏà±È£¬ÈÜÒºÖдóÁ¿¼õÉÙµÄÑôÀë×ÓÊÇ__(Ìî×Öĸ)£¬´óÁ¿Ôö¼ÓµÄÀë×ÓÓÐ___£¨Ìѧʽ£©¡£
A.Mg2+ B.Fe2+ C.Al3+ D.Cu2+
£¨4£©4Fe(NO3)2µÄ¸ß´¿¶È½á¾§ÌåÊÇÒ»ÖÖ×ÏÉ«³±½âÐÔ¹ÌÌ壬ÊʺÏÓÚÖÆ´ÅÐÔÑõ»¯ÌúÄÉÃ×·ÛÄ©£¬¹¤ÒµÉÏÓ÷ÏÌúм(º¬FeºÍFe2O3¼°ÆäËü²»·´Ó¦µÄÎïÖÊ)ÖÆÈ¡Fe(NO3)2¾§ÌåµÄ·½·¨Èçͼ£º
¢ÙµÚ1²½·´Ó¦Èôζȹý¸ß£¬½«µ¼ÖÂÏõËá·Ö½â£¬Å¨ÏõËáÊÜÈÈ·Ö½âµÄ»¯Ñ§·½³ÌʽΪ___¡£
¢Ú´ÅÐÔÑõ»¯ÌúµÄ»¯Ñ§Ê½Îª___£¬ÊÔд³öFe2O3ÓëÏõËá·´Ó¦µÄÀë×Ó·½³Ìʽ___¡£
¢Û²Ù×÷1µÄÃû³ÆΪ___£¬²Ù×÷2µÄ²½ÖèΪ£º___¡¢___£¬¹ýÂËÏ´µÓ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¿§·È÷·Ëá¾ßÓнϹ㷺µÄ¿¹¾ú×÷Óã¬Æä½á¹¹¼òʽÈçÏÂͼËùʾ£º
¹ØÓÚ¿§·È÷·ËáµÄÏÂÁÐÐðÊöÕýÈ·µÄÊÇ(¡¡¡¡)
A£®·Ö×ÓʽΪC16H13O9
B£®1 mol¿§·È÷·Ëá¿ÉÓ뺬8 mol NaOHµÄÈÜÒº·´Ó¦
C£®ÄÜʹËáÐÔKMnO4ÈÜÒºÍÊÉ«£¬ËµÃ÷·Ö×ӽṹÖк¬ÓÐ̼̼˫¼ü
D£®ÓëŨäåË®ÄÜ·¢ÉúÁ½ÖÖÀàÐ͵ķ´Ó¦
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com