¹¤ÒµÉÏÓú¬Ð¿ÎïÁÏ(º¬FeO¡¢CuOµÈÔÓÖÊ)¿ÉÖƵûîÐÔZnO£¬Á÷³ÌÈçÏ£º
(1)ÉÏÊöÁ÷³ÌÖУ¬½þ³öÓõÄÊÇ60%H2SO4(1.5 g¡¤cm£3)£¬ÅäÖÆÕâÖÖH2SO4 100 mLÐèÒª18.4 mol¡¤L£1µÄŨH2SO4________ mL(±£ÁôһλСÊý)¡£
(2)¼ÓÈëÑõ»¯¼ÁH2O2ºó£¬ÓÐFe(OH)3³Áµí³öÏÖ£¬Ã»ÓÐCu(OH)2³Áµí³öÏÖ£¬ÈôÈÜÒºÖÐc(Fe3£«)£½2.6¡Á10£18 mol¡¤L£1£¬ÔòÈÜÒºÖÐc(Cu2£«)µÄÈ¡Öµ·¶Î§ÊÇ________mol¡¤L£1¡£(ÒÑÖªKsp[Fe(OH)3]£½2.6¡Á10£39£¬
Ksp[Cu(OH)2]£½2.2¡Á10£20)
(3)¼ÓÈëNH4HCO3ºóÉú³ÉµÄ³ÁµíÊÇÐÎ̬¾ùΪZna(OH)b(CO3)c(a¡¢b¡¢cΪÕýÕûÊý)µÄÁ½ÖÖ¼îʽ̼ËáпAºÍBµÄ»ìºÏÎAÖÐa£½5¡¢b£½6£¬ÔòÉú³É¼îʽ̼ËáпAµÄ»¯Ñ§·½³ÌʽΪ__________________________________________________¡£
(4)È¡Ï´µÓ¡¢ºæ¸ÉºóµÄ¼îʽ̼ËáпAºÍBµÄ»ìºÏÎï49.70 g£¬ÆäÎïÖʵÄÁ¿Îª0.10 mol£¬¸ßαºÉÕÍêÈ«·Ö½âµÃµ½37.26 g ZnO¡¢3.584 L CO2(±ê×¼×´¿öÏÂ)ºÍË®£¬Í¨¹ý¼ÆËãÇó³ö¼îʽ̼ËáпBµÄ»¯Ñ§Ê½¡£
(1)49.9(50.0Ò²¸ø·Ö) (2)¡Ü2.2¡Á10£6
(3)5ZnSO4£«10NH4HCO3=Zn5(OH)6(CO3)2¡ý£«5(NH4)2SO4£«8CO2¡ü£«2H2O
(4)ÓÉÌâÒâ0.1 mol»ìºÏÎïÍêÈ«·Ö½âµÃµ½ZnO¡¢CO2¡¢H2OµÄÎïÖʵÄÁ¿·Ö±ðΪ0.46 mol¡¢0.16 mol¡¢0.3 mol
¿ÉÖª1 mol»ìºÏÎïÖÐƽ¾ùº¬4.6 mol Zn¡¢1.6 mol C¡¢6 mol H£¬ÓÖÖª1 mol AÖк¬HΪ6 mol¡¢º¬CΪ2 mol£¬Ôò1 mol BÖк¬HΪ6 mol¡¢º¬CΪ1 mol£»ËùÒÔBµÄ»¯Ñ§Ê½¿ÉÒÔ±íʾΪZnx(OH)6CO3£¬ÓÉ»¯ºÏ¼Û´úÊýºÍΪÁãµÃ³öx£½4£¬¼´BµÄ»¯Ñ§Ê½ÎªZn4(OH)6CO3(ÆäËûºÏÀí½â·¨¾ù¿É)
½âÎö
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£ºÌî¿ÕÌâ
£¨8·Ö£©Ä³ÒºÌ廯ºÏÎïX2Y4£¬³£ÓÃ×ö»ð¼ýȼÁÏ¡£16g X2Y4ÔÚÒ»¶¨Á¿µÄO2ÖÐÇ¡ºÃÍêȫȼÉÕ£¬·´Ó¦·½³ÌʽΪX2Y4(l)£«O2(g)===X2(g)£«2Y2O(l)¡£ÀäÈ´ºó±ê×¼×´¿öϲâµÃÉú³ÉÎïµÄÌå»ýΪ11.2 L£¬ÆäÃܶÈΪ1.25g/L£¬Ôò£º
£¨1£©·´Ó¦Ç°O2µÄÌå»ýV(O2)Ϊ________¡£
£¨2£©X2µÄĦ¶ûÖÊÁ¿Îª________£»YÔªËصÄÃû³ÆÊÇ________¡£
£¨3£©Èô·´Ó¦Éú³É0.1mol X2£¬ÔòתÒƵç×ÓµÄÎïÖʵÄÁ¿Îª________mol¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÏÖÓÐÓɵÈÖÊÁ¿µÄNaHCO3ºÍKHCO3×é³ÉµÄ»ìºÏÎïa g£¬Óë100 mLÑÎËá·´Ó¦¡£(ÌâÖÐÉæ¼°µÄÆøÌåÌå»ý¾ùÒÔ±ê×¼×´¿ö¼Æ£¬Ìî¿Õʱ¿ÉÒÔÓôø×ÖĸµÄʽ×Ó±íʾ¡£)
(1)¸Ã»ìºÏÎïÖÐNaHCO3ÓëKHCO3µÄÎïÖʵÄÁ¿Ö®±ÈΪ ¡£
(2)Èç̼ËáÇâÑÎÓëÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¬ÔòÑÎËáÖÐHClµÄÎïÖʵÄÁ¿Îª mol¡£
(3)Èç¹ûÑÎËá¹ýÁ¿£¬Éú³ÉCO2µÄÌå»ýΪ L£»
(4)Èç¹û·´Ó¦ºó̼ËáÇâÑÎÓÐÊ£Ó࣬ÑÎËá²»×ãÁ¿£¬Òª¼ÆËãÉú³ÉCO2µÄÌå»ý£¬»¹ÐèÒªÖªµÀ ¡£
(5)ÈôNaHCO3ºÍKHCO3²»ÊÇÒÔµÈÖÊÁ¿»ìºÏ£¬Ôòa g¹ÌÌå»ìºÏÎïÓë×ãÁ¿µÄÑÎËáÍêÈ«·´Ó¦Ê±Éú³ÉCO2µÄÌå»ý[V(CO2)]·¶Î§ÊÇ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
(1)ÔÚͬΡ¢Í¬Ñ¹Ï£¬ÊµÑé²âµÃCO¡¢N2ºÍO2ÈýÖÖÆøÌåµÄ»ìºÏÆøÌåµÄÃܶÈÊÇH2µÄ14.5±¶£¬ÆäÖÐO2µÄÖÊÁ¿·ÖÊýΪ ¡£ÈôÆäÖÐCOºÍN2µÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º1£¬Ôò»ìºÏÆøÌåÖÐÑõÔªËصÄÖÊÁ¿·ÖÊýΪ ¡£
(2)ÏàͬÌõ¼þÏ£¬Ä³Cl2ÓëO2»ìºÏÆøÌå100 mLÇ¡ºÃÓë150 mL H2»¯ºÏÉú³ÉHClºÍH2O£¬Ôò»ìºÏÆøÌåÖÐCl2ÓëO2µÄÌå»ý±ÈΪ £¬»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿Îª ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ijУ»¯Ñ§Ð¡×éѧÉú½øÐС°ÆøÌåÏà¶Ô·Ö×ÓÖÊÁ¿µÄ²â¶¨¡±µÄʵÑé¡£²Ù×÷ÈçÏ£ºÓÃÖÊÁ¿ºÍÈÝ»ý¶¼ÏàµÈµÄÉÕÆ¿ÊÕ¼¯ÆøÌ壬³ÆÁ¿ÊÕ¼¯ÂúÆøÌåµÄÉÕÆ¿ÖÊÁ¿£¬Êý¾Ý¼ûϱí(ÒÑ»»Ëã³É±ê×¼×´¿öϵÄÊýÖµ)¡£
ÆøÌå | ÉÕÆ¿ºÍÆøÌåµÄ×ÜÖÊÁ¿(g) |
A | 48.4082 |
B | 48.4082 |
C | 48.4082 |
D | 48.4342 |
E | 48.8762 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ͼ°ÆäºÏ½ðÊÇÈËÀà×îÔçʹÓõĽðÊô²ÄÁÏ¡£
(1)½ðÊôͲÉÈ¡ÏÂÁÐÄÄÖÖ·½Ê½¶Ñ»ý(¡¡¡¡)
(2)ÔÚ1¸öCu2O¾§°ûÖÐ(½á¹¹ÈçÉÏͼËùʾ)£¬CuÔ×ÓÅäλÊýΪ__________¡£
(3)¿Æѧ¼Òͨ¹ýXÉäÏßÍƲ⵨·¯ÖмȺ¬ÓÐÅäλ¼ü£¬ÓÖº¬ÓÐÇâ¼ü£¬Æä½á¹¹Ê¾Òâͼ¿É¼òµ¥±íʾÈçÏ£º
¢Ùµ¨·¯µÄ»¯Ñ§Ê½ÓÃÅäºÏÎïµÄÐÎʽ±íʾΪ____________¡£
¢Úµ¨·¯ÖÐSO42¡ªµÄ¿Õ¼ä¹¹ÐÍΪ________£¬H2OÖÐOÔ×ÓµÄÔÓ»¯ÀàÐÍΪ________¡£
¢ÛijÐËȤС×é³ÆÈ¡2.500 gµ¨·¯¾§Ì壬Öð½¥ÉýÎÂʹÆäʧˮ£¬²¢×¼È·²â¶¨²»Í¬Î¶ÈÏÂÊ£Óà¹ÌÌåµÄÖÊÁ¿£¬µÃµ½ÈçͼËùʾµÄʵÑé½á¹ûʾÒâͼ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)
A£®¾§Ìå´Ó³£ÎÂÉýµ½105 ¡æµÄ¹ý³ÌÖÐÖ»ÓÐÇâ¼ü¶ÏÁÑ |
B£®µ¨·¯¾§ÌåÖÐÐγÉÅäλ¼üµÄ4¸öË®·Ö×Óͬʱʧȥ |
C£®120 ¡æʱ£¬Ê£Óà¹ÌÌåµÄ»¯Ñ§Ê½ÊÇCuSO4¡¤H2O |
D£®°´µ¨·¯¾§ÌåʧˮʱËù¿Ë·þµÄ×÷ÓÃÁ¦´óС²»Í¬£¬¾§ÌåÖеÄË®·Ö×Ó¿ÉÒÔ·ÖΪ3ÖÖ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÁòËáµÄ¹¤ÒµÖƱ¸ÊÇÒ»¸öÖØÒªµÄ»¯¹¤Éú²ú¹ý³Ì£¬µ«ÔÚÉú²ú¹ý³ÌÖлá²úÉú´óÁ¿ÎÛȾ£¬ÐèÒªÔÚÉú²ú¹¤ÒÕÖп¼Âǵ½ÂÌÉ«¹¤ÒÕ¡£
IβÆøµÄÎüÊÕºÍ×ÛºÏÀûÓá£
ÒÔ¹¤ÒµÖÆÁòËáµÄβÆø¡¢°±Ë®¡¢Ê¯»Òʯ¡¢½¹Ì¿¡¢Ì¼ËáÂÈ狀ÍKCIΪÔÁÏ¿ÉÒԺϳÉÁò»¯¸Æ¡¢ÁòËá¼Ø¡¢ÑÇÁòËá淋ÈÎïÖÊ¡£ºÏ³É·ÏßÈçÏ£º
£¨1£©·´Ó¦IIIÖÐÑõ»¯¼ÁÓ뻹ԼÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ ¡£
£¨2£©·´Ó¦¢ôµÄ»¯Ñ§·½³ÌʽΪ ¡£
£¨3£©·´Ó¦VÔÚ25¡æ¡¢40%µÄÒÒ¶þ´¼ÈÜÒºÖнøÐУ¬¸Ã·´Ó¦ÄÜ˳Àû½øÐеÄÔÒòΪ ¡£
¢ò´ß»¯¼ÁµÄ»ØÊÕÀûÓá£
SO2µÄ´ß»¯Ñõ»¯ËùʹÓõĴ߻¯¼ÁΪV2O5£¬Êµ¼ÊÉú²úÖУ¬´ß»¯¼ÁÔÚʹÓÃÒ»¶Îʱ¼äºó£¬»áº¬ÓÐV2O5¡¢VOSO4ºÍSiO2µÈ£¬ÆäÖÐVOSO4¡£ÄÜÈÜÓÚË®¡£»ØÊÕV2O5£¬µÄÖ÷ÒªÁ÷³ÌÈçÏ£º
£¨4£©Èô·´ÝÍȡʹÓõÄÁòËáÓÃÁ¿¹ý´ó£¬½øÒ»²½´¦Àíʱ»áÔö¼Ó____ µÄÓÃÁ¿¡£
£¨5£©½þÈ¡»¹Ô¹ý³ÌµÄ²úÎïÖ®Ò»ÊÇVOSO4£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ ¡£
Ñõ»¯¹ý³ÌµÄ»¯Ñ§·½³ÌʽΪKClO3+6VOSO4+3H2SO4= 2(VO)2(SO4)3+KCl+3H2O£»ÈôÁ½²½ËùÓÃÊÔ¼ÁNa2SO3ÓëKC1O3µÄÎïÖʵÄÁ¿Ö®±ÈΪ12£º7£¬Ôò¸Ã´ß»¯¼ÁÖÐV2O5¡¢VOSO4µÄÎïÖʵÄÁ¿Ö®±ÈΪ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
Ëæ×ÅÄÜÔ´ÎÊÌâµÄ½øÒ»²½Í»³ö£¬ÀûÓÃÈÈ»¯Ñ§Ñ»·ÖÆÇâµÄÑо¿Êܵ½Ðí¶à·¢´ï¹ú¼ÒµÄÇàíù¡£×î½üµÄÑо¿·¢ÏÖ£¬¸´ºÏÑõ»¯ÎïÌúËáÃÌ(MnFe2O4)Ò²¿ÉÒÔÓÃÓÚÈÈ»¯Ñ§Ñ»··Ö½âË®ÖÆÇ⣬MnFe2O4µÄÖƱ¸Á÷³ÌÈçÏ£º
(1)ÔÁÏFe(NO3)nÖÐn£½________£¬Í¶ÈëÔÁÏFe(NO3)nºÍMn(NO3)2µÄÎïÖʵÄÁ¿Ö®±ÈӦΪ________¡£
(2)²½Öè¶þÖС°Á¬Ðø½Á°è¡±µÄÄ¿µÄÊÇ__________________________________________
²½ÖèÈýÖÐÏ´µÓ¸É¾»µÄ±ê×¼ÊÇ________________________________________________
(3)ÀûÓÃMnFe2O4ÈÈ»¯Ñ§Ñ»·ÖÆÇâµÄ·´Ó¦¿É±íʾΪ£º
MnFe2O4MnFe2O4£x£«O2¡ü£»
MnFe2O4£x£«xH2OMnFe2O4£«xH2¡ü
ÇëÈÏÕæ·ÖÎöÉÏÊöÁ½¸ö·´Ó¦²¢»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÈôMnFe2O4£xÖÐx£½0.8£¬ÔòMnFe2O4£xÖÐFe2£«Õ¼È«²¿ÌúÔªËصİٷÖÂÊΪ________¡£
¢Ú¸ÃÈÈ»¯Ñ§Ñ»·ÖÆÇâ·¨µÄÓŵãÓÐ_____________________¡¢________________________ (´ðÁ½µã¼´¿É)¡£
¸ÃÈÈ»¯Ñ§Ñ»··¨ÖÆÇâÉÐÓв»×ãÖ®´¦£¬½øÒ»²½¸Ä½øµÄÑо¿·½ÏòÊÇ___________________________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
¹¤ÒµÉÏÀûÓý¹Ì¿ÔÚʯ»ÒÒ¤ÖÐȼÉÕ·ÅÈÈ£¬Ê¹Ê¯»Òʯ·Ö½âÉú²úCO2¡£Ö÷Òª·´Ó¦ÈçÏ£º
C+O2¡úCO2 ¢Ù£¬ CaCO3¡úCO2¡ü+CaO ¢Ú
£¨1£©º¬Ì¼Ëá¸Æ95%µÄʯ»Òʯ2.0 t°´¢ÚÍêÈ«·Ö½â£¨ÉèÔÓÖʲ»·Ö½â£©£¬¿ÉµÃ±ê×¼×´¿öÏÂCO2µÄÌå»ýΪ_________________m3¡£
£¨2£©´¿¾»µÄCaCO3ºÍ½¹Ì¿°´¢Ù¢ÚÍêÈ«·´Ó¦£¬µ±Ò¤ÄÚÅä±ÈÂÊ=2.2ʱ£¬Ò¤ÆøÖÐCO2µÄ×î´óÌå»ý·ÖÊýΪ¶àÉÙ£¿£¨Éè¿ÕÆøÖ»º¬N2ÓëO2£¬ÇÒÌå»ý±ÈΪ4¡Ã1£¬ÏÂͬ£©
£¨3£©Ä³´ÎÒ¤Æø³É·ÖÈçÏ£ºO2 0.2%£¬CO 0.2%£¬CO2 41.6%£¬ÆäÓàΪN2¡£Ôò´Ë´ÎÒ¤ÄÚÅä±ÈÂÊΪºÎÖµ£¿
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