¡¾ÌâÄ¿¡¿
(1)48gÑõÆøºÍ48g³ôÑõ(O3)£¬ËüÃÇËùº¬µÄÔ×ÓÊý_______ (Ìî¡°Ïàͬ¡±»ò¡°²»Í¬¡±£¬ËüÃÇÔÚͬÎÂͬѹϵÄÌå»ý±ÈÊÇ________¡£0.2 mol NH3·Ö×ÓÓë________¸öCH4º¬Óеĵç×ÓÊýÏàͬ£¬Óë_______g H2Oº¬ÓеÄÇâÔ×ÓÊýÏàͬ£¬Óë±ê×¼×´¿öÏÂ____ L COº¬ÓеÄÔ×ÓÊýÏàµÈ¡£
(2)483g Na2SO4¡¤10H2OÖÐËùº¬µÄNa2SO4¡¤10H2OµÄÎïÖʵÄÁ¿ÊÇ_______£» Na2SO4¡¤10H2OµÄĦ¶ûÖÊÁ¿ÊÇ________£¬Ëùº¬Na+µÄÎïÖʵÄÁ¿ÊÇ________¡£º¬0.4 mol Al3£«µÄAl2(SO4)3ÖÐËùº¬µÄSO42-µÄÎïÖʵÄÁ¿ÊÇ________¡£
(3)ʵÑéÊÒ³£ÓõÄŨÑÎËáÃܶÈΪ1.17 g¡¤mL-1£¬ÖÊÁ¿·ÖÊýΪ36.5 %¡£
¢Ù´ËŨÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________________¡£
¢ÚÈ¡´ËŨÑÎËá50mL£¬ÓÃÕôÁóˮϡÊÍΪ200mL£¬Ï¡ÊͺóÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ_____¡£
¡¾´ð°¸¡¿Ïàͬ 3£º2 0.2NA 5.4 8.96 1.5mol 322g/mol 3.0mol 0.6mol 11.7mol/L 2.925mol/L
¡¾½âÎö¡¿
(1) O2ºÍO3ÊôÓÚͬËØÒìÐÎÌ壬ÆäÖÊÁ¿Ïàͬ¡¢ÑõÔ×ÓĦ¶ûÖÊÁ¿Ïàͬ£¬µ¼ÖÂÆäÔ×Ó¸öÊýÏàµÈ£»¸ù¾ÝV=n¡¤Vm=Vm¼ÆËã¶þÕßµÄÌå»ýÖ®±È£»Ã¿¸ö°±Æø·Ö×Ӻͼ×Íé·Ö×ÓÖеç×ÓÊýÏàµÈ¶¼ÊÇ10¸ö£¬ÒªÊ¹¶þÕߵĵç×ÓÊýÏàµÈ£¬Ôò¶þÕßµÄÎïÖʵÄÁ¿ÏàµÈ£¬¸ù¾ÝN=n¡¤NA¼ÆËã¼×Íé·Ö×Ó¸öÊý£»°±ÆøºÍË®·Ö×ÓÖÐHÔ×Ó¸öÊýÏàµÈ£¬Ôò¶þÕßµÄÎïÖʵÄÁ¿Ö®±ÈΪ2£º3£¬ÔÙ¸ù¾Ým=n¡¤M¼ÆËãË®µÄÖÊÁ¿£»½áºÏÆøÌå·Ö×ÓÖк¬ÓеÄÔ×Ó¸öÊý¹Øϵ¼ÆËãCOµÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾ÝV=n¡¤Vm¼ÆËãÆäÌå»ý£»
(2)¸ù¾Ýn=½áºÏÎïÖʵÄ×é³É¼ÆËãÏà¹ØÎïÀíÁ¿£¬×¢ÒâĦ¶ûÖÊÁ¿ÔÚÊýÖµÉϵÈÓÚÏà¶Ô·Ö×ÓÖÊÁ¿£»
(3)¢Ù¸ù¾ÝÎïÖʵÄÁ¿Å¨¶ÈÓëÖÊÁ¿·ÖÊý»»Ëãʽc=¼ÆË㣻
¢ÚÀûÓÃÈÜÒºÔÚÏ¡ÊÍÇ°ºóÈÜÖʵÄÎïÖʵÄÁ¿²»±ä¼ÆËãÏ¡ÊͺóÈÜÒºµÄŨ¶È¡£
(1)O2ºÍO3ÊôÓÚͬËØÒìÐÎÌ壬ÆäÖÊÁ¿Ïàͬ¡¢ÑõÔ×ÓĦ¶ûÖÊÁ¿Ïàͬ£¬µ¼ÖÂÆäÔ×Ó¸öÊýÏàµÈ£»
¸ù¾ÝV=¡¤Vm¿ÉÖª£ºÏàͬÖÊÁ¿Ê±¶þÕßµÄÌå»ýÖ®±ÈµÈÓÚÆäĦ¶ûÖÊÁ¿µÄ·´±È=48g/mol£º32g/mol=3£º2£»
ÿ¸öNH3ºÍÿ¸öCH4·Ö×ÓÖеç×ÓÊýÏàµÈ¶¼ÊÇ10¸ö£¬ÒªÊ¹¶þÕߵĵç×ÓÊýÏàµÈ£¬Ôò¶þÕßµÄÎïÖʵÄÁ¿ÏàµÈ£¬n(NH3)=0.2mol£¬ËùÒÔn(CH4)=0.2mol£»¸ù¾ÝN=nNAµÃ¼×Íé·Ö×Ó¸öÊýN(CH4)=0.2mol¡ÁNA/mol=0.2NA£»
NH3ºÍH2O·Ö×ÓÖÐHÔ×Ó¸öÊýÏàµÈ£¬ÓÉÓÚ1¸öNH3Öк¬ÓÐ3¸öHÔ×Ó£¬1¸öH2O·Ö×ÓÖк¬ÓÐ2¸öHÔ×Ó£¬Ôò¶þÕßµÄÎïÖʵÄÁ¿µÄ±ÈÓë·Ö×ÓÖк¬ÓеÄHÔ×Ó¸öÊý³Ê·´±È£¬ËùÒÔn(NH3)£ºn(H2O)=2£º3£»n(NH3)=0.2mol£¬n(H2O)=0.3mol£¬ËùÒÔË®µÄÖÊÁ¿m(H2O)=n¡¤M=0.3mol¡Á18g/mol=5.4g£»
NH3·Ö×ÓÖк¬ÓÐ4¸öÔ×Ó£¬COÊÇË«Ô×Ó·Ö×Ó£¬Á½ÖÖÆøÌ庬ÓеÄÔ×ÓÊýÄ¿ÏàµÈ£¬ÔòCOÆøÌåµÄÎïÖʵÄÁ¿ÊÇ°±ÆøÎïÖʵÄÁ¿µÄ2±¶£¬ËùÒÔn(CO)=n(NH3)=0.4mol£¬ÔòV(CO)=nVm=0.4mol¡Á22.4mol/L=8.96L£»
(2) M(Na2SO410H2O)=322g/mol£¬n(Na2SO410H2O)= 483g¡Â322g/mol=1.5mol£»¸ù¾ÝÑεÄ×é³É¿ÉÖªn(Na+)=2n(Na2SO410H2O)=2¡Á1.5mol=3.0mol£»
ÑÎAl2(SO4)3µçÀë²úÉúµÄAl3+¡¢SO42-¸öÊý±ÈΪ2:3£¬ÆäµçÀë²úÉúµÄAl3+µÄÎïÖʵÄÁ¿n(Al3+)=0.4 mol£¬ÔòÆäÖк¬ÓеÄSO42-µÄÎïÖʵÄÁ¿n(SO42-)= n(Al3+)=¡Á0.4 mol=0.6mol£»
(3)ʵÑéÊÒ³£ÓõÄŨÑÎËáÃܶÈΪ1.17 g¡¤mL-1£¬ÖÊÁ¿·ÖÊýΪ36.5 %¡£
¢Ù¸ù¾ÝÎïÖʵÄÁ¿Å¨¶ÈÓëÖÊÁ¿·ÖÊý»»Ëã¹Øϵ,¿ÉµÃ´ËŨÑÎËáµÄÎïÖʵÄÁ¿Å¨¶Èc=mol/L=11.7mol/L¡£
¢ÚÓÉÓÚÔÚÏ¡ÊÍÇ°ºóÈÜÖʵÄÎïÖʵÄÁ¿²»±ä£¬ËùÒÔ¸ù¾ÝÏ¡Ê͹«Ê½c1¡¤V1=c2¡¤V2¼ÆË㡣ȡ´ËŨÑÎËá50mL£¬ÓÃÕôÁóˮϡÊÍΪ200mL£¬Ï¡ÊͺóÑÎËáµÄÎïÖʵÄÁ¿Å¨¶Èc(Ï¡ÊÍ)= (11.7mol/L¡Á0.05L)¡Â0.2L=2.925mol/L¡£
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¿¡¿³£ÎÂÏ£¬ÓÃ0.1000mol¡¤L£1 NaOHÈÜÒºµÎ¶¨20.00mL 0.1000 mol¡¤L£1ijËᣨHA£©ÈÜÒº£¬ÈÜÒºÖÐHA¡¢A£µÄÎïÖʵÄÁ¿·ÖÊý¦Ä(X)ËæpHµÄ±ä»¯ÈçͼËùʾ¡££ÛÒÑÖª¦Ä(X)£½£ÝÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A.Ka(HA)µÄÊýÁ¿¼¶Îª10£5
B.ÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H£«)£ºaµã£¾bµã
C.µ±pH£½4.7ʱ£¬c(A£)£«c(OH£)£½c(HA)£«c(H£«)
D.µ±pH£½7ʱ£¬ÏûºÄNaOHÈÜÒºµÄÌå»ýΪ20.00mL
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÐÂÐÍпµâÒºÁ÷µç³Ø¾ßÓÐÄÜÁ¿Ãܶȸߡ¢Ñ»·ÊÙÃü³¤µÈÓÅÊÆ£¬Æ乤×÷ÔÀíÈçͼËùʾ¡£ÏÂÁÐ˵·¨´íÎóµÄÊÇ
A. ·ÅµçʱµçÁ÷´Óʯīµç¼«Á÷Ïòпµç¼«
B. ³äµçʱÑô¼«·´Ó¦Ê½Îª£º3I££2e££½I3£
C. Èô½«ÑôÀë×Ó½»»»Ä¤»»³ÉÒõÀë×Ó½»»»Ä¤£¬·ÅµçʱÕý¸º¼«Ò²ËæÖ®¸Ä±ä
D. ·Åµçʱ×ó²àµç½âÖÊ´¢¹ÞÖеÄÀë×Ó×ÜŨ¶ÈÔö´ó
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÈçͼΪʵÑéÊÒÖÆÈ¡ÕôÁóË®µÄ×°ÖÃʾÒâͼ,¸ù¾Ýͼʾ»Ø´ðÏÂÁÐÎÊÌâ¡£
(1)Ö¸³öͼÖÐÁ½´¦Ã÷ÏԵĴíÎó
¢Ù______________________________________________,
¢Ú_____________________________________________¡£
(2)AÒÇÆ÷µÄÃû³ÆÊÇ_______________,BÒÇÆ÷µÄÃû³ÆÊÇ_______________¡£
(3)ʵÑéʱAÖгý¼ÓÈëÉÙÁ¿×ÔÀ´Ë®Í⣬»¹Ðè¼ÓÈëÉÙÁ¿Ëé´ÉƬ£¬Æä×÷ÓÃÊÇ_______________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐʾÒâͼÖУ¬°×Çò´ú±íÇâÔ×Ó£¬ºÚÇò´ú±íº¤Ô×Ó£¬·½¿ò´ú±íÈÝÆ÷£¬ÈÝÆ÷ÖмäÓÐÒ»¸ö¿ÉÒÔÉÏÏ»¬¶¯µÄ¸ô°å£¨ÆäÖÊÁ¿ºöÂÔ²»¼Æ£©¡£ÆäÖÐÄܱíʾµÈÖÊÁ¿µÄÇâÆøÓ뺤ÆøµÄÊÇ£¨¡¡¡¡£©
A. B. C. D.
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÓÐËÄλͬѧ·Ö±ð¶ÔËÄÖÖÈÜÒºÖÐËùº¬µÄÀë×Ó½øÐмì²â£¬½á¹ûÈçÏ£¬ÆäÖдíÎóµÄÊÇ
A.Ca2+HCO3-Cl-K+B.OH-CO32-Cl-K+
C.Ba2+Na+OH-NO3-D.Cu2+ NO3-OH-Cl-
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿I.ÀûÓÃÏÂͼװÖòⶨÖкÍÈȵÄʵÑé²½ÖèÈçÏ£º
¢ÙÓÃÁ¿Í²Á¿È¡50 mL 0.25 mol/LÁòËáµ¹ÈëСÉÕ±ÖУ¬²â³öÁòËáζȣ»
¢ÚÓÃÁíÒ»Á¿Í²Á¿È¡50 mL 0.55 mol/L NaOHÈÜÒº£¬²¢ÓÃÁíһζȼƲâ³öÆäζȣ»¢Û½«NaOHÈÜÒºµ¹ÈëСÉÕ±ÖУ¬É跨ʹ֮»ìºÏ¾ùÔÈ£¬²â³ö»ìºÏÒº×î¸ßζȡ£
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)д³öÏ¡ÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦±íʾÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ(ÖкÍÈÈÊýֵΪ57.3 kJ/mol)£º_______________________________________________________________
(2)ѧÉú¼×ÓÃÏ¡ÁòËáÓëÏ¡ÉÕ¼îÈÜÒº²â¶¨ÖкÍÈÈ×°ÖÃÈçͼ¡£
¢ÙʵÑéʱËùÐèÒªµÄ²£Á§ÒÇÆ÷³ýÉÕ±¡¢Á¿Í²Í⻹ÐèÒª£º_________¡£
¢Ú¸Ã×°ÖÃÖÐÓÐÒ»´¦´íÎóÊÇ£º______________________£¬
(3)ÓÃÏàͬŨ¶ÈºÍÌå»ýµÄ°±Ë®(NH3¡¤ H2O)´úÌæNaOHÈÜÒº½øÐÐÉÏÊöʵÑ飬²âµÃµÄÖкÍÈȵÄÊýÖµ»á________(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£
¢ò.(1)ÒÑÖª³ä·ÖȼÉÕÒ»¶¨ÖÊÁ¿µÄ¶¡Íé(C4H10)ÆøÌåʱÉú³É1mol¶þÑõ»¯Ì¼ÆøÌåºÍҺ̬ˮ£¬²¢·Å³öÈÈÁ¿bkJ£¬Ôò±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ______________________
(2)ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
C(s£¬Ê¯Ä«)+O2(g)¨TCO2(g)¡÷H=-393.5kJmol-1
2H2(g)+O2(g)¨T2H2O(l)¡÷H=-571.6kJmol-1
2C2H2(g)+5O2(g)¨T4CO2(g)+2H2O(l)¡÷H=-2599kJmol-1
Çëд³öC(s£¬Ê¯Ä«)ºÍH2(g)Éú³É1mol C2H2(g)µÄÈÈ»¯Ñ§·½³Ìʽ____________________
(3)ÒÑÖª¼¸ÖÖ¹²¼Û¼üµÄ¼üÄÜÊý¾ÝÈçÏÂ±í£º
¹²¼Û¼ü | N¡ÔN | H¡ªH | N¡ªH |
¼üÄÜ (kJ/mol) | 946 | 436 | 390.8 |
д³öºÏ³É°±·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º ____________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ä³ÐËȤС×éµÄͬѧÉè¼ÆʵÑéÖƱ¸CuBr(°×É«½á¾§ÐÔ·ÛÄ©£¬Î¢ÈÜÓÚË®£¬²»ÈÜÓÚÒÒ´¼µÈÓлúÈܼÁ)£¬ÊµÑé×°ÖÃ(¼Ð³Ö¡¢¼ÓÈÈÒÇÆ÷ÂÔ)ÈçͼËùʾ¡£
(1)ÒÇÆ÷MµÄÃû³ÆÊÇ________¡£
(2)Èô½«MÖеÄŨÁòËá»»³É70%µÄH2SO4£¬ÔòÔ²µ×ÉÕÆ¿ÖеĹÌÌåÊÔ¼ÁΪ______(Ìѧʽ)¡£
(3)BÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______£¬ÄÜ˵Ã÷BÖз´Ó¦ÒÑÍê³ÉµÄÒÀ¾ÝÊÇ_____¡£ÈôBÖÐCu2+ÈÔδÍêÈ«±»»¹Ô£¬ÊÊÒ˼ÓÈëµÄÊÔ¼ÁÊÇ_______(Ìî±êºÅ)¡£
a.Òºäå b.Na2SO4 c.Ìú·Û d.Na2S2O3
(4)ÏÂÁйØÓÚ¹ýÂ˵ÄÐðÊö²»ÕýÈ·µÄÊÇ_______ (Ìî±êºÅ)¡£
a.©¶·Ä©¶Ë¾±¼â¿ÉÒÔ²»½ô¿¿ÉÕ±±Ú
b.½«ÂËÖ½Èóʪ£¬Ê¹Æä½ôÌù©¶·ÄÚ±Ú
c.ÂËÖ½±ßÔµ¿ÉÒԸ߳ö©¶·¿Ú
d.Óò£Á§°ôÔÚ©¶·ÖÐÇáÇá½Á¶¯ÒÔ¼Ó¿ì¹ýÂËËÙÂÊ
(5)Ï´µÓʱ£¬ÏÈÓÃ×°ÖÃCÖеÄÎüÊÕÒºÇåÏ´£¬ÆäÄ¿µÄÊÇ_______£¬ÔÙÒÀ´ÎÓÃÈܽâSO2µÄÒÒ´¼¡¢ÒÒÃÑÏ´µÓµÄÄ¿µÄÊÇ________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿»¯Ñ§ÊµÑéÊǽøÐпÆѧ̽¾¿µÄÖØҪ;¾¶£¬¹Û²ìÏÂͼµÄʵÑé×°Ö㬻شðÏÂÃæµÄÎÊÌâ¡£
(1)д³öÏÂÁÐÒÇÆ÷µÄÃû³Æ£ºa£º________________£¬b£º________________£¬c£º_________________¡£
(2)ÒÇÆ÷cÔÚʹÓÃÇ°±ØÐë½øÐеIJÙ×÷ÊÇ____________________________________________________¡£
(3)ÈôÓÃ×°ÖÃIÖÆÈ¡ÉÙÁ¿ÕôÁóË®£¬ÊµÑéʱaÖгý¼ÓÈë×ÔÀ´Ë®Í⣬»¹Ðè¼ÓÈë_________________________£¬Æä×÷ÓÃÊÇ________________________________________________________________________¡£
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