5¡¢³ÆÈ¡Ìú¿óʯÊÔÑùms=0.3669g£¬ÓÃHClÈÜÒºÈܽâºó£¬¾Ô¤´¦ÀíʹÌú³ÊFe״̬£¬ÓÃK2Cr2O7±ê×¼ÈÜ
ÒºµÎ¶¨£¬ÏûºÄK2Cr2O7Ìå»ý28.62mL£¬¼ÆËãÒÔFe¡¢Fe2O3¡¢Fe3O4±íʾµÄÖÊÁ¿·ÖÊý¸÷Ϊ¶àÉÙ£¿ [c(1/6K2Cr2O7)=0.1200mol/L]
6¡¢¼ÆËãÏÂÁÐÈÜÒºµÄµÎ¶¨¶È£¬ÒÔg/mL±íʾ£º
£¨1£©0.2615mol/L HClÈÜÒº£¬²â¶¨Ba(OH)2ºÍCa(OH)2 £¨2£©0.1032mol/L NaOHÈÜÒº£¬²â¶¨H2SO4ºÍCH3COOH
7¡¢³ÆÈ¡²ÝËáÄÆ»ù×¼Îï0.2178g±ê¶¨KMnO4ÈÜÒºµÄŨ¶È£¬ÓÃÈ¥KMnO4ÈÜÒºÌå»ý25.48mL£¬¼ÆËãKMnO4ÈÜÒºµÄŨ¶ÈΪ¶àÉÙ£¿
8¡¢ÓÃÅðɰ£¨Na2B4O7¡¤10H2£©0.4709g±ê¶¨HClÈÜÒº£¬µÎ¶¨ÖÁ»¯Ñ§¼ÆÁ¿µãʱÏûºÄHClÈÜÒº25.20mL£¬ÇóHClÈÜҺŨ¶È¡£
9¡¢ÒÑÖªH2SO4ÖÊÁ¿·ÖÊýΪ96%£¬Ïà¶ÔÃܶÈΪ1.84g/mL£¬ÓûÅäÖÆ0.5L0.10mol/LµÄH2SO4ÈÜÒº£¬ÊÔ¼ÆËãÐèH2SO4¶àÉÙºÁÉý£¿
10¡¢³ÆÈ¡CaCO3ÊÔÑù0.2500g£¬ÈܽâÓÚ25.00mL0.2006mol/LµÄHClÈÜÒºÖУ¬¹ýÁ¿µÄHClÓÃ15.50mL0.2050mol/LµÄNaOHÈÜÒº½øÐзµµÎ¶¨£¬Çó´ËÊÔÑùÖÐCaCO3µÄÖÊÁ¿·ÖÊý¡£ 11¡¢Ó¦³ÆÈ¡¶àÉÙ¿ËÁÚ±½¶þ¼×ËáÇâ¼ØÒÔÅäÖÆ500mL0.1000mol/LµÄÈÜÒº£¿ÔÙ×¼È·ÒÆÈ¡ÉÏÊöÈÜÒº25.00mLÓÃÓڱ궨NaOHÈÜÒº£¬ÏûºÄNaOHÈÜÒºÌå»ý24.84mL£¬ÔòNaOHÈÜҺŨ¶ÈÊǶàÉÙ£¿ 12¡¢³ÆÈ¡0.4830gNa2B4O7 ¡¤10H2»ù×¼Î±ê¶¨H2SO4ÈÜÒºµÄŨ¶È£¬ÒÔ¼×»ùºì×÷ָʾ¼Á£¬ÏûºÄH2SO4ÈÜÒº20.84mL£¬Çóc(1/2 H2SO4)ºÍc(H2SO4)¡£
13¡¢·ÖÎö²»´¿µÄCaCO3£¨ÆäÖв»º¬¸ÉÈÅÎ£¬³ÆÈ¡ÊÔÑù0.3000g£¬¼ÓÈëŨ¶ÈΪ0.2500mol/LµÄHCl±ê×¼ÈÜÒº25.00mL£¬Öó·Ð³ýÈ¥CO2£¬ÓÃ0.2012mol/LµÄNaOHÈÜÒº·µµÎ¶¨¹ýÁ¿µÄHClÈÜÒº£¬ÏûºÄNaOHÈÜÒº5.84mL£¬¼ÆËãÊÔÑùÖÐCaCO3µÄÖÊÁ¿·ÖÊý¡£
14¡¢²â¶¨µª·ÊÖÐNH3µÄº¬Á¿¡£³ÆÈ¡ÊÔÑù1.6160g£¬ÈܽâºóÔÚ250mLÈÝÁ¿Æ¿Öж¨ÈÝ£¬ÒÆÈ¡25.00mL£¬¼ÓÈë¹ýÁ¿µÄNaOHÈÜÒº£¬½«²úÉúµÄNH3µ¼Èë40.00mLc(1/2H2SO4)Ϊ0.1020mol/LµÄH2SO4±ê×¼ÈÜÒºÎüÊÕ£¬Ê£ÓàµÄH2SO4Ðè17.00mL0.09600mol/LµÄNaOHÈÜÒºÖкͣ¬¼ÆË㵪·ÊÖÐNH3µÄÖÊÁ¿·ÖÊý¡£
15¡¢³ÆÈ¡´óÀíʯÊÔÑù0.2303gÈÜÓÚËáÖУ¬µ÷½ÚËá¶Èºó¼ÓÈë¹ýÁ¿µÄ(NH4)2C2O4ÈÜÒº£¬Ê¹¸Æ³ÁµíΪCaC2O4¡£¹ýÂË¡¢Ï´¾»£¬½«³ÁµíÈÜÓÚÏ¡H2SO4ÖС£ÈܽâºóÓÃc(1/5KMnO4)Ϊ0.2012mol/LµÄKMnO4±ê×¼ÈÜÒºµÎ¶¨£¬ÏûºÄ22.30mL£¬¼ÆËã´óÀíʯÖÐCaCO3µÄÖÊÁ¿·ÖÊý¡£
16¡¢ÓÃ0.2369gÎÞˮ̼ËáÄÆ±ê¶¨ÑÎËáÈÜÒºµÄŨ¶È£¬ÏûºÄ22.35mLÑÎËáÈÜÒº£¬ÊÔ¼ÆËã¸ÃÑÎËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È¡£
17¡¢ÖкÍ30.00mL NaOHÈÜÒº£¬ÓÃÈ¥38.40mL c(1/2 H2SO4)=0.1000mol/LµÄËáÈÜÒº£¬ÇóNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È¡£
18¡¢Á¿È¡0.1020mol/LµÄNaOH±ê×¼µÎ¶¨ÈÜÒº30.00mL£¬¼Ó50mLÎÞCO2µÄÕôÁóË®¼°2µÎ·Óָ̪ʾ¼Á£¬ÓÃÑÎËá±ê×¼µÎ¶¨ÈÜÒºµÎ¶¨£¬ÏûºÄÑÎËá30.52mL¡£Í¬Ê±Á¿È¡50mLÎÞCO2µÄÕôÁóË®×÷¿Õ°×ʵÑ飬ÏûºÄÑÎËá0.02mL£¬¼ÆËã¸ÃÑÎËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È¡£
2+
19£®³ÆÈ¡3.1015gÓÚ105¡«110¡æÏºæ¸ÉÖÁÖÊÁ¿ºã¶¨µÄ»ù×¼ÎïÁÚ±½¶þ¼×ËáÇâ¼Ø£¬ÈÜÓÚ80mlÎÞCO2µÄÕôÁóË®ÖУ¬¼Ó2µÎ·Óָ̪ʾ¼Á£¬ÓÃÅäÖÆºÃµÄÇâÑõ»¯ÄƱê×¼µÎ¶¨ÈÜÒºµÎ¶¨ÖÁ·ÛºìÉ«£¬ÏûºÄNaOHÈÜÒº30.40mL£¬Í¬Ê±×÷¿Õ°×ʵÑ飬ÏûºÄNaOHÈÜÒº0.01mL£¬ÊÔ¼ÆËãNaOH±ê×¼µÎ¶¨ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È¡£
20£®³ÆÈ¡»ù×¼Îï̼ËáÄÆ1.6098g£¬ÔÚ800¡æ×ÆÉÕ³ÉΪNa2CO3ºó£¬ÓÃË®ÈܽⲢϡÊÍ100.0mL£¬×¼È·ÎüÈ¡25.00mLÈÜÒº£¬ÒÔ¼×»ù³ÈΪָʾ¼Á£¬ÓÃHClÈÜÒºµÎ¶¨ÖÁÖÕµãʱÏûºÄ30.00mL£¬¼ÆËãHCl±ê×¼µÎ¶¨ÈÜÒºµÄŨ¶È¡£
21.ÊÐÊÛÑÎËáµÄÃܶÈΪ1.18g/mL£¬HClº¬Á¿Îª37%£¬ÓûÓôËÑÎËáÅäÖÆ500mL0.1mol/LµÄHClÈÜÒº£¬Ó¦Á¿È¡ÊÐÊÛÑÎËá¶àÉÙºÁÉý£¿
22.ÓÐÒ»NaOHÈÜÒº£¬ÆäŨ¶ÈΪ0.5450mol/L£¬È¡¸ÃÈÜÒº100.0mL£¬Ðè¼ÓË®¶àÉÙºÁÉý·½ÄÜÅä³É0.5000mol/LµÄÈÜÒº£¿
23. T(NaOH/ HCl) =0.003462 g/mL HClÈÜÒº£¬Ï൱ÓÚÎïÖʵÄÁ¿Å¨¶Èc£¨HCl£©Îª¶àÉÙ£¿ 24.³ÆÈ¡º¬ÂÁÊÔÑù0.2000gÈܽâºó¼ÓÈëc(EDTA)=0.02082mol/LµÄEDTA±ê×¼ÈÜÒº30.00mL¡£¿ØÖÆÌõ¼þʹAlÓëEDTAÅäλÍêÈ«£¬È»ºóÒÔc(Zn)=0.02000±ê×¼ÈÜÒº·µµÎ¶¨£¬ÏûºÄZn±ê×¼ÈÜÒº7.20mL¡£¼ÆËãAl2O3µÄÖÊÁ¿·ÖÊý¡£
25.º¬SÓлúÊÔÑù0.4710g£¬ÔÚÑõÆøÖÐȼÉÕ£¬Ê¹SÑõ»¯ÎªSO2£¬ÓÃÔ¤ÏÈÖк͹ýµÄH2O2½«SO2ÎüÊÕ£¬È«²¿×ª»¯ÎªH2SO4£¬ÒÔ (KOH)=0.1080mol/LµÄKOH±ê×¼µÎ¶¨ÈÜÒºµÎ¶¨ÖÁ»¯Ñ§¼ÆÁ¿µã£¬ÏûºÄ28.20mL¡£ÇóÊÔÑùÖÐSµÄº¬Á¿¡£
1. ½â£ºm = cVM = (1 ¡Á 500 ¡Á 10 ¡Á 40)g = 20g
2. ½â£ºc = m / MV = 4.18 /(105.99¡Á75.0¡Á10) = 0.5258mol/L 3. ½â£º·´Ó¦Îª Na2CO3 +2HCl === 2NaCl + H2CO3
c = 2m / MV =£¨2¡Á0.1580£©/£¨105.99¡Á24.80¡Á10£©= 0.1202mol/L 4. ½â£º·´Ó¦Îª H2C2O42H20 + 2NaOH === Na2C2O4 + 4H2O c = 2m / MV =£¨2¡Á0.3280£©/£¨126.07¡Á25.78¡Á10£©= 0.2018mol/L 5.½â£º·´Ó¦Îª Cr2O7 + 6Fe + 14H === 2Cr + 6Fe + 7H2O ¦Ø(Fe)= 0.1200¡Á28.62¡Á10¡Á55.85/0.3669 = 52.28%
¦Ø(Fe203)= 3¡Á0.1200¡Á28.62¡Á10¡Á159.69/(0.3669¡Á6) = 74.74% ¦Ø(Fe3O4)= 2¡Á0.1200¡Á28.62¡Á10¡Á231.54/(0.3669¡Á6) = 72.25% 6.¼ÆËãÏÂÁÐÈÜÒºµÄµÎ¶¨¶È£¬ÒÔg/mL±íʾ£º
£¨1£©½â£ºT Ba(OH)2/HCl = 0.2615¡Á10¡Á171.35/2 = 0.02240g/mL T Ca(OH)2/HCl = 0.2615¡Á10¡Á74.09/2 = 0.009687g/mL
£¨2£©½â£ºT H2SO4/NaOH = 0.1032¡Á10¡Á98.08/2 = 0.005061g/mL T CH3COOH/NaOH = 0.1032¡Á10¡Á60.05 = 0.006197g/mL
7. ½â£º·´Ó¦Îª 2MnO4 + 5C2O4 + 16H === 2Mn + 10CO2¡ü + 8H2O c =(2/5)m/MV =(2/5)¡Á0.2178/(134.00¡Á25.48¡Á10)=0.02552mol/L
-3
-2-+
2+
-3
-3
-3
-3
-3-3
-3
2-2+
+
3+
3+
-3-3
-3
-3
3+
2+
2+
8. ½â£ºc = 2m/MV = 2¡Á0.4709/(381.37¡Á25.20¡Á10)=0.09800mol/L 9. ½â£ºV = 0.5¡Á0.10¡Á98.08/(1.84¡Á96%) = 2.8mL 10. ½â£º·´Ó¦Îª CaCO3 + 2HCl === CaCl2 + H2CO3 HCl + NaOH === NaCl + H20
¦Ø(CaCO3)=(1/2)¡Á(0.2006¡Á25.00-0.2050¡Á15.50)¡Á10¡Á100.09/0.2500 = 36.78% 11. ½â£º£¨1£©m = cVM = 0.1000¡Á500¡Á10¡Á204.23 = 10.21g £¨2£©c = 0.1000¡Á25.00/24.84 = 0.1006mol/L
12. ½â£ºc(H2SO4) = 0.4830/(381.37¡Á20.84¡Á10) = 0.06077mol/L c(1/2H2SO4) = 2c(H2SO4) = 2¡Á0.06077 = 0.1215mol/L
13. ½â£º¦Ø(CaCO3)=(1/2)¡Á(0.2500¡Á25.00-0.2012¡Á5.84)¡Á10¡Á100.09/0.3000 = 84.66% 14.½â£º¦Ø(NH3)=(0.1020¡Á40.00-0.09600¡Á17.00)¡Á10¡Á17.03/[1.6160¡Á(25/250)]=25.80% 15.½â£º¦Ø(CaCO3)=[(1/2)¡Á0.2012¡Á22.30¡Á10¡Á100.09/0.2303 = 97.50% 16£®½â£º
-3
-3
-3
-3
-3
-3
-3
17£® ½â£º ¸ù¾Ý
18£®½â£º
19£®½â
20£®½â£º
21.½â:
ÍÆ³öV2=4.18mL
22.½â:C1V1=C2(V1+V), ÍÆ³öV=9.0mL
23.½â: C(HCl)=T NaOH / HCl ¡Á1000/M NaOH=0.08653 mol/L 24.½â:
25.½â:
·µ»Ø´ð°¸