ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿³£ÎÂÏÂÏÂÁÐÈÜÒºÖеÄÖ¸¶¨Àë×ÓÓë100 mL 3.9 mol/L Ba(NO3)2ÈÜÒºËùº¬NO3£­ÎïÖʵÄÁ¿Å¨¶ÈÏàͬµÄÊÇ

A.390 mL 0.1 mol/L MgCl2ÈÜÒºÖеÄCl£­

B.200 g Ũ¶ÈΪ26%£¬ÃܶÈΪ1.2g/mLNaOHÈÜÒºÖеÄOH£­

C.50 mL 7.8 mol/L Al2(SO4)3ÈÜÒºÖеÄSO42£­

D.200 mL 3.9 mol/L CaCl2ÈÜÒºÖеÄCa2£«

¡¾´ð°¸¡¿B

¡¾½âÎö¡¿

3.9 mol/L Ba(NO3)2ÈÜÒºËùº¬NO3£­ÎïÖʵÄÁ¿Å¨¶ÈΪ3.9 mol/L¡Á2=7.8 mol/L¡£

A. 390 mL 0.1 mol/L MgCl2ÈÜÒºÖеÄc(Cl£­)=0.1 mol/L¡Á2=0.2 mol/L£¬¹Ê²»Ñ¡A£»

B. 200 g Ũ¶ÈΪ26%£¬ÃܶÈΪ1.2g/mLNaOHÈÜÒºÖÐc(OH£­)= c(NaOH)= 7.8 mol/L£¬¹ÊÑ¡B£»

C. 50 mL 7.8 mol/L Al2(SO4)3ÈÜÒºÖÐc(SO42£­)= 7.8 mol/L¡Á3=23.4 mol/L£¬¹Ê²»Ñ¡C£»

D. 200 mL 3.9 mol/L CaCl2ÈÜÒºÖеÄc(Ca2£«)= 3.9 mol/L¡Á1=3.9 mol/L£¬¹Ê²»Ñ¡D¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿¼×´¼£¨CH3OH£©ÊÇÖØÒªµÄÈܼÁºÍÌæ´úȼÁÏ£¬¹¤ÒµÉÏÓÃCOºÍH2ÔÚÒ»¶¨Ìõ¼þÏÂÖƱ¸CH3OHµÄ·´Ó¦ÎªCO(g)+2H2(g) CH3OH(g)£» ¡÷H¡£

£¨1£©ÔÚÌå»ýΪ1LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬³äÈë2molCOºÍ4molH2£¬Ò»¶¨Ìõ¼þÏ·¢ÉúÉÏÊö·´Ó¦£¬²âµÃCO(g)ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼äµÄ±ä»¯Èçͼ¼×Ëùʾ¡£

¢Ù´Ó·´Ó¦¿ªÊ¼µ½5min£¬ÓÃÇâÆø±íʾµÄƽ¾ù·´Ó¦ËÙÂÊv(H2)=________¡£

¢ÚÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________£¨ÌîÐòºÅ£©¡£

A. ´ïµ½Æ½ºâʱ£¬H2µÄת»¯ÂÊΪ75%

B. 5minºóÈÝÆ÷ÖÐѹǿ²»Ôٸıä

C. ´ïµ½Æ½ºâºó£¬ÔÙ³äÈëë²Æø£¬·´Ó¦ËÙÂÊÔö´ó

D. 2minÇ°v(Õý)£¾v(Äæ)£¬2minºóv(Õý)£¼v(Äæ)

£¨2£©Ä³Î¶ÈÏ£¬ÔÚÒ»ºãѹÈÝÆ÷Öзֱð³äÈë1.2molCOºÍ1molH2£¬´ïµ½Æ½ºâʱÈÝÆ÷Ìå»ýΪ2L£¬ÇÒº¬ÓÐ0.4molCH3OH(g)£¬Ôò¸Ã·´Ó¦Æ½ºâ³£ÊýµÄֵΪ__________¡£

£¨3£©¼×´¼ÊÇÒ»ÖÖÐÂÐ͵ÄÆû³µ¶¯Á¦È¼ÁÏ¡£ÒÑÖªH2(g)¡¢CO(g)¡¢CH3OH(l)µÄȼÉÕÈÈ·Ö±ðΪ285.8kJ/mol¡¢283.0kJ/molºÍ726.5kJ/mol£¬Ôò¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽΪ________¡£

£¨4£©ÏÖÓÐÈÝ»ý¾ùΪ1LµÄa¡¢b¡¢cÈý¸öÃܱÕÈÝÆ÷£¬ÍùÆäÖзֱð³äÈë1molCOºÍ2molH2µÄ»ìºÏÆøÌ壬¿ØÖÆζȣ¬½øÐз´Ó¦£¬²âµÃÏà¹ØÊý¾ÝµÄ¹ØϵÈçͼËùʾ¡£bÖм״¼Ìå»ý·ÖÊý´óÓÚaÖеÄÔ­ÒòÊÇ____________¡£´ïµ½Æ½ºâʱ£¬a¡¢b¡¢cÖÐCOµÄת»¯ÂÊ´óС¹ØϵΪ___________¡£

£¨5£©¼×´¼×÷ΪһÖÖȼÁÏ»¹¿ÉÓÃÓÚȼÁϵç³Ø¡£ÔÚζÈΪ650¡æµÄÈÛÈÚÑÎȼÁϵç³ØÖÐÓü״¼¡¢¿ÕÆøÓëCO2µÄ»ìºÏÆøÌå×÷·´Ó¦ÎÄø×÷µç¼«£¬ÓÃLi2CO3ºÍNa2CO3»ìºÏÎï×÷µç½âÖÊ¡£¸Ãµç³ØµÄ¸º¼«·´Ó¦Ê½Îª___________________________¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø