ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿³£ÎÂÏ£¬ÓÃ0.100 mol¡¤L£­1 NaOHÈÜÒºµÎ¶¨10 mL 0.100 mol¡¤L£­1 H3PO4ÈÜÒº£¬ÇúÏßÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A.µÎ¶¨ÖÕµãa¿ÉÑ¡Ôñ·Ó̪×÷ָʾ¼Á

B.cµãÈÜÒºÖÐc(Na£«)>3c(PO43£­)£«2c(HPO42£­)£«c(H2PO4£­)

C.bµãÈÜÒºÖÐc(HPO42£­)>c(PO43£­)>c(H2PO4£­)

D.a¡¢b¡¢cÈýµãÖÐË®µÄµçÀë³Ì¶È×îСµÄÊÇc

¡¾´ð°¸¡¿B

¡¾½âÎö¡¿

NaOHµÎ¶¨Á×Ëá¹ý³ÌÖÐÒÀ´Î·¢Éú·´Ó¦£ºNaOH+H3PO4=NaH2PO4+H2O¡¢NaOH+ NaH2PO4=Na2HPO4+H2O¡¢NaOH+Na2HPO4=Na3PO4+H2O£¬ÒÀ´Î¶ÔÓ¦¼ÓÈë10mLNaOH¡¢20mLNaOH¡¢30mLNaOH¡£

A£®¾Ýͼ¿ÉÖªaµãµÄpH=4£¬·Ó̪µÄ±äɫΪ8.2~10£¬ËùÒÔÓ÷Ó̪×÷ָʾ¼Á²»ºÏÊÊ£¬¹ÊA´íÎó£»

B£®cµã¼ÓÈë30mLNaOH£¬ÈÜÒºÖеÄÈÜÖÊΪNa3PO4£¬¸ù¾ÝÎïÁÏÊغã¿ÉÖªc(Na£«)=3c(PO43£­)£«3c(HPO42£­)£«3c(H2PO4£­)+3c(H3PO4)£¬ËùÒÔc(Na£«)>3c(PO43£­)£«2c(HPO42£­)£«c(H2PO4£­)£¬¹ÊBÕýÈ·£»

C£®bµã¼ÓÈë20mLNaOH£¬ÈÜÒºÖеÄÈÜÖÊΪNa2HPO4£¬´ËʱÈÜÒºÏÔ¼îÐÔ£¬ËµÃ÷HPO42-µÄË®½â³Ì¶È´óÓÚÆäµçÀë³Ì¶È£¬ËùÒÔc(H2PO4£­)> c(PO43£­)£¬¹ÊC´íÎó£»

D£®Ëá»ò¼îµÄµçÀëÒÖÖÆË®µÄµçÀ룬ÑÎÀàµÄË®½â´Ù½øË®µÄµçÀ룬ÔÚcµã´¦ÈÜÖÊÖ»ÓÐNa3PO4£¬´ËʱֻÓÐÁ×Ëá¸ùµÄË®½â´Ù½øµçÀ룬ˮµÄµçÀë³Ì¶È×î´ó£¬¹ÊD´íÎó£»

¹Ê´ð°¸ÎªB¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ç⻯ï®(LiH)ÔÚ¸ÉÔïµÄ¿ÕÆøÖÐÄÜÎȶ¨´æÔÚ£¬ÓöË®»òËáÄܹ»ÒýÆðȼÉÕ¡£Ä³»î¶¯Ð¡×é×¼±¸Ê¹ÓÃÏÂÁÐ×°ÖÃÖƱ¸LiH¹ÌÌå¡£

¼×ͬѧµÄʵÑé·½°¸ÈçÏ£º

£¨1£©ÒÇÆ÷µÄ×é×°Á¬½Ó£ºÉÏÊöÒÇÆ÷×°ÖýӿڵÄÁ¬½Ó˳ÐòΪ___£¬¼ÓÈëÒ©Æ·Ç°Ê×ÏÈÒª½øÐеÄʵÑé²Ù×÷ÊÇ___(²»±Øд³ö¾ßÌåµÄ²Ù×÷·½·¨)£»ÆäÖÐ×°ÖÃBµÄ×÷ÓÃÊÇ___¡£

£¨2£©Ìí¼ÓÒ©Æ·£ºÓÃÄ÷×Ó´ÓÊÔ¼ÁÆ¿ÖÐÈ¡³öÒ»¶¨Á¿½ðÊôï®(¹ÌÌåʯÀ¯ÃÜ·â)£¬È»ºóÔÚ¼×±½£¨Ò»ÖÖÓлú»¯ºÏÎÖнþÏ´Êý´Î£¬¸Ã²Ù×÷µÄÄ¿µÄÊdzýȥﮱíÃæµÄʯÀ¯£¬È»ºó¿ìËÙ°Ñï®·ÅÈ뵽ʯӢ¹ÜÖС£Í¨ÈëÒ»¶Îʱ¼äÇâÆøºó¼ÓÈÈʯӢ¹Ü£¬ÔÚ¼ÓÈÈD´¦µÄʯӢ¹Ü֮ǰ£¬±ØÐë½øÐеÄʵÑé²Ù×÷ÊÇ___¡£

£¨3£©¼ÓÈÈÒ»¶Îʱ¼äºóÍ£Ö¹¼ÓÈÈ£¬¼ÌÐøͨÇâÆøÀäÈ´£¬È»ºóÈ¡³öLiH£¬×°È뵪·âµÄÆ¿À±£´æÓÚ°µ´¦¡£²ÉÈ¡ÉÏÊö²Ù×÷µÄÄ¿µÄÊÇΪÁ˱ÜÃâLiHÓë¿ÕÆøÖеÄË®ÕôÆø½Ó´¥¶ø·¢ÉúΣÏÕ¡£·ÖÎö¸Ã·´Ó¦Ô­Àí£¬Íê³ÉLiHÓëË®·´Ó¦µÄ»¯Ñ§·½³Ìʽ___¡£

£¨4£©×¼È·³ÆÁ¿ÖƵõIJúÆ·0.174g£¬ÔÚÒ»¶¨Ìõ¼þÏÂÓë×ãÁ¿Ë®·´Ó¦ºó£¬¹²ÊÕ¼¯µ½ÆøÌå470.4mL(ÒÑ»»Ëã³É±ê×¼×´¿ö)£¬Ôò²úÆ·ÖÐLiHÓëLiµÄÎïÖʵÄÁ¿Ö®±ÈΪ___¡£

¡¾ÌâÄ¿¡¿¸ß´¿MnCO3ÔÚµç×Ó¹¤ÒµÖÐÓÐÖØÒªµÄÓ¦Ó㬹¤ÒµÉÏÀûÓÃÈíÃÌ¿ó(Ö÷Òª³É·ÖÊÇMnO2£¬»¹º¬ÓÐFe2O3¡¢CaCO3¡¢CuOµÈÔÓÖÊ)ÖÆȡ̼ËáÃ̵ÄÁ÷³ÌÈçͼËùʾ£º

ÒÑÖª£º»¹Ô­±ºÉÕÖ÷·´Ó¦Îª2MnO2£«C2MnO£«CO2¡ü¡£

¿ÉÄÜÓõ½µÄÊý¾ÝÈçÏ£º

¸ù¾ÝÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÔÚʵÑéÊÒ½øÐв½ÖèA£¬»ìºÏÎïÓ¦·ÅÔÚ__________ÖмÓÈÈ£»²½ÖèCÖеÄÂËÔüΪ__________¡£

(2)²½ÖèDÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ__________¡£

(3)²½ÖèEÖе÷½ÚpHµÄ·¶Î§Îª____________£¬ÆäÄ¿µÄÊÇ______________________________¡£

(4)²½ÖèG£¬Î¶ȿØÖÆÔÚ35¡æÒÔϵÄÔ­ÒòÊÇ____________________________________£¬ÈôMn2+Ç¡ºÃ³ÁµíÍêȫʱ²âµÃÈÜÒºÖÐCO32-µÄŨ¶ÈΪ2.2¡Á10£­6mol/L£¬ÔòKsp(MnCO3)£½____________¡£

(5)Éú³ÉµÄMnCO3³ÁµíÐè¾­³ä·ÖÏ´µÓ£¬¼ìÑéÏ´µÓÊÇ·ñ¸É¾»µÄ·½·¨ÊÇ_____________________¡£

(6)ÏÖÓõζ¨·¨²â¶¨²úÆ·ÖÐÃÌÔªËصĺ¬Á¿¡£ÊµÑé²½Ö裺³ÆÈ¡3.300 gÊÔÑù£¬ÏòÆäÖмÓÈëÉÔ¹ýÁ¿µÄÁ×ËáºÍÏõËᣬ¼ÓÈÈʹ²úÆ·ÖÐMnCO3Íêȫת»¯Îª[Mn(PO4)2]3£­(ÆäÖÐNO3-Íêȫת»¯ÎªNO2£­)£»¼ÓÈëÉÔ¹ýÁ¿µÄÁòËá泥¬·¢Éú·´Ó¦NO2-£«NH4£«£½N2¡ü£«2H2OÒÔ³ýÈ¥NO2-£»¼ÓÈëÏ¡ÁòËáËữ£¬ÔÙ¼ÓÈë60.00 mL 0.500 mol¡¤L£­1ÁòËáÑÇÌúï§ÈÜÒº£¬·¢ÉúµÄ·´Ó¦Îª[Mn(PO4)2]3£­£«Fe2£«£½Mn2£«£«Fe3£«£«2PO43£­£»ÓÃ5.00 mL 0.500 mol¡¤L£­1ËáÐÔK2Cr2O7ÈÜҺǡºÃ³ýÈ¥¹ýÁ¿µÄFe2£«¡£

¢ÙËáÐÔK2Cr2O7ÈÜÒºÓëFe2+·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________________________¡£

¢ÚÊÔÑùÖÐÃÌÔªËصÄÖÊÁ¿·ÖÊýΪ____________¡£

¡¾ÌâÄ¿¡¿´ß»¯¼ÁÔÚÏÖ´ú»¯Ñ§¹¤ÒµÖÐÕ¼Óм«ÆäÖØÒªµÄµØ룬Ö÷ÒªÉæ¼°¹ý¶ÉÔªËؼ°Æ仯ºÏÎï¡¢¹è¡¢ÂÁ»¯ºÏÎïµÈ¡£Öйú¿Æѧ¼Ò´´ÔìÐԵع¹½¨Á˹軯Îᄃ¸ñÏÞÓòµÄµ¥ÌúÖÐÐÄ´ß»¯¼Á£¬³É¹¦µØʵÏÖÁ˼×ÍéÔÚÎÞÑõÌõ¼þÏÂÑ¡Ôñ»î»¯£¬Ò»²½¸ßЧÉú²úÒÒÏ©¡¢·¼ÌþºÍÇâÆøµÈ»¯Ñ§Æ·¡£

(1)¹è¡¢Ì¼Î»ÓÚͬһÖ÷×壬Óá°>¡±¡°<¡±»ò¡°£½¡±Ìî¿Õ£º

ÐÔÖÊ

Ô­×Ӱ뾶

µÚÒ»µçÀëÄÜ

ÈÛµã

¼üÄÜ

씀
¢ÙSi__________C

¢ÚC_________Si

¢ÛCO2______SiO2

¢ÜH-Si_______H-C

(2)CN£­ÄÜÓëFe3£«ÐγÉÅäºÏÎÓëCN£­»¥ÎªµÈµç×ÓÌåµÄ·Ö×ÓÓÐ_____________(ÈÎдһÖÖ)£»1mol[Fe(CN)6]3£­Öк¬____________mol ¦Ò¼ü¡£

(3)ÒÑÖª£º·´Ó¦2CH4CH2£½CH2£«2H2£¬Ì¼Ô­×ÓµÄÔÓ»¯ÀàÐÍת»¯¹ý³ÌΪ___________£»

(4)¾ÛËÄ·úÒÒÏ©ÊÇÒ»ÖÖ×¼¾§Ì壬׼¾§ÌåÊÇÒ»ÖÖÎÞƽÒÆÖÜÆÚÐò£¬µ«ÓÐÑϸñ×¼ÖÜÆÚλÖÃÐòµÄ¶ÀÌؾ§Ì壬¿Éͨ¹ý___________·½·¨Çø·Ö¾§Ìå¡¢×¼¾§ÌåºÍ·Ç¾§Ìå¡£

(5)»ù̬FÔ­×ӵļ۲ãµç×ÓÅŲ¼Í¼Îª___________¡£

(6)[H2F]£«[SbF6]£­(·úÌàËá)ÊÇÒ»ÖÖ³¬Ç¿Ëᣬ´æÔÚ[H2F]£«£¬¸ÃÀë×ӵĿռ乹ÐÍΪ____________¡£

(7)CuClµÄÈÛµãΪ426¡æ£¬ÈÛ»¯Ê±¼¸ºõ²»µ¼µç£»CuFµÄÈÛµãΪ908¡æ£¬ÃܶÈΪ7.1g¡¤cm£­3¡£

¢ÙCuFµÄÈÛµã±ÈCuClµÄ¸ß£¬Ô­ÒòÊÇ____________________________________________¡£

¢ÚÒÑÖªNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬CuFµÄ¾§°û½á¹¹ÈçͼËùʾ£¬ÔòCuFµÄ¾§°û²ÎÊýa£½__________nm(Áгö¼ÆËãʽ)¡£

¡¾ÌâÄ¿¡¿KAl(SO4)2¡¤12H2O£¨Ã÷·¯£©ÊÇÒ»ÖÖ¸´ÑΣ¬ÔÚÔìÖ½µÈ·½ÃæÓ¦Óù㷺¡£ÊµÑéÊÒÖУ¬²ÉÓ÷ÏÒ×À­¹Þ£¨Ö÷Òª³É·ÖΪAl£¬ÓÐÉÙÁ¿µÄFe¡¢MgÔÓÖÊ£©ÖƱ¸Ã÷·¯µÄ¹ý³ÌÈçͼËùʾ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)²Ù×÷1ÖÐÓõ½µÄ²£Á§ÒÇÆ÷ÓÐ________________¡£

(2)Ϊ¾¡Á¿ÉÙÒýÈëÔÓÖÊ£¬´ÓÒ×À­¹ÞÈܽâÖÁÉú³ÉAl(OH)3£¬¹ý³ÌÖУ¬Ö÷Òª·¢Éú·´Ó¦µÄÀë×Ó·´Ó¦·½³ÌʽΪ______________£¬__________________________£»ÊÔ¼Á¢ÚÊÇ_______________________________¡£

(3)ÒÑÖª£º³£ÎÂÏÂKW=1.0¡Á10£­14 £¬Al(OH)3ÈÜÓÚNaOHÈÜÒº·´Ó¦µÄƽºâ³£ÊýµÈÓÚ20¡£ÔòAl(OH)3+H2O[Al(OH)4]£­+H+ƽºâ³£ÊýK=_________________¡£

(4)ÌìȻˮÔÚ¾»»¯´¦Àí¹ý³ÌÖмÓÈëÃ÷·¯×÷»ìÄý¼Á£¬Ë®µÄ¾»»¯ºÍÈí»¯µÄÇø±ðÊÇ______________¡£

(5)ÆÕֽͨÕŵÄÖ÷Òª³É·ÖÊÇÏËάËØ£¬ÔÚÔçÆÚµÄÖ½ÕÅÉú²úÖУ¬³£²ÉÓÃÖ½±íÃæÍ¿·óÃ÷·¯µÄ¹¤ÒÕ£¬ÒÔÌî²¹Æä±íÃæµÄ΢¿×£¬·Àֹī¼£À©É¢¡£ÈËÃÇ·¢ÏÖÖ½ÕŻᷢÉúËáÐÔ¸¯Ê´¶ø±ä´à¡¢ÆÆËð£¬ÑÏÖØÍþвֽÖÊÎÄÎïµÄ±£´æ¡£¾­·ÖÎö¼ìÑ飬·¢ÏÖËáÐÔ¸¯Ê´Ö÷ÒªÓëÔìÖ½ÖÐÍ¿¸²Ã÷·¯µÄ¹¤ÒÕÓйأ¬Çë˵Ã÷ÀíÓÉ£º_____________________£»Îª±£»¤ÕâЩֽÖÊÎÄÎÓÐÈ˽¨Òé²ÉÈ¡ÅçÈ÷Zn(C2H5)2µÄ·½·¨£¬Æä¿ÉÒÔÓëË®·´Ó¦Éú³ÉÑõ»¯Ð¿ºÍÒÒÍé¡£Ó÷´Ó¦·½³Ìʽ±íʾ¸Ã·½·¨Éú³ÉÑõ»¯Ð¿¼°·ÀÖ¹ËáÐÔ¸¯Ê´µÄÔ­Àí________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø