15£®ÏÖÓÐ0.1mol•L-1µÄNa2SO4ºÍ0.1mol•L-1µÄH2SO4»ìºÏÈÜÒº100mL£¬ÏòÆäÖÐÖðµÎ¼ÓÈë0.2mol•L-1µÄ Ba£¨OH£©2ÈÜÒº£¬²¢²»¶Ï½Á°è£¬Ê¹·´Ó¦³ä·Ö½øÐУ®£¨ºöÂÔ»ìºÏ¹ý³ÌÖеÄÌå»ý±ä»¯£©
£¨1£©µ±¼ÓÈë50mLBa£¨OH£©2ÈÜҺʱ£¬ËùµÃÈÜÒºÖеÄÈÜÖÊÊÇNa2SO4£¬ÆäÎïÖʵÄÁ¿Å¨¶ÈΪ0.067 mol•
L-1£®
£¨2£©µ±ÈÜÒºÖгÁµíÁ¿´ïµ½×î´óʱ£¬Ëù¼ÓBa£¨OH£©2ÈÜÒºµÄÌå»ýΪ100mL£¬ËùµÃÈÜÒºÖÐÈÜÖÊΪNaOH£¬Ôò¸ÃÈÜÖÊÎïÖʵÄÁ¿Å¨¶ÈΪ0.1mol•L-1£®

·ÖÎö £¨1£©·´Ó¦Ï൱ÓÚBa£¨OH£©2ÏÈÓëH2SO4·´Ó¦£¬È»ºóÔÙNa2SO4Óë·´Ó¦£¬50mLBa£¨OH£©2ÈÜÒºÖÐn[Ba£¨OH£©2]=0.05L¡Á0.2moL•L-1=0.01mol£¬100mLÈÜÒºÖÐn£¨H2SO4£©=0.1L¡Á0.1moL•L-1=0.01mol£¬¹ÊÇâÑõ»¯±µÓëÁòËáÇ¡ºÃ·´Ó¦£¬ÁòËáÄƲ»·´Ó¦£¬¸ù¾Ýn=cV¼ÆËãÁòËáÄƵÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾Ým=nM¼ÆËãÁòËáÄƵÄÖÊÁ¿£»
£¨2£©ÈÜÒºÖгÁµíÁ¿´ïµ½×î´óʱ£¬·¢ÉúBa2++SO42-=BaSO4¡ý£¬¹Ên[Ba£¨OH£©2]=n£¨SO42-£©£¬ÔÙ¸ù¾ÝV=$\frac{n}{V}$¼ÆËãÇâÑõ»¯±µµÄÌå»ý£®ÈÜÒºÖÐÈÜÖÊΪNaOH£¬¸ù¾ÝÄÆÀë×ÓÊغã¿ÉÖªn£¨NaOH£©=n£¨Na+£©£¬ÔÙ¸ù¾Ýc=$\frac{n}{V}$¼ÆË㣮

½â´ð ½â£º£¨1£©·´Ó¦Ï൱ÓÚBa£¨OH£©2ÏÈÓëH2SO4·´Ó¦£¬È»ºóÔÙNa2SO4Óë·´Ó¦£¬100mLBa£¨OH£©2ÈÜÒºÖÐn[Ba£¨OH£©2]=0.1L¡Á0.2moL•L-1=0.02mol£¬100mLÈÜÒºÖÐn£¨H2SO4£©=0.1L¡Á0.2moL•L-1=0.02mol£¬¹ÊÇâÑõ»¯±µÓëÁòËáÇ¡ºÃ·´Ó¦£¬ÁòËáÄƲ»·´Ó¦£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2H++SO42-+Ba2++2OH-=Ba SO4¡ý+2 H2O£¬ÈÜÒºÖÐÈÜÖÊΪNa2SO4£¬Na2SO4µÄÎïÖʵÄÁ¿Îª0.1L¡Á0.1moL•L-1=0.01mol£¬Na2SO4µÄŨ¶È=$\frac{0.01mol}{0.15L}$=0.067molL£¬
¹Ê´ð°¸Îª£ºNa2SO4£»0.067£»
£¨2£©ÈÜÒºÖгÁµíÁ¿´ïµ½×î´óʱ£¬ÁòËá¸ùÍêÈ«·´Ó¦£¬·¢ÉúBa2++SO42-=BaSO4¡ý£¬¹Ên[Ba£¨OH£©2]=n£¨SO42-£©=0.01mol+0.01mol=0.02mol£¬¹ÊÇâÑõ»¯±µÈÜÒºµÄÌå»ýΪ$\frac{0.02mol}{0.2mol/L}$=0.1L=100mL£®
´ËʱÈÜÒºÖÐÈÜÖÊΪNaOH£¬¸ù¾ÝÄÆÀë×ÓÊغã¿ÉÖªn£¨NaOH£©=n£¨Na+£©=0.01mol¡Á2=0.02mol£¬¹ÊÈÜÒºÖÐNaOHŨ¶ÈΪ=$\frac{0.02mol}{0.2L}$=0.1mol/L£¬
¹Ê´ð°¸Îª£º100£»NaOH£¬0.1£®

µãÆÀ ±¾Ì⿼²é»ìºÏÎïµÄÓйؼÆËã¡¢³£Óû¯Ñ§¼ÆÁ¿µÄÓйؼÆËãµÈ£¬ÄѶÈÖеȣ¬Àí½â·¢Éú·´Ó¦µÄ±¾ÖʺÍÔªËØÊغãÊǽâÌâµÄ¹Ø¼ü£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìºþÄÏÊ¡³¤É³ÊиßÈýÉÏѧÆÚµÚ13´ÎÖܲâÀí¿Æ×ۺϻ¯Ñ§ÊÔ¾í ÌâÐÍ£ºÊµÑéÌâ

¼îʽ̼ËáîÜ£ÛCo4£¨OH£©,£¨CO3£©4£Ý³£ÓÃ×÷µç×Ó²ÄÁÏ¡¢´ÅÐÔ²ÄÁϵÄÌí¼Ó¼Á£¬ÄÑÈÜÓÚË®£¬ÊÜÈÈʱ¿É·Ö½âÉú³ÉÈýÖÖÑõ»¯ÎΪÁËÈ·¶¨Æä×é³É£¬Ä³»¯Ñ§ÐËȤС×éͬѧÉè¼ÆÁËÈçͼËùʾµÄ×°Ö㨲»ÍêÕû£©½øÐÐÊÔÑé¡£

ʵÑé²½ÖèÈçÏ£º

¢Ù³ÆÈ¡3.65gÑùÆ·ÖÃÓÚÓ²Öʲ£Á§¹ÜÄÚ£¬³ÆÁ¿ÒÒ¡¢±û×°ÖõÄÖÊÁ¿£»

¢Ú°´ÈçͼËùʾװÖÃ×é×°ºÃÒÇÆ÷£¬¡­¡­ £»

¢Û¼ÓÈÈÓ²Öʲ£Á§¹Ü£¬µ±ÒÒ×°ÖÃÖÐ ¡­¡­£¬Í£Ö¹¼ÓÈÈ£»

¢Ü´ò¿ª»îÈûa£¬»º»ºÍ¨Èë¿ÕÆøÊý·ÖÖӺ󣬳ÆÁ¿ÒÒ¡¢±û×°ÖõÄÖÊÁ¿£»

¢Ý¼ÆËã¡£

£¨1£©´ÓÏÂÁÐͼʾѡ³öºÏÀíµÄ×°ÖÃÌîÓÚ·½¿òÖУ¬Ê¹ÕûÌ×ʵÑé×°ÖÃÍêÕû£¨Ñ¡Ìî×ÖĸÐòºÅ£¬¿ÉÖظ´Ñ¡£©

¼×£º ÒÒ£º ±û£º

¼××°ÖõÄ×÷ÓÃÊÇ ¡£

£¨2£©²½Öè¢ÚÖÐÊ¡ÂÔµÄʵÑé²Ù×÷Ϊ £»

²½Öè¢ÛÖÐÒÒ×°ÖõÄÏÖÏóΪ £»

²½Öè¢ÜÖлº»ºÍ¨Èë¿ÕÆøÊý·ÖÖÓµÄÄ¿µÄÊÇ ¡£

£¨3£©Èô°´ÕýÈ·×°ÖýøÐÐʵÑ飬²âµÃÈçÏÂÊý¾Ý¡£

ÒÒ×°ÖõÄÖÊÁ¿/g

±û×°ÖõÄÖÊÁ¿/g

¼ÓÈÈÇ°

80.00

62.00

¼ÓÈȺó

80.36

62.88

Ôò¸Ã¼îʽ̼ËáîܵĻ¯Ñ§Ê½Îª_____________¡£

£¨4£©CO2ºÍSO2¾ùΪËáÐÔÆøÌ壬ÐÔÖÊÏàËÆ¡£ÎªÁ˱ȽÏÑÇÁòËáºÍ̼ËáµÄËáÐÔÇ¿Èõ£¬Ä³Í¬Ñ§ÓÃÈçÏÂ×°ÖýøÐÐʵÑé¡£

¢Ùд³ö¸ÃʵÑéÄܴﵽʵÑéÄ¿µÄµÄʵÑéÏÖÏó____________¡£

¢ÚÈô½«SO2ͨÈëË®ÖÐÖÁ±¥ºÍ£¬ÇëÉè¼ÆʵÑéÖ¤Ã÷ÑÇÁòËáÊÇÈõËᣬʵÑé·½°¸Îª____________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

6£®ÒÔʯÓÍÁѽâÆøXºÍÓлú»¯ºÏÎïBΪԭÁÏ¿ÉÒԺϳÉijÓлú¸ß¾ÛÎïJ£¬XÔÚ±ê×¼×´¿öϵÄÃܶÈΪ1.25g•L-1£¬X·´Ó¦Éú³ÉA£¬BµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª32£®ºÏ³É¾ÛºÏÎïJµÄת»¯¹ØϵÈçͼËùʾ£¨²¿·Ö·´Ó¦Ìõ¼þÂÔÈ¥£©£º

ÒÑÖª£º¢ñ¡¢R-CHO+R¡ä-CH2-CHO $¡ú_{¡÷}^{C_{2}H_{5}ONa}$
¢ò¡¢
×¢£ºR¡¢R¡äΪÌþ»ù»òÇâÔ­×Ó£¬R¡äΪÌþ»ù
£¨1£©EÖеĺ¬Ñõ¹ÙÄÜÍŵÄÃû³ÆÊÇ£ºÈ©»ù£»MµÄ½á¹¹¼òʽÊÇ£ºHOCH2CH2OH£»
£¨3£©Ð´³öAת»¯ÎªCµÄ»¯Ñ§·½³Ìʽ£º2CH3CH2OH+O2$¡ú_{¡÷}^{Cu}$2CH3CHO+2H2O£®
£¨3£©ÓйØÉÏÊöת»¯¹ØϵͼµÄÎïÖÊÖл¥ÎªÍ¬ÏµÎïµÄÊÇ¢Ù¢Ú
¢ÙAºÍB¢ÚCºÍD¢ÛEºÍF¢ÜHºÍI
£¨4£©Ð´³öI-JµÄ»¯Ñ§·´Ó¦·½³ÌʽnCH3CH=C£¨CH3£©COOH$\stackrel{Ò»¶¨Ìõ¼þÏÂ}{¡ú}$£®
£¨5£©½áºÏÌâÒ⣬д³öÒÔÌâÖеÄÓлúÎïCºÍDΪԭÁϺϳɻ¯ºÏÎïCH2=CHCOOHµÄÏß·Á÷³Ì£®£¨ºÏ³É·ÏßÁ÷³Ìͼʾ
£ºCH3CH2OH$¡ú_{170¡æ}^{ŨÁòËá}$CH2=CH2$\stackrel{Br_{2}}{¡ú}$BrCH2CH2Br£©CH3CHO+HCHO$¡ú_{¡÷}^{C_{2}H_{5}Na}$CH2=CHCHO$¡ú_{¡÷}^{O_{2}/´ß»¯¼Á}$CH2=CHCOOH£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

3£®£¨1£©Ò»¶¨ÖÊÁ¿µÄÌúÓ뺬1molHNO3µÄÏ¡ÏõËáÇ¡ºÃÍêÈ«·´Ó¦£¬Ï¡ÏõËáµÄ»¹Ô­²úÎïÊÇNO£®
¢ÙÉú³ÉµÄNOÔÚ±ê¿öϵÄÌå»ýÊÇ£º5.6 L£®
¢ÚÉèÒ»¶¨ÖÊÁ¿µÄÌúΪmg£¬ÔòmµÄ·¶Î§ÊÇ£º14¡Üm¡Ü21£®
¢Û½«·´Ó¦½áÊøºóµÄÈÜÒº¼ÓˮϡÊ͵½500mL£¬¸ÃÈÜÒºÖÐc£¨NO3-£©=1.5mol/L£®
£¨2£©ÎªÁË̽¾¿Ä³Cu£¬CuO£¬Cu2O»ìºÏÎïµÄ×é³ÉÇé¿ö£¬Ïò»ìºÏÎïÖмÓÈë100mL 0.6mol/LµÄÏõËáÈÜÒº£¬Ç¡ºÃʹÆäÈܽ⣬ͬʱÊÕ¼¯µ½224mLNO£¨±ê¿ö£©
¢Ùд³öCu2OÓëÏ¡ÏõËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º3Cu2O+14HNO3=6Cu£¨NO3£©2+2NO¡ü+7H2O
¢Ú²úÎïÖÐCu£¨NO3£©2µÄÎïÖʵÄÁ¿£º0.025mol
¢ÛÈô»ìºÏÎïÖк¬Í­0.01mol£¬Ôòn£¨Cu2O£©=0.005 mol£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

10£®ÔìÖ½¹¤ÒµÖг£ÓÃCl2Ư°×Ö½½¬£¬Æ¯°×ºóµÄÖ½½¬ÓÃNaHSO3³ýÈ¥²ÐÁôµÄCl2£¬Æ䷴ӦΪ£ºCl2+NaHSO3+H2O=NaCl+H2SO4+HCl£¬ÔÚÕâ¸ö·´Ó¦ÖУ¬Ñõ»¯²úÎïÓ뻹ԭ²úÎïµÄÎïÖʵÄÁ¿Ö®±ÈΪ£¨¡¡¡¡£©
A£®2£º3B£®3£º1C£®2£º1D£®1£º2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

20£®ÔËÓÃÏà¹Ø»¯Ñ§ÖªÊ¶½øÐÐÅжϣ¬ÏÂÁнáÂÛ´íÎóµÄÊÇ£¨¡¡¡¡£©
A£®Ôö´ó·´Ó¦ÎïŨ¶È¿É¼Ó¿ì·´Ó¦ËÙÂÊ£¬Òò´ËÓÃŨÁòËáÓëÌú·´Ó¦ÄÜÔö´óÉú³ÉH2µÄËÙÂÊ
B£®NH4FË®ÈÜÒºÖк¬ÓÐHF£¬Òò´ËNH4FÈÜÒº²»ÄÜ´æ·ÅÓÚ²£Á§ÊÔ¼ÁÆ¿ÖÐ
C£®¿Éȼ±ùÖ÷ÒªÊǼ×ÍéÓëË®ÔÚµÍθßѹÏÂÐγɵÄË®ºÏÎᄃÌ壬Òò´Ë¿É´æÔÚÓÚº£µ×
D£®Ä³ÎüÈÈ·´Ó¦ÄÜ×Ô·¢½øÐУ¬Òò´Ë¸Ã·´Ó¦ÊÇìØÔö·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

7£®¸ù¾ÝµçÀëƽºâ³£Êý£¨ÓÃKa±íʾ£©µÄ´óС¿ÉÒÔÅжϵç½âÖʵÄÏà¶ÔÇ¿Èõ£®25¡æʱ£¬ÓйØÎïÖʵĵçÀëƽºâ³£ÊýÈç±íËùʾ£º
ÎïÖÊHFH2CO3HClO
µçÀëƽºâ³£Êý£¨Ka£©7.2¡Á10-4Kal=4.4¡Á10-7
Ka2=4.7¡Á10-11
3.0¡Á10-8
£¨1£©ÒÑÖª25¡æʱ£¬¢ÙHF£¨aq£©+OH-£¨aq£©¨TF-£¨aq£©+H2 O£¨l£©¡÷H=-67.7kJ•
mol-1¡¡¢ÚH+£¨aq£©+OH-£¨aq£©¨TH2O£¨l£©¡÷H=-57.3kJ•mol-1
Çâ·úËáµÄµçÀë·½³Ìʽ¼°ÈÈЧӦ¿É±íʾΪHF£¨aq£©?H+£¨aq£©+F-£¨aq£©¡÷H=-10.4KJ•mol-1£®
£¨2£©½«Å¨¶ÈΪ0.1mol•L-1µÄHFÈÜÒº¼ÓˮϡÊÍÒ»±¶£¨¼ÙÉèζȲ»±ä£©£¬ÏÂÁи÷ÏîÖеÄÖµ½«Ôö´óµÄÊÇCD£®
A£®c£¨H+£©     B£®c£¨H+£©•c£¨OH-£©   C.$\frac{c£¨H+£©}{c£¨HF£©}$       D.$\frac{c£¨OH-£©}{c£¨H+£©}$
£¨3£©ÏÂÁз½·¨ÖУ¬¿ÉÒÔʹ0.10mol•L-1 HFÈÜÒºÖÐHFµçÀë³Ì¶ÈÔö´óµÄÊÇd£®£¨ÇëÌî×ÖĸÇÒ×¢ÒâÇø·Ö´óСд£¬´óСдÌî´í²»µÃ·Ö£©
a£®Éý¸ßζȠ            
b£®ÏòÈÜÒºÖеÎÈë2µÎŨÑÎËá
c£®¼ÓÈëÉÙÁ¿NaF¹ÌÌå     
d£®¼ÓˮϡÊÍ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

4£®¢ÙMgCl2  ¢Ú¸É±ù ¢Û±ù  ¢ÜNa2O2¢Ý°×Á×£¨P4£© ¢ÞÁò»Ç¢ßNa2CO3  ¢à½ð¸Õʯ    ¢áNaOH ¢âSiO2
£¨1£©ÒÔÉÏÎïÖÊÖÐÊôÓÚÀë×Ó¾§ÌåµÄÊǢ٢ܢߢ᣻
£¨2£©ÊôÓÚ·Ö×Ó¾§ÌåµÄÊǢڢۢݢޣ®
£¨3£©ÊôÓÚÔ­×Ó¾§ÌåµÄÊÇ¢à¢â£¨ÌîÐòºÅ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

2£®ÔÚ¿ÕÆøÖзÅÖÃÒ»¶Îʱ¼äµÄKOH¾­·ÖÎöÖªÆäÖк¬Ë®a%£¬º¬K2CO3b%£¬ÆäÓàΪKOH£®È¡´ËÑùÆ·mgÈÜÓÚ100mLŨ¶ÈΪ1mol/LµÄH2SO4ÖУ¬ËùµÃÈÜÒºÖÐÊ£ÓàµÄH2SO4Ðè¼ÓngKOH²ÅÄÜÍêÈ«Öкͣ¬È»ºó½«ÈÜÒºÕô¸É£¬¿ÉµÃ¹ÌÌåÎïÖÊ£¨²»º¬½á¾§Ë®£©µÄÖÊÁ¿Îª£¨¡¡¡¡£©
A£®3.1£¨m+n£©gB£®14.2gC£®17.4gD£®20.6g

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸