如图表示配制100mL0.100mol′L-1Na2CO3溶液几个关键实验步骤和操作,据图回答下列问题: (1)步骤E中将

\u5982\u4f55\u914d\u5236100mL0.100mol/L\u7684SnCl2\u6eb6\u6db2\uff1f

\u89e3\uff1a \u79f0\u53d61.896 g\u7684SnCl2\u56fa\u4f53\uff0c\u6eb6\u4e8e\u9002\u91cf\u7684 \u6d53\u76d0\u9178 \u4e2d\uff0c\u5c06\u6eb6\u6db2\u79fb\u5165100mL\u7684 \u5bb9\u91cf\u74f6 \u4e2d\uff0c\u7136\u540e\u7528\u5c11\u91cf\u7684 \u84b8\u998f\u6c34\u6d17\u6da4 \u70e7\u676f2~3\u6b21\uff0c\u5e76\u5c06 \u6d17\u6da4\u6db2 \u4e5f\u6ce8\u5165100mL\u5bb9\u91cf\u74f6\u4e2d\uff0c\u52a0\u6c34\u81f3\u79bb\u523b\u5ea6\u7ebf2~3cm\u5904\u5e76\u6447\u5300\uff0c\u518d\u6362\u7528 \u80f6\u5934\u6ef4\u7ba1 \u52a0\u6c34\u81f3\u523b\u5ea6\u7ebf\uff0c\u5373\u53ef\u914d\u6210100mL 0.100mol/L\u7684SnCl2\u6eb6\u6db2\uff0c\u6700\u540e\u79fb\u81f3\u8bd5\u5242\u74f6\u4e2d\uff0c\u5e76\u52a0\u5165\u51e0\u7c92\u950c\u7c92\u4ee5\u9632\u6b62\u5176\u6c27\u5316\u3002(\u6700\u540e\u8d34\u4e86\u6807\u7b7e\uff0c\u786e\u5b9a\u4e0b\u6d53\u5ea6\u4ee5\u5907\u4e0b\u6b21\u53d6\u7528\u65f6\u77e5\u9053\u662f\u4ec0\u4e48\u7269\u8d28\u4ee5\u53ca\u6d53\u5ea6\u7684\u5927\u5c0f)

0.100mol/l\u7684NaOH\u6eb6\u6db2500ml\u4e2d\u542b\u706b\u78b1\u7684\u8d28\u91cf\u662f0.1\u00d70.5\u00d740=2\u514b
0.05000mol/l\u7684NaOH\u6eb6\u6db2400ml\u4e2d\u542b\u706b\u78b1\u7684\u8d28\u91cf\u662f0.05\u00d70.4\u00d740=0.8\u514b
\u5373\u57280.05000mol/l\u7684NaOH\u6eb6\u6db2400ML\u4e2d\u52a01.2\u514b\u706b\u78b1\uff0c\u5145\u5206\u6405\u62cc\u540e\uff0c\u518d\u52a0\u6c34\u81f3500ml\u5373\u53ef\u3002

(1)配制一定物质的量浓度溶液的仪器是容量瓶,所以该仪器的名称是容量瓶,故答案为:容量瓶;
(2)加水至液面距离刻度线1~2cm时,用胶头滴管向容量瓶中滴加溶液,该操作的名称是定容,故答案为:定容;
(3)操作步骤有计算、称量、溶解、移液、洗涤移液、定容、摇匀等操作,所以实验过程先后次序排列为:DCBFAE,故答案为:DCBFAE;
(4)根据C=
n
V
知,定容时仰视刻度线,导致溶液偏大,则配制溶液的浓度偏小,所以所配制溶液浓度 小于0.1mol/L,故答案为:小于;
(5) C=
1 0 3 ρW
M
mol/L=
1 0 3 ×1.84×98%
98
mol/L
=18.4mol/L,
设所需浓硫酸的体积为V,则有V×18.4mol/L=O.1L×2.3mol/L,v=0.0125L=12.5mL,选取的量筒应等于或稍大于量取溶液的体积,所以应选25mL量筒,故答案为:12.5,25.


  • 瑕閰嶅埗0.10鐨凥CL婧舵恫1000ml闇瑕佸瘑搴︿负1.19g/l璐ㄩ噺鍒嗘暟0.37鐨勬祿鐩愰吀澶...
    绛旓細鍥炵瓟锛氫竴涓2浠1G鍐呭瓨鏉¤100鍑哄ご,涓嶈繃涓嶈兘涓嶅湪浣犵殑鐢佃剳涓,寤鸿鍗囩骇涓涓嬫暣涓钩鍙,鐜板湪缁勪釜2浠i厤缃篃涓嶅お璐,缁欎綘涓浜閰嶅埗鍗曠湅鐪 900--1000鍏冪殑瀹跺涵閰嶇疆: CPU: AMD Sempron LE-1200(鐩)-165 涓绘澘: 鏄犳嘲 MCP6P-M2-295 鏄傝揪C61-299 鍐呭瓨: 濞佸垰1G DDR2 800(涓囩传鍗冪孩)-90 閲戝+椤1G DDR鈪 800-90...
  • 閰嶅埗pH=7.21鐨勭紦鍐叉憾娑500ml, 0.10MOL.NAH2PO4鍜0.20MOL.LNAOH鍚勫灏ML...
    绛旓細2-]/[H2PO4 -]) 瀵规暟鐨勮繍绠楋紝涓嶈В閲 瑕佹眰锛歱H=7.21=pKa2 鍒檒g([HPO4 2-]/[H2PO4 -])=0锛沎HPO4 2-]/[H2PO4 -]=1 涔熷氨鏄姹傚姞鍏ョ⒈鍚庯紝[HPO4 2-]涓嶽H2PO4 -]鐨勬祿搴︾浉绛夈傝嚜宸辫В涓涓嬫柟绋嬶紝涓嶅垪浜 绛旀锛400ml, 0.10mol/L.NaH2PO4锛100ml 0.20mol/L.NaOH ...
  • 鏌愬悓瀛︽閰嶅埗95mL 0.10mol/L CuSO4婧舵恫.(1)搴旂О鍙栬儐鐭炬櫠浣撶殑璐ㄩ噺涓...
    绛旓細锛1锛夊疄楠屽娌℃湁90mL鐨勫閲忕摱锛屽簲璇ラ夌敤100mL瀹归噺鐡讹紝閰嶅埗100mL 0.10mol/L鐨勭~閰搁摐婧舵恫锛岄渶瑕佺~閰搁摐鐨勭墿璐ㄧ殑閲忎负锛歯锛圕uSO4锛=0.10mol/L脳0.1L=0.01mol锛岄渶瑕佷簲姘寸~閰搁摐鐨勮川閲忎负锛歮锛圕uSO4?5H2O锛=250g/mol脳0.01mol=2.5g锛屾晠绛旀涓猴細2.5锛涳紙2锛夊疄楠屽娌℃湁90mL鐨勫閲忕摱锛屽簲璇ラ夌敤100...
  • ...鑳借揪鍒板疄楠岀洰鐨勬槸( )A.绉伴噺姘㈡哀鍖栭挔鍥轰綋B.閰嶅埗100mL 0.
    绛旓細A锛庣О閲忚嵂鍝佹椂锛屽簲宸︾墿鍙崇爜锛屾阿姘у寲閽犳湁鑵愯殌鎬с佹槗娼В锛屽簲鏀惧湪灏忕儳鏉腑绉伴噺锛屾晠A閿欒锛汢锛庡悜瀹归噺鐡堕噷杞Щ婧舵恫鏃跺簲璇ョ敤鐜荤拑妫掑紩娴侊紝闃叉娑蹭綋娴佸埌瀹归噺鐡跺闈紝鏁匓閿欒锛汣锛庣敤鑷潵姘村埗澶囧皯閲忚捀棣忔按锛岀敤钂搁鐑х摱锛屾俯搴﹁浣嶇疆锛屽喎鍗存按鐨勬祦鍚戝潎姝g‘锛屾晠C姝g‘锛汥锛庨厭绮惧拰姘翠簰婧朵笉鍒嗗眰锛屾晠涓嶈兘鐢ㄥ垎娑茬殑...
  • 鐢0.10mol/LHAc婧舵恫鍜0.10mol/LNaAc婧舵恫閰嶅埗PH=4.95鐨勭紦鍐叉憾娑100m
    绛旓細缂撳啿婧舵恫涓細pH=pKa + lg c锛圓c-锛/c锛圚Ac锛塴g c锛圓c-锛/c锛圚Ac锛=4.95-4.74=0.21 c锛圓c-锛/c锛圚Ac锛=1.32 闇瑕侀唻閰哥殑浣撶Н锛100/2.23=43.1ml 闇瑕侀唻閰搁挔鐨勪綋绉細100-43.1=56.9ml
  • 100mL0.10mol.L¯¹ AgNO3婧舵恫鍜100mL0.30mol.L¯¹NaCN婧舵恫娣 ...
    绛旓細AgCN鐨刱sp涓2.2*10^-12 100mL0.10mol.L¯¹ AgNO3婧舵恫鍜100mL0.30mol.L¯¹NaCN婧舵恫娣峰悎锛屽拷鐣ユ憾娑蹭綋绉殑鏀瑰彉锛屽垯寰楀埌200ml婧舵恫锛岀敓鎴0.01molAgCN锛屾憾娑蹭腑鍓╀綑0.1mol/LCN-.鎵浠g+娴撳害涓簁sp/[CN-]=2.2*10^-11mol/L ...
  • ...鐨勬祿搴﹀湪瀹為獙瀹や腑杩涜瀹為獙銆傝瀹屾垚涓嬪垪濉┖:(1)閰嶅埗100
    绛旓細锛1锛夆憼杞Щ锛涘畾瀹癸紱鎽囧寑锛涒憽0.4锛涚儳鏉(鎴栬〃闈㈢毧)锛涜嵂鍖欙紙2锛夆憼鏈鍚庝竴婊碞aOH婧舵恫鍔犲叆锛屾憾娑茬敱鏃犺壊鎭板ソ鍙樻垚娴呯矇绾㈣壊锛屼笖鍗婂垎閽熷唴涓嶈お鑹诧紱鈶0.11 mol路L -1 锛涒憿涓欙紱鈶
  • 鐢ㄥ凡鍑嗙‘绉伴噺鐨1.06g Na2
    绛旓細閰嶅埗80mL纰抽吀閽犳憾娑诧紝瀹為獙瀹ゆ病鏈80mL瀹归噺鐡讹紝闇瑕侀夌敤100mL瀹归噺鐡堕厤鍒讹紝瀹為檯涓婇厤鍒剁殑鏄100mL 0.10mol/L鐨勭⒊閰搁挔婧舵恫锛屾牴鎹厤鍒舵楠ゅ彲鐭ワ紝闇瑕佷娇鐢ㄧ殑浠櫒涓猴細鐑ф澂銆佺幓鐠冩銆侀噺绛掋100mL瀹归噺鐡躲佽兌澶存淮绠$瓑锛岃繕缂哄皯鐨勪华鍣ㄤ负锛氱幓鐠冩銆100mL瀹归噺鐡躲佽兌澶存淮绠★紝鏁呯瓟妗堜负锛氱幓鐠冩銆100mL瀹归噺鐡躲佽兌澶存淮绠...
  • 鈪.鎬ラ渶95mL 0.1mol/L鐨凬a2CO3婧舵恫,鏍规嵁瀹為獙瀹や华鍣ㄨ姹閰嶅埗Na2CO3婧舵恫...
    绛旓細鈪狅紟锛1锛夊疄楠屽娌℃湁95mL瀹归噺鐡讹紝闇瑕侀夌敤100mL瀹归噺鐡讹紝瀹為檯涓婇厤鍒剁殑婧舵恫涓100mL 0.1mol/L鐨勭⒊閰搁挔婧舵恫锛岄渶纰抽吀閽犵殑璐ㄩ噺涓簃=0.1L脳0.1mol?L-1脳106g/mol=1.06g锛屾墭鐩樺ぉ骞冲彧鑳借鍒0.1g锛屽垯闇瑕佺О閲1.1g纰抽吀閽狅紱閰嶅埗100ml 0.10mol?L-1鐨凬a2CO3婧舵恫锛岄渶瑕100mL瀹归噺鐡讹紱鎿嶄綔姝ラ鏈夎绠椼...
  • 娆蹭娇100mL0.10mol/l鐨凥Cl婧舵恫鐨刾H浠1澧炲姞鍒4.44,闇瑕佸姞鍏ュ浐浣揘aAc澶氬皯鍏...
    绛旓細HAc=H+Ac Ka=1.75*10鐨-5鏂 鍗砙H]*[Ac]/[HAc]=1.75*10鐨-5娆℃柟 [H]=10鐨勶紙-4.44锛夋鏂 [HAc]=0.1 鎵浠Ac]=1.75*10鐨勶紙-5锛夋鏂*0.1/10鐨勶紙-4.44锛夋鏂=0.0482mol/L 鍏盢aAc0.1482mol*0.1=82*0.1482*0.1=8.60鍏 ...
  • 扩展阅读:配制0.1mol每l氢氧化钠 ... 配制0.1moll草酸溶液100ml ... 1比20兑水配比 ... 0.9%的氯化钠溶液100ml ... 0.1%甲基橙的配置方法 ... 0.1 0.2肥皂水怎么配 ... 配制100ml0.1mol的氯化钠 ... 100ml稀盐酸怎么配制 ... 800-1000倍液怎么兑水 ...

    本站交流只代表网友个人观点,与本站立场无关
    欢迎反馈与建议,请联系电邮
    2024© 车视网