在标准状况下,224mL某气体的质量为0.64g,求该气体的相对分子质量

在标准状况下224mL某气体的物质的量为0.01mol
该气体的摩尔质量M=0.64/0.01=64g/mol
该气体的相对分子质量为64

鍦ㄦ爣鍑嗙姸鍐典笅,224mL鏌愭皵浣鐨勮川閲忎负0.64g,姹傝姘斾綋鐨勭浉瀵瑰垎瀛愯川閲廮鐧惧害...
绛旓細鍦ㄦ爣鍑嗙姸鍐典笅224mL鏌愭皵浣鐨勭墿璐ㄧ殑閲忎负0.01mol 璇ユ皵浣撶殑鎽╁皵璐ㄩ噺M=0.64/0.01=64g/mol 璇ユ皵浣撶殑鐩稿鍒嗗瓙璐ㄩ噺涓64

224mL鏌愭皵浣撳湪鏍囧噯鐘跺喌涓鐨勮川閲忎负0.32g,璇ユ皵浣撶殑鎽╁皵璐ㄩ噺绾︿负? A.32g...
绛旓細A 鎽╁皵璐ㄩ噺鐨勫崟浣嶄负g路mol-1锛岀浉瀵瑰垎瀛愯川閲忕殑鍗曚綅涓1.鏍囧噯鐘跺喌涓嬶紝224mL姘斾綋涓0.01mol锛岃姘斾綋鐨勬懇灏旇川閲忎负 =32g路mol-1鏁呴堿銆傛湰棰樿娉ㄦ剰鍖哄垎鎽╁皵璐ㄩ噺涓庣浉瀵瑰垎瀛愯川閲忕殑鍗曚綅銆

鍦ㄦ爣鍑嗙姸鍐典笅,224mL鏌愭皵浣鐨勮川閲忎负0.64g,姹傝姘斾綋鐨勭浉瀵瑰垎瀛愯川閲廮鐧惧害...
绛旓細鏍囧噯鐘舵佷笅涓鎽╁皵浠讳綍姘斾綋鐨勪綋绉兘鏄22.4L 锛屾墍浠224mL姘斾綋鐨勬懇灏旀暟涓 0.224/22.4=0.01mol 銆傚姝ゅ彲浠ユ眰鍑烘懇灏旇川閲=0.64/0.01=64锛屾墍浠ヨ姘斾綋鐩稿鍒嗗瓙璐ㄩ噺涓64

224ML鏌愭皵浣撳湪鏍囧噯鐘跺喌涓鐨勮川閲忎负0.32g,姘斾綋鐨勬懇灏旇川閲忔槸澶氬皯_鐧惧害鐭 ...
绛旓細鏍囧喌涓嬶紝224ml姘斾綋锛堟棦0.224l锛夌殑鐗╄川鐨勯噺n=0.224/22.4=0.01mol 鏍规嵁n=m/m 寰梞=m/n=0.32g/0.01mol=32g/mol 鐢辨鐭ヨ姘斾綋鏄哀姘攐2

璁＄畻棰( 鍐欏嚭璁＄畻杩囩▼)(1)鍦ㄦ爣鍑嗙姸鍐典笅,224mL鏌愭皵浣鐨勮川閲忎负0.40g.璇...
绛旓細锛1锛塶=VVm=0.224L/22.4L?mol-1=0.01 mol锛孧=mn=0.40g/0.01mol=40 g/mol锛岀瓟锛氭懇灏旇川閲忎负40 g/mol锛涳紙2锛夎闇VL鐨0.5mol/L AlCl3鏍规嵁 n=C脳V锛宯锛圕l-锛=1 mol/L脳2脳0.5L=0.5mol/L脳3脳VV=23L锛屾晠绛旀涓猴細23锛涳紙3锛塶锛圕uSO4锛=CVn锛圕uSO4锛=2.0mol/L脳0.1L=0...

楂樹竴鍖栧棰 1.鍦ㄦ爣鍑嗙姸鍐典笅,224ml鏌愭皵浣鐨勮川閲忎负0.64g,鍒欒姘斾綋鐨勬懇灏...
绛旓細鏍囧喌涓嬩綋绉224ml鍗0.224L鍗0.01mol,0.64g/0.01mol,鎵浠ユ爣鍐典笅姝姘斾綋鎽╁皵閲忎负64g/mol,

鍦ㄦ爣鍑嗙姸鍐典笅,灏224L HCL姘斾綋婧朵簬635ml姘翠腑,鎵寰楃洂閰哥殑瀵嗗害涓1.18_鐧惧害...
绛旓細n=224/22.4=10mol 婧舵恫浣撶Н涓篤 V=m梅瀵嗗害 m=婧惰川+婧跺墏=10*36.5+635*1.0=1000g c=n/V=10梅[锛1000梅1.18锛*10^-3]=11.8mol/L 鐢ㄥ瘑搴︾畻鍑烘潵鐨勪綋绉崟浣嶆槸ml 鎹㈢畻鎴怢 鎵浠ヤ箻浠10^-3

鍖栧棰樺皢鏍囧噯鐘跺喌涓鐨勭敳鐑峰拰姘ф皵鐨勬贩鍚姘斾綋224mL,鐐圭噧鍚庡畬鍏ㄧ噧鐑,灏...
绛旓細鎵浠ユ湁涓ょ鎯呭喌 1.姘㈡哀鍖栭挋杩囬噺锛0.001mol纰抽吀閽欏嵆涓轰簩姘у寲纰崇殑閲忥紝涔熷氨鏄敳鐑风殑閲忎负0.001mol.鎵浠ョ敳鐑风殑浣撶Н涓22.4ml,鍗犳讳綋绉殑1/10,鎵浠ョ敳鐑峰拰姘ф皵鐨勪綋绉瘮涓1/9.2.浜屾哀鍖栫⒊杩囬噺,鍒氬ソ灏嗘阿姘у寲閽欏弽搴旀垚,纰抽吀閽欎笌纰抽吀姘㈤挋,鍙互鑲畾鐨勬槸纰抽吀閽欐槸0.001.纰抽吀姘㈤挋鏄0.002-0.001=0.001.鏍规嵁纰抽吀...

鍦ㄦ爣鍑嗙姸鍐典笅,灏224L姘寲姘姘斾綋婧朵簬635mL鐨勬按(瀵嗗害涓1.00g/mL)涓,鎵...
绛旓細婧惰川姘寲姘㈢殑鎽╁皵鏁版槸锛224L/(22.4L/mol=10.0mol 锛涙隘鍖栨阿婧朵簬姘村悗寰楀埌鐨勬憾娑茶川閲忔槸锛10.0mol*36.5g/mol+635ml*1.00g/ml =365g+635g=1000g 锛涙隘鍖栨阿姘存憾娑茬殑浣撶Н鏄細1000g/(1.18g/ml)=847ml 锛涙墍寰楃洂閰哥殑鐗╄川閲忔祿搴︽槸锛(10.0mol/847ml)*1000ml/L=11.8mol/L 銆

鍦ㄦ爣鍑嗙姸鍐典笅,灏224LHCI姘斾綋婧朵簬635ml姘翠腑,鎵寰楃洂閰哥殑瀵嗗害涓1.20g\cm...
绛旓細mH2O=635g锛宮婧舵恫=1000g锛岄偅涔堢洂閰哥殑鎬讳綋绉 V=1000/1.2mL鍚1/1.2L c=10/锛1/1.2锛=12mol/L 鈶100mL鐨勮瘽nHCl=1.2mol NaHCO3+HCl=NaCl+H2O+CO2鈫 鐭ラ亾浜咰O2鏄1.2mol锛屼綋绉槸1.2脳22.4=26.88L 鈶O2鏄1.2mol 2Na2O2+2CO2=2Na2CO3+O2 2涓狢O2璧颁簡1涓狾2鐩稿綋浜庡鍔犱簡2涓狢O...