一道高一的化学题,我实在是不会写,可是明天就要交了,有谁会的,帮帮忙可以吗??拜托了~~~~ 急急急!!!!高一一道化学题,会的人请帮忙!拜托了
\u4e00\u9053\u9ad8\u4e00\u7684\u5316\u5b66\u9898\u76ee\uff0c\u9a6c\u4e0a\u5c31\u8981\u7684\uff0c\u90a3\u4f4d\u5929\u624d\u5e2e\u5e2e\u5fd9\u554a...?\u52a0\u5165H2SO4\u540e\u6eb6\u6db2\u4e2dK2SO4\u7269\u8d28\u7684\u91cf\u4e0d\u53d8
\u52a0\u5165H2SO4\u524dn(K2SO4) = 0.5*0.1 = 0.05mol
\u6240\u4ee5\u52a0\u5165H2SO4\u6eb6\u6db2\u4f53\u79ef = 0.05/0.2 = 0.25L = 250ml
\u6240\u4ee5\u52a0\u5165H2SO4\u540eH2SO4\u7269\u8d28\u7684\u91cf = 2*0.25 = 0.5mol
\u9700\u8981\u52a0\u5165H2SO4\u7269\u8d28\u7684\u91cf = 0.5 - 1*0.1 = 0.4mol
\u9700\u8981\u52a0\u5165\u786b\u9178\u7684\u4f53\u79ef = 0.4*98/(98%*1.84) = 21.7ml
\u7136\u540e\u52a0\u6c34\u7a00\u91ca\u5230250ml
\u89e3\u6790\uff1a
\uff081\uff09\u70c3\u7684\u71c3\u70e7\u65b9\u7a0b\u5f0f\uff1a
CxHy + \uff08x+y/4\uff09O2 \u2015\uff08\u70b9\u71c3\uff09\u2192 xCO2 + \uff08y/2\uff09H2O
\u6d53H2SO4\u4e3a\u5e72\u71e5\u5242\uff0c\u589e\u91cd9g\u4e3a\u6c34\u7684\u8d28\u91cf\uff0c
\u4ece\u800c\u5f97\u5230\u6c34\u7684\u7269\u8d28\u7684\u91cf = 9g / 18g/mol = 0.5mol
\u6c34\u4e2d\u7684H\u5143\u7d20\u4e0e\u6c34\u7269\u8d28\u7684\u91cf\u6bd4 = 2:1\uff0c\u6240\u4ee5H\u5143\u7d20\u7269\u8d28\u7684\u91cf = 0.5mol*2 = 1mol
\u6c22\u5143\u7d20\u5168\u90e8\u6765\u6e90\u4e8e\u70c3\u3002
\u78b1\u77f3\u7070\u589e\u91cd17.6g\u4e3aCO2\u8d28\u91cf\uff0c
\u4ece\u800c\u5f97\u5230CO2\u7684\u7269\u8d28\u7684\u91cf = 17.6g / 44g/mol = 0.4mol
CO2\u4e2d\u7684C\u5143\u7d20\u4e0eCO2\u7269\u8d28\u7684\u91cf\u6bd4 = 1:1\uff0c\u6240\u4ee5C\u5143\u7d20\u7269\u8d28\u7684\u91cf = 0.4mol*1 = 0.4mol
\u78b3\u5143\u7d20\u5168\u90e8\u6765\u6e90\u4e8e\u70c3\uff0c
\u6240\u4ee5C\uff1aH\u7269\u8d28\u7684\u91cf\u6bd4 = 0.4mol\uff1a1mol = 4:10
\u6240\u4ee5\u8be5\u70c3\u4e3aC4H10
\uff082\uff09\u53ef\u80fd\u7684\u7ed3\u6784\u7b80\u5f0f\uff1a
\u6b63\u4e01\u70f7\uff1aCH3-CH2-CH2-CH3
\u5f02\u4e01\u70f7\uff082-\u7532\u57fa\u4e19\u70f7\uff09\uff1aCH3-CH-CH3
|
CH3
2NH4Cl+Ca(OH)2=CaCl2+2NH3+2H2O
2 2
2 x
x=2*2/2=2
产生的NH3的物质的量=2mol
标况下这些气体的体积=2*22.4=44.8升
2)氨水的物质的量浓度=2/0.1=20mol/L
(1)解:设所产生的氨气标准状态的体积为,则。
2NH4Cl+Ca(OH)2=2NH3↑+2H2O+CaCl2
107 44.8L
107g x
x=44.8L
答:所产生的氨气标准状态的体积为44.8L。
(2)n(NH3)=V(NH3)/(22.4L/mol)=2mol
NH3 ~ NH3·H2O
1 1
2mol n(NH3·H2O)
n(NH3·H2O)=2mol
c(NH3·H2O)=n(NH3·H2O)/V(NH3·H2O)=2mol/0.1L=20mol/L
答:所得氨水的物质的量浓度是20mol/L。
2NH4Cl+Ca(OH)2=2NH3+2H2O+CaCl2
107g xg
53.5g/mol 17g/mol
因为我们算体积用的是物质的量,所以氨气的物质的量=氯化铵的物质的量=2mol
所以氨气的体积是:2*22.4=44.8L
如果溶于水后溶液体积不变:2mol/0.1L=20mol/L
标况下是8.736L。
氯化铵的物质的量是n=m/M=21/53.5=0.39mol
2NH4Cl+Ca(OH)2=CaCl2+2NH3↑+2H2O
2 2
0.39mol n
n=0.39mol
NH3在标况下的体积是V=n*Vm=0.39*22.4=8.736L
44.8L
20mol/L
绛旓細2锛夋皑姘寸殑鐗╄川鐨勯噺娴撳害=2/0.1=20mol/L
绛旓細澶畝鍗曚簡锛屼笉鍥炵瓟浜
绛旓細璐ㄩ噺姣:K:O=13:8,鎽╁皵鏁版瘮锛2锛3锛岃K2O2涓簒锛孠O2涓簓 2x+y=2 2锛坸+y锛=3 x/y=1/2 w(K2O2)=m(K2O2)/(m鎬=39脳2+32/110+71=43.7
绛旓細1銆14.51g鐧借壊娌夋穩鏄~閰搁挕鍜岀⒊閰搁挕锛屽姞鍏ヨ繃閲忕█纭濋吀澶勭悊锛屾矇娣鍙墿涓嬬~閰搁挕銆侼a2SO4+Ba(OH)2=BaSO4+2NaOH 142 233 x 4.66 x*142/142=4.66/233 x=0.02mol 2銆佺⒊閰搁挕璐ㄩ噺涓14.51-4.66=9.85g BaCO3+2HNO3=CO2+Ba(NO3)2 197 44 9.85 y y/44=9.85/197 y=0....
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