化学作业帮一下？
发布网友
共5个回答
热心网友
1、解：n（HNO3）=1L * 1mol/L =1mol
m（HNO3）=1mol * 63g/mol =63g
m[HNO3(aq)]=63g / 65%=96.9g
答：略。
2、解：【因为质量分数不确定，所以不能按照第一题的思路，只能先求出溶液体积，再通过密度求溶液质量】
设需要12mol/L的盐酸体积为x
12mol/L * x= 1L * 1mol/L
x=83.3mL
m[HCl(aq)]=83.3mL * 1.19g/mL=99.1g
答：略
热心网友
需要溶质HNO₃的质量是:
1mol/L×1L×63g/mol=63g
需要市售*的质量是:
63g÷65%≈97g追答需要浓盐酸的体积是:
1mol/L×1L÷12molL
=(1/12)L
需要取市售浓盐酸的质量是:
（1/12×1000×1.19
≈99.2（克）
热心网友
练习1
设需要Xml，则根据稀释定律
C1V1＝C2V2
65％×1.4X/＝1
X＝70ml
练习2
由C＝1000p·w/M得:
w＝C·M/1000p=(12mol/L ×36.5g/mol)/(1000×1.19g/ml)=36.8%
则 m(液)=m(HCl)/w=99.18g
热心网友
（1+14+16x3）/65%
=63/0.65
=96.9克
热心网友
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