14.13 分子分数

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  分子分数

  

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  The mixed blessing of sulfur dioxide
  ::混合的二氧化硫

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  Sulfur dioxide is a by-product of many processes, both natural and human-made. Massive amounts of this are released during volcanic eruptions such as the one seen above on the Big Island (Hawaii). Humans produce sulfur dioxide by burning coal. The gas has a cooling effect when in the atmosphere by reflecting sunlight back away from the earth. However, sulfur dioxide is also a component of smog and rain, both of which are harmful to . Many efforts have been made to reduce SO ₂ levels to lower acid rain production. An unforeseen complication: as we lower the of this gas in the atmosphere, we lower its ability to cool and then we have global warming concerns.
  ::二氧化硫是许多自然和人为过程的副产品,其中大量是在火山爆发期间释放的,如上文提到的大岛(哈瓦伊岛)火山爆发时释放出来的。人类通过燃烧煤炭生产二氧化硫。气体在大气中通过反射阳光从地球反射而产生冷却效应。然而,二氧化硫也是烟雾和雨的成分,两者都有害。许多努力都是为了降低二氧化硫水平,降低酸雨的产量。一个意外的复杂因素:随着大气中这种气体的减少,我们降低其降温能力,然后我们降低其降温能力,我们就会关注全球变暖问题。

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  Mole Fraction
  ::分子分数

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  One way to express relative amounts of substances in a is with the mole fraction. Mole fraction   $\begin{array}{r}(X)\end{array}$ is the ratio of moles of one substance in a mixture to the total number of moles of all substances. For a mixture of two substances,  $\begin{array}{r}A\end{array}$ and $\begin{array}{r}B\end{array}$ , the mole fractions of each would be written as follows:
  ::一种表示一种物质相对数量的方法是用摩尔分数表示。分子分数(X)是混合物中一种物质的摩尔与所有物质的摩尔总数的比率。

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  $$\begin{array}{r}{X}_A=\frac{\text{mol}A}{\text{mol}A+\text{mol}B}\text{and}{X}_B=\frac{\text{mol}B}{\text{mol}A+\text{mol}B}\end{array}$$

  
  ::XA=摩尔 Amol A+摩尔 B=摩尔 B+摩尔 B

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  If a mixture consists of 0.50 mol  $\begin{array}{r}A\end{array}$ and 1.00 mol $\begin{array}{r}B\end{array}$ , then the mole fraction of  $\begin{array}{r}A\end{array}$ would be $\begin{array}{r}{X}_A=\frac{0.5}{1.5}=0.33\end{array}$ .  Similarly, the mole fraction of  $\begin{array}{r}B\end{array}$ would be $\begin{array}{r}{X}_B=\frac{1.0}{1.5}=0.67\end{array}$ .
  ::如果混合物由0.50摩尔 A和1.00摩尔 B组成,那么A的摩尔部分为XA=0.51.5=0.33。同样,B的摩尔部分为XB=1.01.5=0.67。

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  Mole fraction is a useful quantity for analyzing gas mixtures in conjunction with Dalton’s law of partial pressures . Consider the following situation: A 20.0 liter vessel contains 1.0 mol of hydrogen gas at a pressure of 600 mmHg. Another 20.0 liter vessel contains 3.0 mol of helium at a pressure of 1800 mmHg. These two gases are mixed together in an identical 20.0 liter vessel. Because each will exert its own pressure according to Dalton’s law , we can express the partial pressures as follows:
  ::分子分数是结合Dalton的部分压力法分析气体混合物的有用数量。 考虑以下情形: A 20. 0 升容器含有1.0 毫升氢气,压力为600毫米Hg。 另一 20. 0 升容器含有3. 0 毫升氢气,压力为1800 毫米Hg。 这两种气体混合在一个相同的20. 0升容器中。 因为根据Dalton 的法律,每个容器将自己施加压力,我们可以表示部分压力如下:

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  $$\begin{array}{r}{P}_{H_2}=X_{H_2}\times P_{\text{Total}}\text{and}{P}_{He}=X_{He}\times P_{\text{Total}}\end{array}$$

  
  ::PH2 = XH2 xPTotalandPhee = XHxPTotolal = XH2 xPTotalandPhee = XHxPTotatal = XH2 xPTotalandPHe = XH2 = XH2 xPTotalandPHe = XHxPTotolal = XH2 = XH2xPTotatal = XH2xPTOPotatal = XH2 = XH2xPTPotalandphe = XExPTotal = XHT = XH2 = XPTPTotal = XH2 = XH2x = XPTPTPO = X = XH2 = XH2 =X =X = XH2x = XPTPTAPAT = X = X = X =X =X = X = XPTOATATAT = X = X = X = X = X = X = X = X = X = X = XPT = X = X = X = X = X = XPT = X = X = X = = = = = = X = X = X = X = X = X = X = X = XPTO = X = XPTTO = = = = = = X = = X

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  The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. For our mixture of hydrogen and helium:
  ::混合物中气体的部分压力等于其摩尔分数乘以总压力乘以其摩尔分数。

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  $$\begin{array}{r}{X}_{H_2}=\frac{1.0\text{mol}}{1.0\text{mol}+3.0\text{mol}}=0.25\text{and}{X}_{He}=\frac{3.0\text{mol}}{1.0\text{mol}+3.0\text{mol}}=0.75\end{array}$$

  
  ::XH2=1.0 mol1.0 mol+3.0 mol=0.25和XHE=3.0 mol1.0 mol+3.0 mol=0.75

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  The total pressure according to Dalton’s law is $\begin{array}{r}600\text{ mmHg}+1800\text{ mmHg}=2400\text{ mmHg}\end{array}$ . So, each partial pressure will be:
  ::根据道尔顿法律,总压力为600毫米Hg+1800毫米Hg=2400毫米Hg。

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  $$\begin{array}{rl}& P_{H_2}=0.25\times 2400\text{ mmHg}=600\text{ mmHg}\\ & P_{He}=0.75\times 2400\text{ mmHg}=1800\text{ mmHg}\end{array}$$

  
  ::PH2=0.25×2400毫米Hg=600毫米HgPhee=0.75×2400毫米Hg=1800毫米Hg

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  The partial pressures of each gas in the mixture don’t change since they were mixed into the same size vessel and the temperature was not changed.
  ::混合物中每种气体的部分压力不会改变,因为它们混合成同一尺寸的容器,温度没有改变。

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  Sample Problem: Dalton’s Law
  ::抽样问题:道尔顿法律

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  A flask contains a mixture of 1.24 moles of hydrogen gas and 2.91 moles of oxygen gas. If the total pressure is 104 kPa, what is the partial pressure of each gas?
  ::如果总压力为104千帕,每种气体的部分压力是多少?

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  Step 1: List the known quantities and plan the problem .
  ::第1步:列出已知数量并规划问题。

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  Known
  ::已知已知

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  • 1.24 mol H ₂
    ::1.24毫摩拉H2
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  • 2.91 mol O ₂
    ::2.91 mol O2
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  • $\begin{array}{r}{P}_{\text{Total}}=104\text{kPa}\end{array}$
    ::PTalal=104千帕

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  Unknown
  ::未知

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  • $\begin{array}{r}{P}_{H_2}=?\text{kPa}\end{array}$
    ::PH2=? kPa
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  • $\begin{array}{r}{P}_{O_2}=?\text{kPa}\end{array}$
    ::PO2=? kPa(千帕)

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  First, the mole fraction of each gas can be determined. Then, the partial pressure can be calculated by multiplying the mole fraction by the total pressure.
  ::首先,每种气体的摩尔分数可以确定。然后,部分压力可以通过将摩尔分数乘以总压力来计算。

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  Step 2: Solve .
  ::步骤2:解决。

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  $$\begin{array}{rlrl}& X_{H_2}=\frac{1.24\text{mol}}{1.24\text{mol}+2.91\text{mol}}=0.299& & X_{O_2}=\frac{2.91\text{mol}}{1.24\text{mol}+2.91\text{mol}}=0.701\\ & P_{H_2}=0.299\times 104\text{ kPa}=31.1\text{ kPa}& & P_{O_2}=0.701\times 104\text{ kPa}=72.9\text{ kPa}\end{array}$$

  
  ::XH2=1.24 mol1.24 mol+2.91 mol=0.2999XO2=2.91 mol1.24 mol+2.91 mol=0.701PH2=0.299×104 kPa=31.1 kPAP2=0.701×104 kPa=72.9 kPa

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  Step 3: Think about your result .
  ::步骤3:想想你的结果。

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  The hydrogen is slightly less than one third of the mixture, so it exerts slightly less than one third of the total pressure.
  ::氢略低于混合物的三分之一,因此,氢的含量略低于总压力的三分之一。

   

   

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  Summary
  ::摘要

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  • Use of the mole fraction allows calculation to be made for mixtures of gases.
    ::使用摩尔分数可以计算气体混合物。

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  Explore More
  ::探索更多

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  Use the resource below to answer the questions that follow.
  ::利用以下资源回答以下问题。

   

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  1. What is mole percent?
     ::什么是摩尔百分率?
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  2. Do the mole fractions add up to 1.00?
     ::摩尔的分数加到1.00吗?
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  3. What other way could you calculate the mole fraction of oxygen once you have the mole fraction of nitrogen?
     ::一旦你拿到了分子的氮分数 你还能用什么其他方法来计算 氧的摩尔分数呢?

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  Review
  ::回顾

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  1. What is mole fraction?
     ::什么是摩尔分数?
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  2. How do you determine partial pressure of a gas when given the mole fraction and the total pressure?
     ::如果给定了摩尔分数和总压力,如何确定气体的局部压力?
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  3. In a gas mixture containing equal numbers of moles of two gases, what can you say about the partial pressures of each gas?
     ::在一种气体混合物中,两种气体的摩尔数量相等, 你能对每种气体的部分压力说些什么?