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4 years ago

How many moles of S are in 35.4 g of (C3H5)2S?

1 answer:

6 0

<span>0.310 moles First, look up the atomic weights of the elements involved. Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight sulfur = 32.065 Molar mass (C3H5)2S = 6 * 12.0107 + 10 * 1.00794 + 32.065 = 114.2086 g/mol Moles (C3H5)2S = 35.4 g / 114.2086 g/mol = 0.309959145 mol Since there's just one sulfur atom per (C3H5)2S molecule, the number of moles of sulfur will match the number of moles of (C3H5)2S which is 0.310 when rounded to 3 significant digits.</span>

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3 years ago

An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle a

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1. 0.0637 moles of nitrogen.

2. The partial pressure of oxygen is 0.21 atm.

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1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):

$V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol$

2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:

$n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol$

$V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21$

$P_{O_2}=X_{O_2} \times P = 1 atm \times 0.21 = 0.21 atm$

As you see, the molar fraction and volume fraction are the same because of the assumption of ideal behaviour.

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