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Bingel [31]
3 years ago
13

Does a part or slice of a substance have a different density than the whole piece? Pls help!

Chemistry
1 answer:
Nostrana [21]3 years ago
3 0
Density is defined as mass per unit volume so even when you cut an object in half unit volume does not change so each part would have a different density even if it’s cut into the same pieces
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A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture
Cerrena [4.2K]

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

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V = 2.50 L

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n=(\frac{1.45*2.50}{0.0821*293})

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Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

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CH_4+2O_2\rightarrow CO_2+2H_2O

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From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

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moles of carbon dioxide = 0.195 mol

Hence, 0.195 = X + 3(0.151 - X)

Let's solve this for X as:

0.195 = X + 0.453 - 3X

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mole fraction of methane = \frac{moles of Methane}{total moles}

mole fraction of methane = \frac{0.129}{0.151}

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Hence, the mole fraction of methane gas in the original gas mixture is 0.854.

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3 years ago
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