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ahrayia [7]
3 years ago
9

At a given temperature the vapor pressures of hexane and octane are 183 mmhg and 59.2 mmhg, respectively. Calculate the total va

por pressure over a solution of haxane and octane with xhexane = 0.580.
Chemistry
1 answer:
Bas_tet [7]3 years ago
4 0

Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

P=X_{A}P_{A}+X_{B}P_{B}

Here, X_{A} is mole fraction of A, X_{B} is mole fraction of B, P_{A} is partial pressure of A and P_{B} is partial pressure of B.

The mole fraction of A and B are related to each other as follows:

X_{A}+X_{B}=1

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

X_{octane}=1-X_{hexane}=1-0.58=0.42

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.

Now, vapor pressure can be calculated as follows:

P=X_{hexane}P_{hexane}+X_{octane}P_{octane}

Putting the values,

P=(0.580)(183 mmHg)+(0.420)(59.2 mmHg)=131 mmHg

Therefore, total vapor pressure over the solution of hexane and octane is 131 mmHg.

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 Ch3 CH(CH3) CH2 CH2 CH2 CH2 CH(CH3) CH3

Equation  is  as follows

 CH3 CH(CH3) CH2  C=C CH2 CH(CH3)  CH3  +  H2
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CH3 CH(CH3)CH2 CH2 CH2 CH2 CH(CH3) CH3

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6 0
3 years ago
What is the answer? Please
Delicious77 [7]

Answer:

Parasitism

Explanation:

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2 years ago
What is the molarity of a solution containing 5 moles of HCl in 2.5 L of solution? Be sure to include the formula for calculatin
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Answer:

\boxed{\text{2 mol/L}}

Step-by-step explanation:

c = \dfrac{ \text{moles} }{ \text{litres}}\\\\c = \dfrac{n }{V }

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V = 2.5 L

Calculation:

c = \dfrac{ \text{5 mol} }{\text{2.5 L}} = \text{2 mol/L}

The molar concentration of the solution is \boxed{\textbf{2 mol/L}}.

6 0
3 years ago
Butane (C4 H10(g), mc031-1.jpgHf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , mc031-2.jpgHf = –393.5 kJ/
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The balanced chemical equation for the combustion of butane is:

2C_{4}H_{10}(g) +13 O_{2}(g)-->8CO_{2}(g)+10H_{2}O(g)

ΔH_{reaction}^{0} = Σn_{products}ΔH_{f}^{0}_{(products)}-Σn_{reactants}ΔH_{f}^{0}_{(reactants)}

                         = [{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]

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Calculating the enthalpy of combustion per mole of butane:

1mol C_{4}H_{10}*( \frac{-5315kJ}{2mol C_{4}H_{10} })=-2657.5 \frac{kJ}{molC_{4}H_{10}}

Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol

Correct answer: -2657.5 kJ/mol

6 0
2 years ago
Can methene exist? Explain yes or no
finlep [7]
Methene (non existent) would be an alkene and therefore would need a carbon to carbon double bond, since it only has one carbon it can't bond to another, this is why the lowest alkene is ethene (C2H4).         i hope this helps  
7 0
3 years ago
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