Moles of Carbon dioxide(CO2) = 2
<h3>Further explanation</h3>
Given
Reaction
2 C2H6 (g) + 7 O2 (g) —+ 4CO2 (g) + 6 H20 (g)
Required
moles of carbon dioxide
Solution
The reaction coefficient shows the mole ratio of the compounds in the reaction equation (reactants and products)
From the equation, mol ratio of C2H6 : CO2 = 2 : 4, so mol CO2 :
mol CO2 = (4/2) x mol C2H6
mol CO2 = 2 x 1
mol CO2 = 2
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Answer: Option (a) is the correct answer.
Explanation:
The given data is as follows.
= 4.19 
= 1.9
Heat of vaporization (
) at 1 atm and 
is 2259 kJ/kg
= 0
Therefore, calculate the enthalpy of water vapor at 1 atm and as follows.
= +
= 0 + 2259 kJ/kg
= 2259 kJ/kg
As the desired temperature is given 
and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and is calculated as follows.
= 
= 334.781 kJ/kg
Hence, enthalpy of water vapor at 10 bar and is calculated as follows.
= 
= 2410.81 kJ/kg
Therefore, calculate the latent heat of vaporization at 10 bar and as follows.
= 
= 2410.81 kJ/kg - 334.781 kJ/kg
= 2076.029 kJ/kg
or, = 2076 kJ/kg
Thus, we can conclude that at 10 bar and latent heat of vaporization is 2076 kJ/kg.
Answer:
The pH of the buffer is 7.0 and this pH is not useful to pH 7.0
Explanation:
The pH of a buffer is obtained by using H-H equation:
pH = pKa + log [A⁻] / [HA]
<em>Where pH is the pH of the buffer</em>
<em>The pKa of acetic acid is 4.74.</em>
<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>
<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>
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Replacing:
pH = 4.74 + log [0.1779mol] / [0.001mol]
<em>pH = 6.99 ≈ 7.0</em>
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The pH of the buffer is 7.0
But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,
this pH is not useful to pH 7.0
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