Answer:
The correct answer is 0.12 grams.
Explanation:
The mass of carbon monoxide or CO collected in the tube can be determined by using the ideal gas equation, that is, PV = nRT.
Based on the given question, P or the pressure of the gas is given as 1 atm, volume of the gas collected in the tube is 117 ml or 0.117 L.
The number of moles or n can be determined by using the equation, mass/molar mass.
R is the universal gas constant, whose value is 0.0821 L atmK^-1mol^-1, and temperature is 55 degree C or 328 K (55+273).
On putting the values we get:
n = PV/RT
= (1 atm*0.117 L) / (0.0821 L atmK^-1mol^-1 * 328 K)
= 0.0043447 mol
Therefore, mass of CO will be moles * molar mass of CO
= 0.0043447 mol * 28 g/mol
= 0.12 g
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Answer:
The standard enthalpy of formation of this isomer of is -220.1 kJ/mol.
Explanation:
The given chemical reaction is as follows.
The expression for the entropy change for the reaction is as follows.
Substitute the all values in the entropy change expression.
Therefore, The standard enthalpy of formation of this isomer of is -220.1 kJ/mol.
Answer:
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