Use the Clausius-Clapeyron equation...
<span>Let T1 be the normal boiling point, which will occur at standard pressure (P1), which is 101.3 kPa (aka 760 torr or 1.00 atm). You know the vapour pressure (P2) at a different temperature (T2). And you are given the enthalpy of vaporization. Therefore, we can use the Clausius-Clapeyron equation.
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![ln(P_1/P_2) = \frac{-\delta H_{vap}}{R} \times [\frac{1}{T_1} - \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28P_1%2FP_2%29%20%3D%20%5Cfrac%7B-%5Cdelta%20H_%7Bvap%7D%7D%7BR%7D%20%20%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
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</span><span>ln(101.3 kPa / 52.7 kPa) = (-29.82 kJ/mol / 8.314x10^{-3} kJ/molK) (1/T - 1/329 K)
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------ some algebra goes here -----
<span>T = 349.99K ...... or ...... 76.8C </span>
Answer:
Explanation
ΔH is directly proportional to the quantity of a substance that reacts or is produced by a reaction. Enthalpy is directly proportional to mass. Therefore, if you double the coefficients in an equation, then the value of ΔH is multiplied by two.
Answer:
1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.
Explanation:
The word equation for the combustion of propane can be obtained from the chemical equation;
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
The word equation is therefore:
1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.
For such a combustion reaction, carbon dioxide and water are produced in the process.
Non-metals. it is fluorine i believe and in the periodic table it is under the non-metal category