Petroleum products<span> are materials derived from crude oil</span><span> as it is processed.
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Answer : The normal boiling point of ethanol will be, 
or 
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of ethanol at 
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = 
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the normal boiling point of ethanol will be, or
Answer: 20 mg Te-99 remains after 12 hours.
Explanation: N(t) = N(0)*(1/2)^(t/t1/2)
N(t) = (80 mg)*(0.5)^(12/6)
N(t) = 20 mg remains after 12 hours
