If the vapor pressure of ethanol at 34.7degree c is 100
mmhg, and its heat of vaporization is 38.6 kj/mol. at what
temperaturewould ethanol boil at sea level?

we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:

 ln (P2/P1) = -)

where

P1 and P2 are the vapour pressures at temperatures  T1 and T2

Δ vapH  = the enthalpy of vaporization of the ETHANOL

R  = the Universal Gas Constant

In this problem,

P 1 = 100 mmHg

;  T 1 = 34.7 °C = 307.07 K

P 2 = 760mmHg

T 2 =T⁻²=?

Δ vap H = 38.6 kJ/mol

R = 0.008314 kJ⋅K -1 mol -1

ln ( 760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1 /0.008314 )

0.0004368=(0.00325 - T⁻²)  

T⁻²=0.002813

T² = 355.47K

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

the answer is c

tell me if wrong