Answer:
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) = ![\frac{ΔvapH}{R}([tex]\frac{1}{T1}](https://tex.z-dn.net/?f=%5Cfrac%7B%CE%94vapH%7D%7BR%7D%28%5Btex%5D%5Cfrac%7B1%7D%7BT1%7D)
-
)
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
No of moles of MgCl2 = weight of MgCl2 / Molecular weight of MgCl2
Weight of MgCl2 =moles of MgCl2 x molecular mass of MgCl2
= 8.90 x 95=845.5 gm
The larger the piece the longer it will take to break down. This is because it has more mass that needs to be broken down.
The standard entropy for the substances are as follows:
C6H12O2(s) = -212
<span>O2(g) = -205 </span>
<span>CO2(g) = -214 </span>
<span>H2O(l) = -70
</span>
We calculate the ∆S°r<span>eaction by the expression:
</span>∆S°rxn = ∆S°products - ∆S° reactants
∆S°rxn = (212+6x205)-(6x214+6x70)
∆S°rxn = -262 J/K ------> OPTION 3
Integer i think but i not sure
