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:
Mass = 0.158 g
Explanation:
Formula used,
P V = n R T
Or,
n = P V / R T
Putting values,
n = 0.948 atm . 0.025 L / 0.0821 L.atm.K⁻¹.mol⁻¹ . 291.45
n = 0.00099 mol
Note: we have changed pressure from mmHg to atm, volume from mL to L and temperature from C to K)
Also,
Mass = n . Molecular Mass
Mass = 0.00099 mol × 159.808 g/mol
Mass = 0.158 g
Oxygen can combine with a metal to produce a compound
Answer:
Q = 0.50
No
Left
Explanation:
At a generic reversible equation
aA + bB ⇄ cC + dD
The reaction coefficient (Q) is the ratio of the substances concentrations:
![Q = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5BC%5D%5Ec%2A%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%2A%5BB%5D%5Eb%7D)
Solids and liquid water are not considered in this calculus.
When the reaction achieves equilibrium (concentrations are constant), the Q value is named as Kc, which is the equilibrium constant of the reaction. If Q > Kc, it indicates that the concentration of the products is higher, so, the reaction must progress to the left and form more reactants; if Q < Kc, than the concentrations of the reactants, are higher, so, the reaction progress to the right.
In this case:
Q = ![\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)

Q = 0.50
So, Q > Kc, the reaction is not at equilibrium and it progresses to the left.