ΔG > 0
is always true for the freezing of water.
Explanation:
- The freezing of water is only spontaneous when the temperature is fairly small. Over 273 K, the higher value of TΔS causes the sign of ΔG to be positive, and there is no freezing point.
- The entropy decreases as water freezes. This does not infringe the Thermodynamics second law. The second law doesn't suggest entropy will never diminish anywhere.
- Entropy will decline elsewhere, provided it increases by at least as much elsewhere.
Answer:
The amount of heat absorbed is <u>5.183889 kJ</u> .
Explanation:
In conversion of water to ice it rejects some heat while in conversion of ice to water it absorbs heat which is called latent heat which is given as 6.02 kJ/mol.
The amount of ice given is 15.5 g.
Converting it to moles as the latent heat is given in per moles:

Molecular mass of Hydrogen (H) and Oxygen (O) is 1 u and 16 u respectively.
Molecular mass of water is 18 g (
⇒ 2*1+16=18 ).
mole = 15.5/18 ≈ 0.8611 moles
Therefore the amount of heat absorbed by 15.5 g of ice ( 0.8611 moles) = <em>Latent heat * moles
</em>
Heat absorbed = 6.02*0.8611
= 6.02*(15.5/18)
≈ 5.183889 kJ
Answer:

Explanation:
Given:
Pressure = 745 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 745 / 760 = 0.9803 atm
Temperature = 19 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (19 + 273.15) K = 292.15 K
Volume = 0.200 L
Using ideal gas equation as:

where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K
⇒n = 0.008174 moles
From the reaction shown below:-

1 mole of
react with 2 moles of 
0.008174 mole of
react with 2*0.008174 moles of 
Moles of
= 0.016348 moles
Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)
So,



Answer:
36.63 Torr
Explanation:
You need to use two expressions, one for pressure and the other with the relation of density and height of the column.
For the pressure:
P = h * d * g (1)
h is height.
d density
g gravity
The second expression put a relation between the densities and height of the column so:
d1/d2 = h1/h2 (2)
let 1 be the phthalate, and 2 the mercury.
Let's calculate first the relation of density:
d1/d2 = 13.53 / 1.046 = 12.93
Now with the first expression, we can calculate the pressure so:
P = hdg
We have two compounds so,
h1d1g = h2d2g ---> gravity cancels out
From here, we can solve for h2:
h2 = h1*(d1/d2)
replacing:
h2 = 459 / 12.53
h2 = 36.63 mm
1 mmHg is 1 torr, therefore the pressure of the gas in Torr would be 36.63 Torr