Explanation:
The given data is as follows.
T =
= (120 + 273.15)K = 393.15 K,
As it is given that it is an equimolar mixture of n-pentane and isopentane.
So,
= 0.5 and
= 0.5
According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.
(393.15 K) = 9.2 bar
(393.15 K) = 10.5 bar
Hence, we will calculate the partial pressure of each component as follows.

= 
= 4.6 bar
and, 
= 
= 5.25 bar
Therefore, the bubble pressure will be as follows.
P =
= 4.6 bar + 5.25 bar
= 9.85 bar
Now, we will calculate the vapor composition as follows.

= 
= 0.467
and, 
= 
= 0.527
Calculate the dew point as follows.
= 0.5,
= 0.5


= 0.101966
P = 9.807
Composition of the liquid phase is
and its formula is as follows.

= 
= 0.5329

= 
= 0.467
Answer:
S/.486 es el valor del anillo
Explanation:
Para hallar el precio del anillo se deben encontrar las moles de oro que contiene este.
Si el anillo es de 90g y solo el 59.1% contiene oro, la cantidad de oro en gramos es:
90g × 59.1% = 53.19g Oro en el anillo
Ahora, para convertir los gramos de oro a moles se debe usar la masa atómica del oro (197g/mol), así:
53.19g × (1mol / 197g) = <em><u>0.27 moles de oro contiene el anillo</u></em>.
Ya que cada mol de oro cuesta S/.1800, 0.27 moles de oro (Y por lo tanto, el anillo) costarán:
0.27mol × (S/.1800 / 1mol oro) =
<h3>S/.486 es el valor del anillo</h3>
Answer:
AgC2H3O2 is: Silver acetate
A. high potential energy and zero kinetic energy.
Answer:
d = 9.5× 10⁻⁴ Kg/cm³
Explanation:
Given data:
Length of cube = 1.02 cm
Mass of cube = 0.001012 Kg
Density of cube = ?
Solution:
The length , height and width of cube is always same. Thus the volume of cube is,
Volume = length × height× width
Volume = 1.02 cm × 1.02 cm × 1.02 cm
Volume = 1.06 cm³
Density of cube:
d = 0.001012 Kg /1.06 cm³
d = 9.5× 10⁻⁴ Kg/cm³