<span>There are 1000 cm3 in 1 liters.
Hence 1 liter of the liquid would weigh:
1000 cm3 x (1.17 g/cm3) = 1170 gm
and there are 1000 gm in 1 kg, so we want enough liters to have a mass of
3.75 kg x 1000 gm/kg = 3750 gm
Hence, # of liters = desired mass / # of gm per liter
= 3750 gm / 1170 gm/liter
= 3.2051282 liters</span>
Answer:
y1 = 0.3162
y2 = 0.6838
Explanation:
ok let us begin,
first we would be defining the parameters;
at 25°C;
1-propanol P1° = 20.90 Torr
2-propanol P2° = 45.2 Torr
From Raoults law:
P(1-propanol) = P⁰ × X(1-propanol)
P(1-propanol) = 20.9 torr × 0.45 = 9.405
P(1-propanol) = 9.405 torr
Also P(2-propanol) = P⁰ × X(2-propanol)
P(2-propanol) = 45.2 torr × 0.45
P(2-propanol) = 20.34 torr
but the total pressure = sum of individual pressures
total pressure = 9.405 + 20.34
total pressure = 29.745 torr
given that y1 and y2 represent the mole fraction of each in the vapor phase
y1 = P1 / total pressure
y1 = 9.405/29.745
y1 = 0.3162
Since y1 + y2 = 1
y2 = 1 - y1
∴ y2 = 1 - 0.3162
y2 = 0.6838
cheers, i hope this helps.
Answer:
C co2 2co enthalpy
2 Answers. Ernest Z. The standard enthalpy of formation of carbon monoxide is -99 kJ/mol.
