How could one rapidly separate red #40 from zinc oxide? indicate every step.
2 answers:
The mixture of Red # 40 and zinc oxide is a heterogeneous mixture which can be separated by filtration
<h3>Further explanation</h3>The mixture is a combination of two or more substances whose constituent properties do not change. The properties of the mixture component are not lost / unchanged as in compounds.
The mixture can be divided into a homogeneous mixture if the composition/ratio of each substance in the mixture is the same and heterogeneous mixture if the ratio of the composition of the substances is not the same in each place.
Mixtures can also be divided into solutions, suspensions, and colloids based mainly on the size of the particles
There are various mixed separation techniques including filtration, distillation, evaporation, extraction or chromatography
Red # 40, is an artificial coloring obtained from mixing hydrocarbons and is used in dyes such as candy or drinks
Zinc oxide or ZnO is an inorganic powder form which is used as an additive in the manufacture of products such as ceramics or glass and is not soluble in water but can dissolve in acids or bases
A heterogeneous mixture can be separated through physical separation techniques
The mixture of Red # 40 and zinc oxide is a heterogeneous mixture which can be separated by filtration because
Red # 40 can dissolve in water while zinc oxide is insoluble
Zinc oxide and red # 40 is filtered from the mixture using a filter paper
The filtering results called filtrate will be zinc oxide and the residual filtering called residue is Red # 40
While the separation of red # 40 and zinc oxide from water using evaporation techniques to obtain the dry solids
<h3><em>Learn more </em></h3>
A heterogeneous mixture
the difference between a solution and a heterogeneous mixture
Keywords: A heterogeneous mixture, red # 40, ZnO, filtration
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Answer:
Explanation:
Given that:
the temperature = 250 °C= ( 250+ 273.15 ) K = 523.15 K
Pressure = 1800 kPa
a)
The truncated viral equation is expressed as:
where; B = - C = -5800
R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹
Plugging all our values; we have
Multiplying through with V² ; we have
V = 2250.06 cm³ mol⁻¹
Z =
Z =
Z = 0.931
b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].
The generalized Pitzer correlation is :
The compressibility is calculated as:
Z = 0.9386
V = 2268.01 cm³ mol⁻¹
c) From the steam tables (App. E).
At
V = 0.1249 m³/ kg
M (molecular weight) = 18.015 gm/mol
V = 0.1249 × 10³ × 18.015
V = 2250.07 cm³/mol⁻¹
R = 729.77 J/kg.K
Z =
Z =
Z = 0.588