Describe a qualitative test for sulfate in alum crystals using ionic reactions of barium chloride (BaCl2)

1 answer:

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A qualitative test for sulfate in alum crystals using ionic reactions of barium chloride (BaCl2) is given Ba²⁺(aq) + SO₄²⁻ (aq) → BaSO₄(s).

<h3>What is qualitative test?</h3>

Qualitative test measures changes in color, melting point, odor, reactivity, radioactivity, boiling point, bubble production, and precipitation of the sample.

<h3>Qualitative test for sulfate in alum crystals </h3>

When an aqueous solution of a barium salt (BaCl₂) is mixed with an aqueous solution containing sulfate, a white precipitate of insoluble BaSO₄ forms according to the net ionic equation given below;

Ba²⁺(aq) + SO₄²⁻ (aq) → BaSO₄(s)

Thus, a qualitative test for sulfate in alum crystals using ionic reactions of barium chloride (BaCl2) is given Ba²⁺(aq) + SO₄²⁻ (aq) → BaSO₄(s).

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You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr

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Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

$W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t$

The power needed to process 50 ton/hor is

$P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP$

2) The density of the packed bed can be expressed as

$\rho=f_v*\rho_v+f_s*\rho_s$

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).

Then we can rearrange

$\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37$