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xxTIMURxx [149]
1 year ago
6

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

Chemistry
1 answer:
Lady bird [3.3K]1 year ago
8 0

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).

Learn more about qualitative test here: brainly.com/question/2109763

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explain why elements lying on the extreme right of the periodic table as from non metals but those lying on the extreme left a s
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Element at Extreme Left In Periodic Table:
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where n is principle quantum number, which shows main energy level or shell. These metals can gain Noble gas configuration (stable configuration) either by loosing one electron or by gaining seven or more electrons. As it is quite reasonable to loose one electron instead of gaining seven or more electrons so these element easily loose one electron to gain noble as configuration. The Metallic character decreases along the period from left to right. So Group II-A (2) are second most metallic elements and so on. These metals at extreme left mainly exist in solid form.

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5 0
3 years ago
What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o
Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

1 mL= 10^{-3} L

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

The formula can be written for the calculation of moles as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Thus,  

Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}

Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles

Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles

The chemical reaction taking place:

AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)

According to reaction stoichiometry:

<u>1 mole</u> of AgNO₃ reacts with <u>1 mole</u> of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

1 mL= 10^{-3} L

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Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}

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<u>Thus, the concentration of NaCl = 0.3374 M</u>

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