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xxTIMURxx [149]
2 years 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]2 years 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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Which of the following is the largest volume? 4.0 x 101 m3 3.8 x 107 cm3 3.8 x 105 dm3 3.8 x 108 mm3
fiasKO [112]
Answer : Option A) 4.0 X 101 m^{3} 

Explanation : As per the measurement chart the greatest metric system is measured in meters.

So the rest values are given in the form of metric values which are obviously less than that of meters. values of centimeters, decimeter and millimeters are less than that of meters.
8 0
3 years ago
Which of the following has to remain constant to allow you to use the combined gas law for calculations?
Mazyrski [523]

Answer:

Amount of gas

Explanation:

The combined gas law has to do with pressure, volume, and temperature. All of them can be a variable that you have to solve for. However, amount of gas is not a variable you can solve for, which means it must remain constant.

4 0
3 years ago
Beginning
Bas_tet [7]

Answer:

yeah it is

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4 0
3 years ago
What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

4 0
3 years ago
How many grams is 20 moles of (nh4)2so4 please show work
ryzh [129]

n=20 mol

(NH)4 SO4

Atomic masses :

N- 14

H- 1

S- 32

O- 16

Therefore M= 14×2 + 1×8 + 32 + 16×4

= 132

m= nM

= 20×132

= 2640g

5 0
2 years ago
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