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SVETLANKA909090 [29]
4 years ago
15

If aqueous solution of lead (ll) nitrate and sodium sulfate, which insoluble precipitate is formed

Chemistry
1 answer:
forsale [732]4 years ago
7 0

Answer:

Lead(II) sulfate

Explanation:

This looks like a double displacement reaction, in which the cations change partners with the anions.

The possible products are

Pb(NO₃)₂ (aq)+ Na₂SO₄(aq) ⟶PbSO₄(?) + 2NaNO₃(?)

To predict the product, we must use the solubility rules. Two important ones for this question are:

  1. Salts containing Group 1 elements are soluble.
  2. Most sulfates are soluble, but PbSO₄ is an important exception.

Thus, NaNO₃ is soluble and PbSO₄ is the precipitate.

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In a chemical reaction, 36 grams of hydrochloric acid reacts with 40 grams of sodium hydroxide to produce sodium chloride and wa
Elena L [17]
The answer is <span>The mass of sodium chloride formed is less than 76 grams.
</span>
Remember Mass is never gained or loss in any chemical reactions

So,

hydrochloric acid  + sodium hydroxide  =>  <span>sodium chloride + water
                          Total 76 gram                         Total will also 76 gram

In this case, the mass of sodium chloride will be less than 76 gram because we need to take out the mass of the water.

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3 0
4 years ago
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denpristay [2]
Hydrogen and oxygen! Happy holidays btw
3 0
3 years ago
Sulfur reacts with fluorine to produce three different compounds. The mass ratio of fluorine to sulfur for each compound is give
NARA [144]

Answer:

Explanation:

Mass of F / Mass of S = 2.962/1 =2.962 X 32 / 32 = 94.78/32

Mass of F / Mass of S = 2.370 /1 = 2.370 X 32 / 32 = 75.84 /32

Mass of F /Mass of S = 3.555/1 = 3.555 x 32 / 32 = 113.76 / 32 .

Now constant mass of S that is 32 g reacts with different mass of F. They are as follows :

94.78 g , 75.84 g , and 113.76 g

Their ratio = 94.78 : 75.84 : 113.76

divide them by 19

their ratio = 5 : 4 : 6

So these data are consistent with law of multiple proportion.

8 0
3 years ago
The periodic table is based on an element's
nasty-shy [4]

Answer:

Atomic number

Explanation:

7 0
3 years ago
Describe a qualitative test for sulfate in alum crystals using ionic reactions of barium chloride (BaCl2)
Lady bird [3.3K]

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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8 0
2 years ago
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