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vladimir2022 [97]
3 years ago
10

A student with an unknown code of Q3A conducts a qualitative analysis of her unknown. Her results indicated the formation of a y

ellow precipitate and a grey precipitate. From her results, what ions can you confirm were not in her unknown sample?
Chemistry
1 answer:
insens350 [35]3 years ago
6 0

Answer:

Grey precipitate implies the presence of silver ions

Yellow precipitate implies the presence of lead II ions

Explanation:

Qualitative analysis provides us a quick method of identifying ions present in a sample by chemical reactions involving simple reagents. Precipitates having a unique colour is formed. The identity of ions in the sample is deduced from the colour of precipitate obtained when particular reagents are added.

In the question, a precipitate containing silver ions upon standing turn into grey colour. Similarly, lead II ions give a yellow precipitate.

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Which equation represents a neutralization reaction?
bulgar [2K]

Answer: Hello, there! your Answer is Below

The answer is letter C

Explanation:

A. 4Fe(s) + 3O2(g) → 2Fe2O3(s), 2H2(g) + O2(g) → 2H2O() and AgNO3(aq) + KCl(aq) → KNO3(aq) + AgCl(s) are not neutralization reaction.

Hope this Helps you!!!

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4 0
3 years ago
Read 2 more answers
How many grams of hydrochloric acid are produced when 15.0 grams NaCl react with excess H2SO4 in the reaction
e-lub [12.9K]

Answer:

11.65 g

Explanation:

Amount of NaCl  = 15.0 grams

amount of H₂SO₄ = Excess

mass of hydrochloric acid (HCl) = ?

Solution:

To solve this problem first we will look for the reaction that NaCl react with  H₂SO₄

Reaction:

         2NaCl + H₂SO₄ --------> 2 HCl + Na₂SO₄

As the  H₂SO₄ is in excess so the amount of hydrochloric acid (HCl) depends on the amount of NaCl as it its act as limiting reactant.

Now if we look at the reaction

         2NaCl + H₂SO₄ --------> 2HCl + Na₂SO₄

           1 mol                              2 mol  

Now convert moles to mass

Then

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

So,

            2NaCl           +      H₂SO₄     -------->       2HCl        +     Na₂SO₄

      2 mol (58.5 g/mol)                                    2 mol (36.5 g/mol)

              94 g                                                           73 g

So if we look at the reaction; 94 g of NaCl gives with 73 g of hydrochloric acid (HCl) in a this reaction, then how many grams of hydrochloric acid (HCl) will produce from 15.0 g of NaCl

For this apply unity formula

           94 g of NaCl  ≅ 73 g of HCl

           15.0 g of NaCl  ≅ X g of HCl

By Doing cross multiplication

           X g of HCl = 73 g x 15.0 g / 94 g

           X g of HCl = 11.65 g

11.65 g of hydrochloric acid (HCl) will produce by 15.0 g of NaCl

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