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Vsevolod [243]
2 years ago
7

Identify the precipitation reactions. drag the appropriate items to their respective bins.

Chemistry
2 answers:
ycow [4]2 years ago
8 0

The precipitation reactions :

2KBr + Pb(NO₃)₂ -----> PbBr₂ + 2KNO₃

Hg₂(C₂H₃O₂)₂ + 2NaI ----> Hg₂I₂ + 2NaC₂H₃O₂

Ba(NO₃)₂ + K₂SO₄ -----> BaSO₄ + 2KNO₃

<h3>Further explanation </h3>

There are two types of chemical reactions that may occur,

namely single-replacement reactions and double-replacement reactions.

1. A single replacement reaction is a chemical reaction in which one element replaces the other elements of a compound to produce new elements and compounds

Not all of these reactions can occur. We can use the activity series, which is a list of elements that can replace other elements below / to the right of them in a single replacement reaction.

2. Double-Replacement reactions. Happens if there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product

To predict whether this reaction can occur or not is to see the precipitation reaction. A precipitation reaction occurs if two ionic compounds which are dissolved reacted to produce one of the products of the ion compound does not dissolve. Formation of these precipitating compounds that cause reactions can occur

Solubility Rules:

  • 1. soluble compound

All compounds of Li +, Na +, K +, Rb +, Cs +, and NH4 +

All compounds of NO₃⁻ and C₂H₃O₂⁻

Compounds of Cl−, Br−, I− except Ag⁺, Hg₂²⁺, Pb²⁺

Compounds of SO₄²⁻ except Hg₂²⁺, Pb²⁺, Sr²⁺, Ba²⁺

  • 2. insoluble compounds

Compounds of CO₃²⁻ and PO₄³⁻ except for Compounds of Li +, Na +, K +, Rb +, Cs +, and NH₄ +

Compounds of OH− except Compounds of Li +, Na +, K +, Rb +, Cs +, NH₄⁺, Sr²⁺, and Ba²⁺

Let's look at possible reactions that are not shown in the question

1. S + O₂ ---> SO₂

SO₂ = gas so this reaction not the precipitation reactions

 2. H₂O₂ + NaIO ---> NaI + H₂O + O₂

NaI = soluble so this reaction not the precipitation reactions

3. 2KBr + Pb(NO₃)₂ -----> PbBr₂ + 2KNO₃

PbBr₂ precipitates = so this reaction the precipitation reactions

4. HCl + KOH ----> KCl + H₂O

KCl soluble so this reaction not the precipitation reactions

5. HBr + KOH ----> KBr + H₂O

KBr soluble so this reaction not the precipitation reactions

6. Hg₂(C₂H₃O₂)₂ + 2NaI ----> Hg₂I₂ + 2NaC₂H₃O₂

Hg₂I₂ precipitates = so this reaction the precipitation reactions

7. Ba(NO₃)₂ + K₂SO₄ -----> BaSO₄ + 2KNO₃

BaSO₄ precipitates = so this reaction the precipitation reactions

<h3>Learn more</h3>

the rate of a chemical reaction

brainly.com/question/807610

the reducing agent

brainly.com/question/6966537

substance loses electrons in a chemical reaction

brainly.com/question/2278247

Keywords: a double-replacement equation, A single-replacement reaction

lana66690 [7]2 years ago
7 0
I can't fully answer this question because it is incomplete. In order for me to help you, I could just define what a precipitation reaction is and give a concrete example.

A precipitation reaction consists of two aqueous solutions that when reacted together, forms an insoluble salt. For example, 

AgNO₃ (aq) + HCl (aq) --> AgCl (s) + HNO₃ (aq)

In this case, the precipitate is AgCl, Silver Chloride, which appears as a white solid.
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Which event is most likely an example of a chemical reaction
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Explanation:

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Answer:

Explanation:

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6 0
3 years ago
In the following reaction, how many grams of NaBr will produce 244 grams of NaNO3? Pb(NO3)2(aq) 2 NaBr(aq) PbBr2(s) 2 NaNO3(aq)
lyudmila [28]

Answer : The mass of NaBr is, 295.323 grams

Solution :

First we have to calculate the moles of NaNO_3.

\text{Moles of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Molar mass of }NaNO_3}=\frac{244g}{85g/mole}=2.87moles

Now we have to calculate the moles of NaBr.

The balanced chemical reaction is,

Pb(NO_3)_2(aq)+2NaBr(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)

From the balanced reaction we conclude that

As, 2 moles of NaNO_3 react with 2 moles of NaBr

So, 2.87 moles of NaNO_3 react with 2.87 moles of NaBr

Now we have to calculate the mass of NaBr.

\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr

\text{Mass of }NaBr=(2.87mole)\times (102.9g/mole)=295.323g

Therefore, the mass mass of NaBr is, 295.323 grams

7 0
2 years ago
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