For each of the following reactions, identify the missing reactant(s) or products(s) and then balance the resulting equation. Note that each empty slot may require one or more substances. a. synthesis: ___ ⟶Li2O
b. decomposition:Mg(ClO3)2⟶___, c. double displacement: HNO3+Ca(OH)2→___, d. combustion: C5H12+O2⟶___

a. Synthesis: ___ → Li₂O

Reactants are: Li and O₂

Balanced chemical reaction: 4Li + O₂ → 2Li₂O

b. Decomposition :Mg(ClO₃)₂→ ___

Products are: MgCl₂ and O₂

Balanced Chemical reaction: Mg(ClO₃)₂→ MgCl₂ + 3O₂

c. Double displacement: HNO₃ + Ca(OH)₂ →___

Products are: Ca(NO₃)₂ and H₂O

Balanced Chemical reaction: 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O

d. Combustion: C₅H₁₂+O₂→ ___

Products are:  CO₂ and H₂O

Balanced Chemical reaction: C₅H₁₂+ 8O₂→ 5CO₂ + 6H₂O

Lets understand these reaction types:

1. Synthesis: Chemical synthesis is the process in which chemical reactions are performed with the idea of converting a reactant into a product or multiple products.

For example: 4Li + O₂ → 2Li₂O

2. Decomposition: In these reactions chemical species break up into simpler parts.

For example: Mg(ClO₃)₂→ MgCl₂ + 3O₂

3. Double displacement: is a type of chemical reaction where two compounds react, and positive ions and the negative ions of the two reactants switch places, forming two new compounds or products.

For example:  2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O

4. Combustion: These reactions occur when oxygen reacts with another substance and gives off heat and light.

For example: C₅H₁₂+ 8O₂→ 5CO₂ + 6H₂O

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A. The reactants are= Li and O2

The balanced equation is:

4Li + O2 —> 2Li2O

B. The products are: MgCl2 and O2

The balanced equation is:

Mg(ClO3)2 —> MgCl2 + 3O2

C. The products are: Ca(NO3)2 and H2O

The balanced equation is:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 +

2H2O

D. The products are: CO2 and H2O

The balanced equation is:

C5H12 + 8O2 —> 5CO2 + 6H2O

Explanation:

A. —> Li2O

The reactants are Li and O2. Thus the equation is given below:

Li + O2 —> Li2O

Thus the equation is balanced as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Li2O as shown below:

Li + O2 —> 2Li2O

Now, we have 4 atoms of Li on the right side and 1 atom on the left. It can be balance by putting 4 in front of Li as shown below:

4Li + O2 —> 2Li2O

Now the equation is balanced

B. Mg(ClO3)2 —> __

The products are MgCl2 and O2

The equation is given below:

Mg(ClO3)2 —> MgCl2 + O2

The equation can be balance as follow:

There are 6 atoms of O on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

Mg(ClO3)2 —> MgCl2 + 3O2

Now the equation is balanced

C. HNO3 + Ca(OH)2 —>___

The products are: Ca(NO3)2 and H2O

The equation is given below:

HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

The equation is balanced as follow:

There are 2 atoms of NO3 on the right side and 1 atom on the left. It can be balance by putting 2 in front of HNO3 as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

There are a total of 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + 2H2O

Now the equation is balanced

D. C5H12 + O2 —>__

The products are: CO2 and H2O

The equation is given below:

C5H12 + O2 —> CO2 + H2O

The equation can be balance as follow:

There are 5 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 5 in front of CO2 as shown below:

C5H12 + O2 —> 5CO2 + H2O

There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:

C5H12 + O2 —> 5CO2 + 6H2O

Now, there are a total of 16 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 8 in front of O2 as shown below:

C5H12 + 8O2 —> 5CO2 + 6H2O

Now we can see that the equation is balanced.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

it is the most commonly accepted one