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pantera1 [17]
11 months ago
8

provide 1 example of a balanced reaction for each of the 6 types of chemical reactions: 1. synthesis 2. single-replacement 3. do

uble-replacement 4. decomposition 5. combustion 6. acid-base neutralization
Chemistry
1 answer:
m_a_m_a [10]11 months ago
3 0

Water is an example of a balanced synthesis equation: 2 H2(g) + O2(g) → 2 H2O(g). The number of atoms in the reactants and the number of atoms in the products must balance in a chemical equation.

<h3>What is balanced equation?</h3>

A chemical reaction equation is said to be balanced if both the reactants and the products have the same total charge and the same number of atoms in the reaction. In other words, both sides of the reaction have an equal mass and charge balance.

The Law of Conservation of Mass states that mass cannot be created or destroyed in a chemical reaction. On the left and right sides of a balanced chemical equation, there are an equal number of atoms of each element.

Therefore,

Water: 2 H2(g) + O2(g) → 2 H2O(g)

2K + 2H2O → 2KOH + H.

NaCN ( a q ) + HBr ( a q ) → NaBr ( a q ) + HCN ( g )

2HgO(s)→2Hg(l)+O2(g)

2CH3OH + 3O2 → 4H2O + 2CO2

HCl(aq)+NaOH(aq)⇌NaCl(aq)+H2O

To learn more about balanced equation, refer to:

brainly.com/question/26694427

#SPJ4

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7 0
2 years ago
Read 2 more answers
A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following
marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

\frac{V_1}{n_1} =\frac{V_2}{n_2}

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}

<em>V₂ = 4,00L</em>

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}

<em>V₂ = 16,00L</em>

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}

<em>V₂ = 12,31L</em>

I hope it helps!

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