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3 years ago

A skeleton equation shows just the reactants and products whereas a balanced equation shows the

Chemistry

1 answer:

Anettt [7]3 years ago

8 0

Answer:

A skeleton equation shows just the reactants and products whereas a balanced equation shows the relative amounts of, or proportion between, reactants and products.

Explanation:

The skeleton equation shows which reactants are being used and which products are being formed.

The reactants are shown on the left and the products are shown on the right side of the equations, separeted by an arrow.

For example, the skeleton equation to obtain water is:

H₂(g) + O₂(g) → H₂O (g)

From it you know that hydrogen and oxygen react to form water, yet you do not know in which ratio they do it.

Then, you balance the equation, adding the appropiate coefficients, to make the number of atoms of each kind on the reactant side equal to the number of the same kind of atoms on the product side.

This is, for the example of water, the number of hydrogen atoms on the left must equal the number of atoms of hygrogen on the right side, and the number of oxygen atoms of the left must equal the number of oxygen atoms on the right.

For the water example that is:

2H₂(g) + O₂(g) → 2H₂O (g)

Showing that 2 molecuies of hydrogen (or 4 atoms) react with 1 molecule of oxygen (or 2 atoms) to produce 2 molecules of water, and that proportion (relative amounts) will always be true for that reaction.

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for example; whale, dolphin, sharks, or turtles etc.

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3 years ago

Which of these statements best explains why pure water has a pH of 7? Pure water contains only H2O molecules, but pH is based on

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Read 2 more answers

If there are 1.55 x 1024 molecules of hydrogen peroxide (H2O2), what is the mass of the sample?

bezimeni [28]

Answer:

87.54 g of H₂O₂

Explanation:

From the question given above, the following data were obtained:

Number of molecules = 1.55×10²⁴ molecules

Mass of H₂O₂ =.?

From Avogadro's hypothesis,

6.02×10²³ molecules = 1 mole of H₂O₂

Next, we shall determine the mass of 1 mole of H₂O₂. This can be obtained as follow:

1 mole of H₂O₂ = (2×1) + (2×16)

= 2 + 32

= 34 g

Thus,

6.02×10²³ molecules = 34 g of H₂O₂

Finally, we shall determine mass of H₂O₂ that contains 1.55×10²⁴ molecules. This can be obtained as follow:

6.02×10²³ molecules = 34 g of H₂O₂

Therefore,

1.55×10²⁴ molecules

= (1.55×10²⁴ × 34)/6.02×10²³

1.55×10²⁴ molecules = 87.54 g of H₂O₂

Thus, 87.54 g of H₂O₂ contains 1.55×10²⁴ molecules.

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3 years ago

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3 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

3 years ago