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timurjin [86]
3 years ago
13

Which of the following best describes an empirical formula?

Chemistry
1 answer:
Monica [59]3 years ago
6 0

Answer:

Option C :

a chemical formula that shows the relative number of each type of atom in a molecule, using the smallest possible ratio

Explanation:

Empirical Formula:

Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.

So,

Tha ration of the molecular formula should be divided by whole number to get the simplest ratio of molecule

For Example

      C₂H₆O₂ Consist of  Carbon (C), Hydrogen (H), and Oxygen (O)

Now

Look at the ratio of these three atoms in the compound

                         C : H : O

                         2 : 6 : 2

Divide the ratio by two to get simplest ratio

                          C : H : O

                         2/2 : 6/2 : 2/2

                             1 : 3 : 1

So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1

So the empirical formula will be

                     Empirical formula of C₂H₆O₂ = CH₃O

So, Option C is correct :

a chemical formula that shows the relative number of each type of atom in a molecule, using the smallest possible ratio

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Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

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[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

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This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

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