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Fynjy0 [20]
3 years ago
5

What is an empirical formula

Chemistry
2 answers:
yarga [219]3 years ago
8 0
A formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.
zlopas [31]3 years ago
6 0

<em>A</em><em>n empirical </em><em> </em><em>shows the simplest </em><em> </em><em>ratio among atoms in a compound</em><em> </em><em>.</em>

<h2><em>More</em><em> </em><em>information</em></h2>

  • <em>an empirical formula</em><em> </em><em>is the simplest formula for a compound </em><em>.</em><em>For</em><em> </em><em>example</em><em>,</em><em> </em><em>H2O2</em><em> </em><em>can be </em><em> </em><em>reduced to a </em><em>si</em><em>mpler</em><em> </em><em>formula</em><em>.</em><em> </em>
  • <em>Determining a Molecular Formula from an Empirical Formula The empirical formula for a compound is P 2 O 5 . Its experimental molar mass is 284 g/mol.</em>
  • <em>For example, the empirical formula of benzene (C5H5) is Qj, - because the mole ratio of carbon am'hydrogen is 1:1.</em>
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Which measurement contains three significant figures?
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8 0
2 years ago
Read 2 more answers
Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
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