The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
<h3>How to calculate empirical formula?</h3>
The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.
The empirical formula of the given compound can be calculated as follows:
- Hafnium = 55.7% = 55.7g
- Chlorine = 44.3% = 44.3g
First, we convert mass values to moles by dividing by the molar mass of each element
- Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
- Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol
Next, we divide each mole value by the smallest
- Hafnium = 0.312 ÷ 0.312 = 1
- Chlorine = 1.25 ÷ 0.312 = 4
Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
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Well, these particles happens to be small, like REALLY small. So microscopically small they aren't picked up or observed my the naked eye. also the vibrations are in an atomic scale which is also VERY tiny This goes for all solids too.
Answer:
that not of i grade sorry
Explanation:
Answer:
C24H50
Explanation:
The empirical fomula's molar mass is 169.25 g/mol.
We know the molecular formula's molar mass is 338 g/mol.
338/169.25= 1.99 or approximately 2
Answer:
Option B. Cation that is smaller than the original atom.
Explanation:
Magnesium is a divalent element. This implies that magnesium can give up 2 electrons to become an ion (cation) as shown below:
Mg —> Mg²⁺ + 2e¯
Next, we shall write the electronic configuration of magnesium atom (Mg) and magnesium ion (Mg²⁺). This can be written as follow:
Mg (12) = 2, 8, 2
Mg²⁺ (10) = 2, 8
From the above illustration, we can see that the magnesium atom (Mg) has 3 shells while the magnesium ion (Mg²⁺) has 2 shells.
This simply means that the magnesium ion (Mg²⁺) i.e cation is smaller that the original magnesium atom (Mg).
