Consider the reaction below:
1 answer:
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The empirical formula of the initial zinc oxide is ZnO.
<h3>What is Empirical Formula?</h3>
The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.
It is the lowest whole number ratio of the element in the compound.
<h3>How to find out the empirical formula?</h3>- Find out the given masses and molar masses of the elements
The molar mass of Zn = 65 gmol⁻¹
Given the mass of Zn = 2.156 g
The molar mass of Oxygen = 16 gmol⁻¹
The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal
= (2.684 - 2.156) g
= 0.528 g
- Find the number of moles of the elements in the compound
The number of moles is given by
where m = given mass and
M = Molar mass
Number of moles of Zinc = = 0.033 moles
Number of moles of Oxygen = = 0.033 moles
- Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.
Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.
Therefore, the empirical formula of zinc oxide is ZnO.
Learn more about the empirical formula:
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Answer:
1. 0.02 M
2. 0.01 M
3. 4×10⁻⁶
Explanation:
We know that V₁S₁ = V₂S₂
1.
Concentration of HCl = 0.05 M
end point comes at = 10 ml
So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M
2.
2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)
[Ca²⁺] = 0.02 ÷ 2 = 0.01 M
3.
= [Ca²⁺(aq)] [OH⁻(aq)]²
Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)
= [0.01 × (0.02)²] = 4×10⁻⁶
4.
If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.
5.
value describes the solubility of a particular ionic compound. The higher the
value, the higher the Solubility will be.
6.
This may be due to uncommon ion effect. The process of other ions (K⁺ or Na⁺) may increase the solubility