Relation between glancing and angle of deviation
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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: 167.2 J
Explanation:
Answer:
Explanation:
1. Write the skeleton equation for the half-reaction
NO₃⁻ ⟶ N₂O
2. Balance all atoms other than H and O
2NO₃⁻ ⟶ N₂O
3. Balance O by adding H₂O molecules to the deficient side.
2NO₃⁻ ⟶ N₂O + 5H₂O
4. Balance H by adding H⁺ ions to the deficient side.
2NO₃⁻ + 10H⁺ ⟶ N₂O + 5H₂O
5. Balance charge by adding electrons to the deficient side.
2NO₃⁻ + 10H⁺ + 8e⁻ ⟶ N₂O + 5H₂O
The amount of charge required to reduce 2 mol of NO₃⁻ is 8 F