Using the periodic table to locate each element, write the electron configuration of(c) Re.
1 answer:
Rhenium is a chemical element with the symbol Re and atomic number 75. The electron configuration of Re is [Xe] .
1. Identify the given element and its atomic number from the periodic table.
2. Write the electron configuration by the energy level and the type of orbital first, then the number of electrons present in the orbital as superscript.
The easiest way to write the electronic configuration for any element is by using a diagonal rule for electron filling order in the different subshells according to the Aufbau principle.
The 3 rules for writing the electron configuration in the orbital box diagram are – the Aufbau rule, the Pauli-exclusion rule, and Hund's Rule.
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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.
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Answer:
The answer to your question is given after the questions so I just explain how to get it.
Explanation:
a)
Get the molecular weight of Phosphoric acid
H₃PO₄ = (3 x 1) + (31 x 1) + (16 x 4)
= 3 + 31 + 64
= 98 g
98 g ----------------- 1 mol
0.045 g --------------- x
x = (0.045 x 1) / 98
x = 0.045 / 98
x = 0.00046 moles or 4.6 x 10 ⁻⁴
b)
Molarity =
Molarity =
Molarity = 0.0013 or 1.31 x 10⁻³
c)
Formula C₁V₁ = C₂V₂
V₁ = C₂V₂ / C₁
Substitution
V₁ = (0.0013)(1) / 0.01
Simplification and result
V₁ = 0.0013 / 0.1
V₁ = 0.13 l = 130 ml