The reason that some of the elements of period three and beyond are steady in spite of not sticking to the octet rule is due to the fact of possessing the tendency of forming large size, and a tendency of making more than four bonds. For example, sulfur, it belongs to period 3 and is big enough to hold six fluorine atoms as can be seen in the molecule SF₆, while the second period of an element like nitrogen may not be big to comprise 6 fluorine atoms.
The existence of unoccupied d orbitals are accessible for bonding for period 3 elements and beyond, the size plays a prime function than the tendency to produce more bonds. Hence, the suggestion of the second friend is correct.
Answer:
the sum of the numbers of protons and neutrons present in the nucleus of an atom.
Answer:
2.52L
Explanation:
Given parameters:
T₁ = 400K
V₁ = 4L
T₂ = 252K
unknown
V₂ = ?
Solution:
To solve this problem, we are going to apply charle's law. The law states that the volume of a fixed mass of gas is directly proportional to temperature provided pressure is constant.
Mathematically,
Substitute and solve for V₂
V₂ = 2.52L
The micromoles of mercury(II) iodide : 0.013 μ moles
<h3>Further explanation</h3>
Given
215.0mL of a 6.0x10⁻⁵mmol/L HgI₂
Required
micromoles of HgI₂
Solution
Molarity(M) = moles of solute per liters of solution
Can be formulated :
M = n : V
n = moles
V = volume of solution
V = 215 mL = 0.215 L
so moles of solution :
n = M x V
n = 6.10 mmol/L x 0.215 L
n = 1.312 . 10⁻⁵ mmol
mmol = 10³ micromol
so 1.312 mmol = 1.312.10⁻⁵ x 10³ = 0.01312 micromoles ⇒ 2 sif fig = 0.013 μ moles
<h3>
Answer:</h3>
1.9 moles
<h3>
Explanation:</h3>
Carbon dioxide (CO₂) is a compound that is made up of carbon and oxygen elements.
It contains 2 moles of oxygen atoms and 1 mole of carbon atoms
Therefore;
We would say, 1 mole of CO₂ → 2 moles of Oxygen atoms + 1 mole of carbon atoms
Thus;
If a sample of CO₂ contains 3.8 moles of oxygen atoms we could use mole ratio to determine the moles of CO₂
Mole ratio of CO₂ to Oxygen is 1 : 2
Therefore;
Moles of CO₂ = 3.8 moles ÷ 2
= 1.9 moles
Hence, the moles of CO₂ present in a sample that would produce 3.8 moles of Oxygen atoms is 1.9 moles
