Identify the spectator ions in the following molecular equation.
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Answer:
Explanation:
Molecular Formula is representation of the chemical compound in terms of the symbols of all the elements that are present in the compound followed by subscripts, which give the count of each element in that compound.
We need to write the molecular formula of Tricarbon nonachloride. Tri means three, so Tricarbon means there are 3 atoms of Carbon. Likewise, nona stands for 9, so nonachloride means there are 9 atoms of chlorine. Therefore, we can represent nonachloride as:
Carbon (3 atoms) Chlorine (9 atoms) =
Thus, molecular formula of Tricarbon nonachloride is
<h3>Answer:</h3>
Longest wavelength = 343.7 nm
<h3>Solution and Explanation:</h3>In this question we need to first use the concept of energy of a photon.
Energy of a photon, E, is given by the formula, E = hf, where h is the plank's constant, f is the frequency.
But since, f is given by dividing speed, c, by wavelength, λ, then;
E = hc/λ
We are given 348 kJ/mol required to break carbon-carbon bonds.
We know that; 1 mole of bonds = 6.022 × 10^23 bonds.
We are required to find the longest wavelength with enough energy to break the C-C bonds.
This can be worked out in simple steps:
Step 1: Energy required to break one bond (kJ/bond)
1 mole of bonds = 6.022 × 10^23 bonds.
Therefore;
348 kJ = 6.022 × 10^23 bonds.
Thus;
1 bond = 348 kJ ÷ 6.022 × 10^23 bonds.
= 5.778 x 10^-22 kJ
But; 1000 joules = 1 kJ
Hence; energy per bond = 5.778 x 10^-19 Joules
Step 2: Energy per photon
Breaking one bond requires energy equivalent to energy of a photon.
Therefore;
1 photon = 5.778 x 10^-19 Joules
= 5.778 x 10^-19 J/photon
Step 3: Calculating the wavelength
From the equation of energy of a photon;
E = hc/λ
h is the plank's constant = 6.626 × 10^-34 J/s
c is the speed of light in vacuum = 2.9998 × 10^8 m/s
E is the energy of a photon = 5.778 x 10^-19 Joules
Therefore, making λ (wavelength) the subject;
= 3.437 x 10^-7 m
But; 1 nm = 10^-9 m
Thus;
wavelength = 343.7 nm
Therefore, the longest wavelength of the radiation will be 343.7 nm