NO2 can react with the NO in smog, forming a bond between the N atoms. Draw the structure of the resulting compound, including f
ormal charges.
1 answer:
First, let's write down the balanced chemical reaction between the given reactants:
NO₂ + NO → N₂O + O₂
The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:
Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)
For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1
NO₂ + NO → N₂O + O₂
The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:
Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)
For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1
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Here is the full question
Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 10 cm. The two liquids do not mix.
What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.
Express your answer in two significant figures and include the appropriate units (in cm)
Answer:
ΔH ≅ 3.73 cm
Explanation:
The pressure inside a liquid is known as hydrostatic pressure and which is represent by the formula:
P = ρ × g × h
where;
ρ is the density of the fluid
g is the gravitational constant
h is the height from the surface
From the question above;
For glycerine; we have:
density of glycerine = 1260 kg/m³
gravitational constant = 9.8 m/s²
height = ???
∴
----- equation (1)
On the other hand for alcohol:
density of alcohol is given as = 790 kg/m³
gravitational constant = 9.8 m/s²
height = 10 cm
∴
----------- equation (2)
if we equate equation 1 and 2 together; we have
Making the subject of the formula, we have :
= 6.269 cm
The difference in the height denoted by ΔH can therefore be calculated as:
ΔH
ΔH
ΔH = 3.731 cm
ΔH ≅ 3.73 cm (to two significant figures)