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MrRissso [65]
2 years ago
8

How many resonance structures can be drawn for N2O5 (no N-N bond) (minimal formal charge)?

Chemistry
1 answer:
Nezavi [6.7K]2 years ago
8 0
There are approximately 4 resonance structures that can be drawn for N205 (no N-N bond) (minimal formal charge). 
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Balance the following and label what type of reaction is taking place:
Vadim26 [7]

4,27,20,18

I hope this helps.

3 0
2 years ago
What are the pH of these solutions?
Tju [1.3M]

Answer:

The answer to your question is below

Explanation:

a)    HCl  0.01 M

      pH = -log [0.01]

      pH = - (-2)

     pH = 2

b)    HCl = 0.001 M

      pH = -log[0.001]

      pH = -(-3)

      pH = 3

c)    HCl = 0.00001 M

       pH = -log[0.00001]

       pH = - (-5)

      pH = 5

d) Distilled water

      pH = 7.0

e) NaOH = 0.00001 M

       pOH = -log [0.00001]

       pOH = -(-5)

       pH = 14 - 5

       pH = 9

f)  NaOH = 0.001 M

      pOH =- log [0.001]

      pOH = 3

      pH = 14 - 3

      pH = 11

g)   NaOH = 0.1 M

       pOH = -log[0.1]

       pOH = 1

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3 0
3 years ago
Which of the following would label the boxes correctly?
choli [55]
The last one 1) exothermic; 2) exothermic
5 0
2 years ago
What is a limitation of using a chemical formula, such as C6H12O6, to represent a compound?
Kaylis [27]
The chemical formula does not show how the atoms are connected to one another.
When we write the chemical formula of any substance, we are not able to understand the spatial arrangement of that substance's atoms. This is extremely important in organic compounds, which exhibit different physical characteristics as well as different chemical characteristics due to the way their atoms are arranged in space. These isomers are known as enantiomers.
5 0
3 years ago
Read 2 more answers
2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0
3 years ago
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