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Readme [11.4K]
2 years ago
13

Nitric oxide occurs in the tropospheric nitrogen cycle, but it destroys ozone in the stratosphere.(a) Write equations for its re

action with ozone and for the reverse reaction.
Chemistry
1 answer:
sukhopar [10]2 years ago
5 0

Nitric oxide occurs in the tropospheric nitrogen cycle, but it destroys ozone in the stratosphere. The equation for its reaction with ozone is :

                                 NO + O3 → NO2 + O2

The equation of reverse reaction is:

                                 NO2 + O2 →  NO + O3

When Nitrogen oxide reaches the  stratosphere, it breaks down to form nitrogen oxides. These reactions causes ozone-depletion. This gas has become the largest ozone destroying factor.

Nitrogen oxide is produced due to fertilizers and at sewage treatment plants. In 1980s, Man-made CFCs started to make holes on the ozone but it was banned due to the treaty of Montreal Protocol. Regular Emission of NO can cause worse situation than CFCs.

If you need to learn more about the production of Nitric oxide, click here

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Which of the following possible outcomes regarding the universe do scientists think is most likey?
hichkok12 [17]
The universe will continue to expand
5 0
3 years ago
Read 2 more answers
What is the difference between a substance at low temperature and one at high temperature
givi [52]

Answer:

Explained below.

Explanation:

A substance at low temperature simply means that the average energy of molecular motion in that substance is low while at higher temperature, the average energy of molecular ml tip in that substance is high.

4 0
3 years ago
Comment Both propane and benzene are hydrocarbons. As a rule,
kozerog [31]

The enthalpy change : -196.2 kJ/mol

<h3>Further explanation  </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

Reaction

2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)

∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

\tt \Delta H_{rxn}=2.(-285.8)-2.(-187.8)\\\\\Delta H_{rxn}=-571.6+375.4=-196.2~kJ/mol\rightarrow \Delta Hf~O_2=0

5 0
3 years ago
Use wedge-bond perspective drawings (if necessary) to sketch the atom positions in a general molecule of formula (not shape clas
yuradex [85]

The atom positions in a general molecule of formula (not shape class) AXn that has shape square pyramidal at the corers of square and one at the above center of the square.

<h3>What is square pyramidal?</h3>

The square pyramidal is a shape geometry of the hybridization in which it consists of one lone pair and 5 bond pairs of electrons that repel each other and due to which the geometry changes from octahedral to square pyramidal.

As atoms are located at the four corners of the planer and one atom at the above center of the planner which is repelled by 4 atoms present at the corner of the planer.

Therefore, the atom positions in a general molecule of formula (not shape class) AXn that has a shape square pyramidal at the corners of the square and one at the above center of the square.

Learn more about square pyramidal, here;

brainly.com/question/8742529

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7 0
1 year ago
What is the half-life of an isotope that decays to 6.25% of its original activity in 18.9 hours?
inessss [21]
Radioactive material obeys 1st order decay kinetics,
For 1st order reaction, we have 
k = \frac{2.303}{t}Xlog \frac{\text{initial conc.}}{\text{final conc.}}
where, k = rate constant of reaction

Given: Initial conc. 100, Final conc. = 6.25, t = 18.9 hours

∴ k = \frac{2.303}{18.9} X log \frac{100}{6.25} = 0.1467 hours^(-1)

Now, for 1st order reactions: half life = \frac{0.693}{k} =  \frac{0.693}{0.1467} = 4.723 hours.


8 0
3 years ago
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