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LiRa [457]
2 years ago
5

Would an increase in pressure favor the formation of ozone or of oxygen?

Physics
1 answer:
max2010maxim [7]2 years ago
8 0
<h3><u>Answer;</u></h3>

<u>An increase in pressure favors the formation of ozone </u>

<h3><u>Explanation;</u></h3>
  • Ozone, O3, decomposes to molecular oxygen in the stratosphere according to the reaction

2O3(g) ⇆ 3O2 (g).

  • There are more moles of product gas than moles of reactant gas. An increase in total pressure increases the partial pressure of each gas, shifting the equilibrium towards the reactants.
  • Therefore; an increase in pressure favors backward reactions towards the formation of ozone.
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Which one of the following represents an acceptable set of quantum numbers for an electron in an atom? (arranged as n, l, m l ,
Vitek1552 [10]

Answer:

The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;

(b) 4, 3, -3, 1/2.

Explanation:

To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;

4, 4, 4, 1/2

4, 3, -3, 1/2

4, 3, 0, 0

4, 5, 7, -1/2

4, 4, -5, 1/2

Let us label them as a to as follows

(a) 4, 4, 4, 1/2

(b) 4, 3, -3, 1/2

(c) 4, 3, 0, 0

(d) 4, 5, 7, -1/2

(e) 4, 4, -5, 1/2

Next we note the rules for the assignment and arrangement of quantum numbers are as follows

Number                                   Symbol                Possible values

Principal Quantum Number  .......n........................1, 2, 3, ......n

Angular momentum quantum

number...............................................l.........................0, 1, 2, .......(n - 1)

Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l  

Spin Quantum Number.................m_s.....................+1/2, -1/2

We are meant to analyze each of the arrangement for acceptability.

Therefore for (a),

we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.

Therefore (a) is not acceptable.

(b) Here we note that

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.

(c) Here we have

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 0 ∉ (+1/2, -1/2) → not acceptable

Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.

(d) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = -1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

(e) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

3 0
1 year ago
When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pull
Mandarinka [93]

Answer:

The answer is "Option b, c, and a".

Explanation:

Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.

We now understand the effects of gravity:

F \propto  M\\\\F\propto  \frac{1}{r^2}\\\\or\\\\F \propto  \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}

The force of the sun is, therefore, 10^{10} times greater and the proper sequence, therefore, option steps are:

b. Pull-on phone from earth

c. Pull-on phone from sun

a. Pull phone from you

5 0
1 year ago
The grooved pulley of mass m is acted on by a constant force F through a cable which is wrapped securely around the exterior of
Sidana [21]

Answer:

Answer; v= 1.2654m/s

T= 110.76N

Explanation:

Apply Momentum Principle

Fdtro - Mgridt = Iow +Mvr

Fdtro - Mgridt = mK2 v/r1 + Mvr1

85 x 3x 0.345 -11 x 9.81 x 0.23 x 3 =30 x 0.25 x 0.25 x v/0.23 + 11 x v x 0.23 =

v = 1.2654m/s

To find the timed average value

Tdt -Mgdt =MV

T x 3 - 11 x 9.81 x 3 = 11 x 0.778

T= 110.76N

3 0
2 years ago
When were Earth’s landmasses first recognizable as the continents we know today? 10 million years ago 135 million years ago 180
Bess [88]

Answer:

b

Explanation:

i took the test

6 0
1 year ago
Read 2 more answers
The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n
qaws [65]
Felectric = q*E 
<span> Ftranslational = m*a 
</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a 
</span><span> Solve for a 
</span><span> a = q/m*E </span>
<span> Our sign convention is "up is positive" 
</span><span> q = 1.6*10^-19 C 
</span><span> m = 1.67*10^-27 kg 
</span><span> E = -150 N/C (- because it is down and up is positive) 
</span> a =<span> -6,4*10^5</span><span> m/s^2 (downward) 
</span> answer
 a = -6,4*10^5 m/s^2 (downward) 
3 0
2 years ago
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