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levacccp [35]
3 years ago
12

Assume that a P orbital is dumbbell shaped the two major lobes in a certain atom contains one electron. In which lobe is electro

n most likely to be found? Explain, indicating the principle on which the answer is based.
Chemistry
1 answer:
Tanya [424]3 years ago
5 0

Answer:

See explanation

Explanation:

The p orbital is threefold degenerate. This implies that the p-sublevel is composed of three orbitals; px, py and pz.

According to Hund's rule, electrons occur singly when filling degenerate orbitals before pairing takes place. Since the three orbitals are degenerate, any of px, py or pz may be first filled.

If one lobe of any of the px, py or pz is first filled, the next electron must go into the next degenerate orbital. When all are filled, pairing of electron spins may now begin.

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The arrangement of particles is least ordered in a sample of
lora16 [44]

Answer:

The arrangement of particles is least ordered is  a > c > b > d

Explanation:

Given that,

The arrangement of particles is least ordered in a sample of NaCl.

We know that,

The least ordered arrangement are.

The form of state of particle in least ordered.

So, The least ordered arrangement will be

aqua > gas > liquid > solid

The arrangement of particles is least ordered in a sample of NaCl

NaCl(aq) > NaCl(g) > NaCl(l) > NaCl(s)

Hence, The arrangement of particles is least ordered is a > c > b > d

4 0
3 years ago
g The most common position for an double bond in an unsaturated fatty acid is delta _________(fill in the number).
Stolb23 [73]

Answer:

The most common position for an double bond in an unsaturated fatty acid is delta 9 (Δ⁹)

Explanation:

Unsaturated fatty acids are carboxylic acids which contains one or more double bonds. The chain length as well as the number of double bonds is written separated by a colon. The positions of the double bonds are specified starting from the carboxyl carbon, numbered as 1, by superscript numbers following a delta (Δ). For example, an 18-carbon fatty acid containing a  single double bond between carbon number 9 and 10 is written as 18:1(Δ⁹).

In most monounsaturated fatty acids, the double bond is between C-9 and C-10 (Δ⁹), and the other double bonds of polyunsaturated fatty acids are generally Δ¹² and Δ¹⁵. This positioning is due to the nature of the biosynthesis of fatty acids. In the mammalian hepatocytes, double bonds are introduced easily into fatty acids at the Δ⁹ position, but cannot introduce additional double bonds between C-10 and the methyl-terminal end. However, plants are able to introduce these additional double bonds at the  Δ¹² and Δ¹⁵ positions.

5 0
4 years ago
Bronze is an example of which of the following:<br><br> element<br> atom<br> solution<br> solvent
Brut [27]
It is a solution, because it's the alloy of tin and copper 
8 0
4 years ago
Choose all the answers that apply.
Marrrta [24]

Well the universe is immensely  Huge and contains Stars, Gas, Dust, Planets,

and Energy (from star and planets).

Hope this helped

4 0
4 years ago
A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,
Dafna1 [17]

Answer : The concentration of A,B\text{ and }C at equilibrium are 0.132 M, 0.232 M  and 0.168 M  respectively.

Explanation :

The given chemical reaction is,

A(aq)+B(aq)\rightleftharpoons C(aq)

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

\Delta G^o=-RT \ln k

where,

\Delta G^o = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = 25^oC=273+25=298K

Now put all the given values in the above relation, we get:

-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k

k=5.45

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

                          A(aq)+B(aq)\rightleftharpoons C(aq)

Initially               0.30      0.40         0  

At equilibrium  (0.30-x) (0.40-x)     x

The expression of equilibrium constant will be,

k=\frac{[C]}{[A][B]}

5.45=\frac{x}{(0.30-x)\times (0.40-x)}

By solving the term x, we get

x=0.168\text{ and }0.716

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of A at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of B at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of C at equilibrium = x = 0.168 M

3 0
4 years ago
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