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3 years ago

The number that describes the main energy level of an electron in an atom is

Physics

1 answer:

Otrada [13]3 years ago

7 0

Answer:

the principal quantum number (n)

Explanation:

Quantum numbers are found in the mathematical description given to each orbital (Schrödinger's wave equation). Schrödinger is the 5th atomic model and it seeks to describe the characteristics of all the electrons in an atom. These digits are represented by letters and each of them indicates the position and energy of each of them.

The main quantum number always has integer and positive numbers and allows us to know the energy level of the orbital as well as its size. N determines the measure of the orbital, so the greater the probability of finding an electron near the nucleus of an atone, the energy of the orbital increases. All those groups of orbitals that share the same value of N are known as level or layer.

The values that the main quantum number (n) can adopt can be positive integers such as: n = 1, 2, 3, 4, 5, 6, 7

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10 kg of R-134a fill a 1.115-m^3 rigid container at an initial temperature of -30∘C. The container is then heated until the pres

Sidana [21]

Answer : The final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa

Explanation :

First we have to calculate the initial pressure of gas.

As we know that, R-134a is 1,1,1,2-Tetrafluoroethane.

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT$

where,

P = pressure of gas = ?

V = volume of gas = $1.115m^3$

T = temperature of gas = $-30^oC=273+(-30)=243K$

R = gas constant = $8.314m^3Pa/mole.K$

w = mass of gas = 10 kg = 10000 g

M = molar mass of R-134a gas = 102.03 g/mole

Now put all the given values in the ideal gas equation, we get:

$P\times 1.115m^3=\frac{10000g}{102.03g/mole}\times (8.314m^3Pa/mole.K)\times (243K)$

$P=177587.9687Pa=177.6kPa$

Thus, the initial pressure of gas is, 177.6 kPa

Now we have to calculate the final temperature of gas.

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$

or,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 177.6 kPa

$P_2$ = final pressure of gas = 200 kPa

$T_1$ = initial temperature of gas = $-30^oC=273+(-30)=243K$

$T_2$ = final temperature of gas = ?

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

$\frac{177.6kPa}{243K}=\frac{200kPa}{T_2}$

$T_2=273.6K$

Thus, the final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa

4 0

3 years ago

A 65-cm segment of conducting wire carries a current of 0.35

algol13

The intensity of the magnetic force exerted on the wire due to the presence of the magnetic field is given by
$F=ILB \sin \theta$
where
I is the current in the wire
L is the length of the wire
B is the magnetic field intensity
$\theta$ is the angle between the direction of the wire and the magnetic field

In our problem, L=65 cm=0.65 m, I=0.35 A and B=1.24 T. The force on the wire is F=0.26 N, therefore we can rearrange the equation to find the sine of the angle:
$\sin \theta= \frac{F}{ILB}= \frac{0.26 N}{(0.35 A)(0.65 m)(1.24 T)}=0.922$

and so, the angle is
$\theta=\arcsin(0.921)=67.1^{\circ}$

6 0

3 years ago

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A ring of diameter 7.70 cm is fixed in place and carries a charge of 5.00 mC uniformly spread over its circumference. (a) How mu

MatroZZZ [7]

Answer:

3.974 Joule

Explanation:

Diameter of ring = 7.7 cm

a = Distance from the center = d/2 = 3.85 cm = 0.0385 m

Q = Charge = 5 mC

q = Charge to move = 3.4 mC

k = Coulomb constant = 9×10⁹ Nm²/C²

Work done will be equal to Potential energy when mass is at center

$U=\frac{kQq}{a}\\\Rightarrow U=\frac{9\times 10^9\times 5\times 10^{-6}\times 3.4\times 10^{-6}}{0.0385}=3.974\ J$

∴ Work to move a tiny 3.4 mC charge from very far away to the center of the ring is 3.974 Joule

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3 years ago

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Which events may occur when ocean salinity increases? Check all that apply.

Pavlova-9 [17]

The answer is number two, number four, and number one

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3 years ago

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The moon is smaller and more dense than the Earth, and has less extreme temperature changes.

Lerok [7]

<span>The moon is smaller and more dense than the Earth, and has less extreme temperature changes. The statement presented is True. In terms of temperature, since there is no atmosphere on the moon, then it has less extreme temperature changes. The moon can reach 253 Fahrenheit in the day and -387 Fahrenheit at night.</span>

8 0

3 years ago

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