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leva [86]
3 years ago
8

Rounded to the nearest whole number, how many electrons are in an atom of zirconium? 40 51 91 131

Physics
2 answers:
alexdok [17]3 years ago
8 0
The correct answer is 40
Gre4nikov [31]3 years ago
4 0

Answer: The number of electrons in an atom of zirconium are 40.

Explanation:

Electrons are one of the subatomic particles present in an atom.

Electrons in an atom is judged by the atomic number of the atom.

Atomic number = Number of protons = Number of electrons

Atomic number of Zirconium = 40

Number of electrons will be equal to 40

Electronic configuration of Zirconium (Z = 40): 1s^22s^22p^63s^23p^64s^23d^{10}4p^65d^24d^2

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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
A what occurs when light changes direction after colliding with particles of matter
Paha777 [63]
I believe scattering is what occurs when light essentially changes direction after colliding with small particles of matter.
7 0
3 years ago
The atmosphere of Jupiter is essentially made up of hydrogen, H2. For H2, the specific gas constant is 4157 J/(kg K). The accele
Alenkinab [10]

Answer:

h=17357.9m

Explanation:

The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.

To calculate this, you need to use the barometric formula:

P=P_0e^{-\frac{Mg}{RT}h}

Where R is the gas constant, M the molar mass of the gas, g the acceleration of gravity, T the temperature and h the height.

Furthermore, the specific gas constant is defined by:

R_{H_2}=\frac{R}{M}

Therefore yo can write the barometric formula as:

P=P_0e^{-\frac{g}{R_{H_2}T}h}

at the surface of the planet (h =0) the pressure is P_0[\tex]. The pressure at the height requested is half of that:[tex]P=\frac{P_0}{2}

applying to the previuos equation:

\frac{P_0}{2} =P_0e^{-\frac{g}{R_{H_2}T}h}

solving for h:

h=17357.9m

3 0
3 years ago
A mass m is tied to an ideal spring with force constant k and rests on a frictionless surface. The mass moves along the x axis.
7nadin3 [17]

Answer:x=\frac{x_m}{\sqrt{2}}

Explanation:

Given

initially mass is stretched to x_m

Let k be the spring Constant of spring

Therefore Total Mechanical Energy is \frac{kx_m^2}{2}

Position at which kinetic Energy is equal to Elastic Potential Energy

K=\frac{mv^2}{2}

U=\frac{kx^2}{2}

it is given

k=U

thus 2U=\frac{kx_m^2}{2}

2\times \frac{kx^2}{2}=\frac{kx_m^2}{2}

2x^2=x_m^2

x=\frac{x_m}{\sqrt{2}}

3 0
4 years ago
In order for us to see the red apple, where do the wavelengths of light have to go?
tankabanditka [31]

Answer:

Explanation:650

colour* wavelength (nm) energy (eV)

red 650 1.91

orange 600 2.06

yellow 580 2.14

green 550 2.25

5 0
3 years ago
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