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3 years ago

8

Please helpppppppppppp

1 answer:

tatuchka [14]3 years ago

4 0

In chemistry, a valence electron is an outer shell electron that is associated with an atom and that can participate in the formation of a chemical bond only the outer shell is not closed. So, the answer to this question wold be outermost energy level.

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How does geosphere interact with biosphere

Marina86 [1]

<span>With a partner, describe interactions in this scene, tracing the movement of materials or energy through all four of Earth's spheres if possible. Plants (biosphere) draw water (hydrosphere) and nutrients from the soil (geosphere) and release water vapor into the atmosphere.</span>

6 0

3 years ago

The heating element of an iron operates at 110 V with a current of 9A. (a) What is the resistance of the iron?

Dmitrij [34]

Answer:

(A) 12.222 ohm (B) 990 W

Explanation:

We have given the voltage of the heating element V = 110 V

The current in the heating element i = 9 A

(a) According to ohm's law V =iR

So $110=9\times R$

$R=\frac{110}{9}=12.222ohm$

(b) The power dissipated in the resistor is given by $P=i^2R=9^2\times 12.222=990W$

So the power dissipated = 990 W

3 0

3 years ago

Light hits ocean during

Lostsunrise [7]

Answer:

no way to tell since the ocean surface is moving too violently it's not possible to determine the reflected angle

4 0

2 years ago

Help me please please help

andrew-mc [135]

Answer:

The answer is C.

Explanation:

If you have ever tried doing so, hair stick to the ballon. Opposites attract as well, so the answer is C.

7 0

3 years ago

An electron is accelerated from rest by a potential difference of 412 V. It then enters a uniform magnetic field of magnitude 18

Salsk061 [2.6K]

Explanation:

Given that,

Potential difference, V = 412 V

Magnitude of magnetic field, B = 188 mT

(a) The potential energy of electron is balanced by its kinetic energy as :

$eV=\dfrac{1}{2}mv^2$

v is speed of the electron

$v=\sqrt{\dfrac{2eV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 412}{9.1\times 10^{-31}}} \\\\v=1.2\times 10^7\ m/s$

(b) When the charged particle moves in magnetic field, it will move in circular path. The radius of the circular path is given by :

$r=\dfrac{mv}{eB}\\\\r=\dfrac{9.1\times 10^{-31}\times 1.2\times 10^7}{1.6\times 10^{-19}\times 188\times 10^{-3}}\\\\r=3.63\times 10^{-4}\ m$

Hence, this is the required solution.

7 0

3 years ago