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Kazeer [188]
3 years ago
5

Which of the following explains how electricity flows? Protons flow from the positive end of a source towards the negative end.E

lectrons flow from the negative end of a source towards the positive end.
Protons flow from the negative end of a source towards the positive end.


Electrons flow from the positive end of a source towards the negative end.

Chemistry
1 answer:
zalisa [80]3 years ago
7 0
I know for a fact the answer would (not) be the last option. This would not be a way that electricity would be able to really flow.

Your correct answer would be of when the "e<span>lectrons flow from the negative end of a source towards the positive end."

This would be because when the electrons would be free, they would have enough energy to then move from the negative side of the end, to the positive side.</span>

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What is the frequency of an X-ray that has a wavelength of 1.5 x 10 -9 m? The speed of electromagnetic radiation is 3.0 x 10 8 m
AlexFokin [52]

Answer:

2.0x10¹⁷ Hz is the frequency of the X-ray

Explanation:

We can find the frequency of a wave of energy from the wavelenght and its speed using the formula:

v = λƒ

<em>Where v is speed (For electromagnetic radiation = 3.0x10⁸m/s)</em>

<em>λ is the wavelength in meters = 1.5x10⁻⁹m</em>

<em>And f is the frequency in s⁻¹ = Hz</em>

<em />

Replacing:

3.0x10⁸m/s = 1.5x10⁻⁹m*ƒ

3.0x10⁸m/s / 1.5x10⁻⁹m = f

f =

<h3>2.0x10¹⁷ Hz is the frequency of the X-ray</h3>

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3 0
2 years ago
Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:
Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

                                   \frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} ),

where  λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from n_1 to n_2 and R_h = 109677 is the Rydberg Constant. Here n_1 and  n_2 represents the transitions.

(a) n_1 =2 to n_2 = infinity

            \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2}  ) = 109677/4     [since 1/infinity = 0] Therefore, \lambda = 4 / 109677 = 0.00003647 m

(b) n_1=4 to  n_2 = 20

           \frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2}  ) = 6580.62

Therefore,  \lambda = 1 / 6580.62 = 0.000152 m

(c) n_1=3 to  n_2 = 10

          \frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2}  ) = 11089.56

Therefore,  \lambda = 1 / 11089.56 = 0.00009 m

(d)  n_1=2 to  n_2 = 1

          \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2}  ) = - 82257.75

Therefore,  \lambda = 1 /82257.75  = - 0.0000121 m  

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

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8 0
1 year ago
What would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the
lutik1710 [3]

Answer:

1.05 V

Explanation:

Since;

E°cell= E°cathode- E°anode

E°cathode= -0.40 V

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Equation of the process;

2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)

n= 8 electrons transferred

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]

Since log 1=0

Ecell= E°cell= 1.05 V

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I think the answer is False.
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