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bezimeni [28]
3 years ago
7

How to find the number of neutrons in an atom

Chemistry
1 answer:
kirill [66]3 years ago
7 0

So, to determine the number of neutrons in atom, we only have to subtract the number of protons from the mass number. Take note that the nucleus of an atom is composed of protons and neutrons. And the number of particles present in the nucleus is referred as mass number (Also, called as atomic mass).

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How does a volcano erupt ?
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3 years ago
A student has four resistors. Each resistor has a resistance of 100 ohms.
ValentinkaMS [17]

Answer:

The minimum resistance is 25 ohms.

Explanation:

Resistance of each resistor is 100 ohms. When resistors are connected in parallel, the equivalent resistance is lowest. For parallel combination, the equivalent resistance is given by :

\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....

Here, all resistors are 100 ohms. So,

\dfrac{1}{R}=\dfrac{1}{100}+\dfrac{1}{100}+\dfrac{1}{100}+\dfrac{1}{100}\\\\R=25\ \Omega

So, the minimum resistance is 25 ohms.

7 0
3 years ago
Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0
1 year ago
Please help me with my science thank you!
Tatiana [17]
I cant see it maybe post it again?
4 0
2 years ago
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