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beks73 [17]
3 years ago
10

What metric measurement is similar to a mile?

Chemistry
2 answers:
Dmitriy789 [7]3 years ago
4 0
A kilometer. It's about .62 miles and is the largest unit of measurement for the metric system.
amm18123 years ago
4 0
Kilómetro is just about a mile
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If winds blow against a mountain from the south, then on which slope would you find the least precipitation?
Aleksandr-060686 [28]

The northern slope will have the least precipitation if winds blow against a mountain from the south.

<h3>What is Slope?</h3>

This is referred to as the degree of steepness on a surface such as rocky areas etc.

When wind blows on the southern part of a mountains, it gets stuck which is why the northern slope have the least precipitation.

Read more about Slope here brainly.com/question/24208141

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Periodic law states that when the elements are arranged according to their atomic numbers, this allows elements with similar pro
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is a mutiply choice answer just to be specific

Explanation:

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NaCl and AgNO3 react in a double replacement reaction. Which is one of the products of this reaction?
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double displacement reaction

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Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr
Alecsey [184]

Answer:

The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:

E_{n} = -\frac{m_{e}e^{4}}{2(4\pi\epsilon_{0})^2\hbar^{2}}\frac{1}{n^2} = E_{1}\frac{1}{n^{2}}, where:

[tex]m_{e}[tex] = electron mass

e = electron charge

[tex]\epsilon_{0}[tex] = vacuum permittivity

[tex]\hbar[tex] = Planck's constant over 2pi

n = quantum number

[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV

Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:

[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]

Now, knowing the energy of the photon, we can calculate its wavelength using the equation:

[tex]E = \frac{hc}{\lambda}[tex], where:

E = Photon's energy

h = Planck's constant

c = speed of light in vacuum

[tex]\lambda[tex] = wavelength

Solving for [tex]\lambda[tex] and substituting the required values:

[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).

5 0
3 years ago
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