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Nataly [62]
3 years ago
13

As an electron in an atom moves from the ground state to the excited state, the electron (1) gains energy as it moves to a highe

r energy level (2) gains energy as it moves to a lower energy level (3) loses energy as it moves to a higher energy level (4) loses energy as it moves to a lower energy level
Chemistry
1 answer:
yawa3891 [41]3 years ago
5 0
Electron requires energy to move from ground state to excited state(higher energy levels). Hence option 1 is correct. Hope this helps!
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DNA is the molecule that carries the genetic code. These are found in the nucleus of cells. They copy themselves during replication when a cell divides and splits.
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Which of the following elements are more reactive than the others.
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The answer is D-sodium!
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3 years ago
The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in
astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}

For this, it is necessary to know the values in meters for any of these diameters:

\\ 1m = 10^{3}mm = 1e+3mm

\\ 1m = 10^{12}pm = 1e+12pm

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m

<h3>Diameter of a biscuit in meters</h3>

\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0
3 years ago
According to the law of definite proportions, if a sample of a compound contains 24.00 grams of carbon and 4.00 grams of hydroge
Inga [223]

Answer: 8.00 grams

Explanation:

The ratio of carbon and hydrogen must be the same.

6 0
2 years ago
Nitrogen is a group 15 element. What does being In this group imply about the structure of the nitrogen atom?
sveticcg [70]

Answer:

Nitrogen has 5 valence electrons

Explanation:

nitrogen has it's attoms form triple bonds which are very hard to break

so non-reactive

4 0
3 years ago
Read 2 more answers
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