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Marianna [84]
3 years ago
6

Describe how light is emitted from an atom. A) As protons absorb energy they travel from the nucleus through the atom emitting l

ight along the way. B) The light is emitted when energy from sunlight is absorbed by the atom to excite a neutron into a higher level. C) When the nucleus absorbs sunlight, photons are energized and move to a higher energy level and are then released. D) Light is emitted from an atom as an electron falls from an excited state to the ground state releasing a photon.
Chemistry
1 answer:
vovangra [49]3 years ago
7 0

Answer is: D) Light is emitted from an atom as an electron falls from an excited state to the ground state releasing a photon.

Electrons can jump from one energy level to another, absorbing or emitting electromagnetic radiation with a frequency ν (energy difference of the levels).

When electron jump from higher to lower energy level (shell), it emitting (releasing) energy.

For example, when the electron changes from n=4 (fouth shell) to n=2 (second shell), the photons are emitted.

Albert Einstein (1879-1955) proposed that a beam of light is a collection of discrete wave packets (photons) with energy hν, where h is Planck constant and ν is frequency.

The photons have a characteristic energy proportional to the frequency of the light.

Minimum frequency or the threshold frequency is energy below which no photoelectrons are emitted.  

Above the minimum frequency, energy depends on the frequency of the light, not on the intensity of the light.

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A chemical process dissolves 500 milligrams of iron oxide every 20 minutes. How long would it take this reaction to dissolve 2 l
stiv31 [10]

Answer:

36290 min = 604.8 hr.

Explanation:

  • Knowing that:

1 lbs = 453.59237 grams.

∴ 2 lbs = 907.18474 grams.

<em><u>Using cross multiplication:</u></em>

500 mg of iron oxide dissolved → 20 minutes.

907184.74 mg of iron oxide dissolved → ??? minutes.

<em>∴ The time needed to dissolve 2 lbs of iron oxide =</em> (907184.74 mg)(20 min)/(500 mg) = <em>36290 min = 604.8 hr.</em>

4 0
3 years ago
In humans, infections by fungi and protists are usually more difficult to treat than bacteria infections. Suggest an explanation
Basile [38]
<span>Fungal diseases are difficult to treat mainly because they are eukaryotic organisms just like us humans, and therefore have less differences for drugs to target without harming the human body as well. Most antibiotics target e.g. the peptidoglycan layer in the bacterial (a prokaryote) cell wall, which is a safe target since eukaryotic cells do not have equivalent structures. Similarly many differences in metabolic pathways between humans and prokaryotes is often targeted by antibiotics, but metabolism of fungi and humans is much more uniform, and hence it is difficult to exclusively target the fungi only.

HOPE THIS HELPS!

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5 0
3 years ago
Read 2 more answers
The best way to explain the relationships in a nuclear reaction is by using:
denis23 [38]
The best and most correct answer among the choices provided by your question is all of the above.

All of the choices given are the best ways to explain a nuclear reaction.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
7 0
3 years ago
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Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
Large crystals are made from the slow cooling of magma. <br> a. True<br> b. False
Step2247 [10]
I would have to say the answer is a. True.
5 0
3 years ago
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