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4 years ago

A south magnetic pole will attract what type of magnetic pole?

1 answer:

7 0

North magnets attract the opposite side in this case south would attract to north and vise versa if it was negative it would attract to positive and vise versa

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Question 11: How does the energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital compare to

hodyreva [135]

Answer:

Explanation:

The energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital is exactly same as the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital

3 0

4 years ago

PLEASE ANSWER NEED BY MIDNIGHT TONIGHT!!!!!!

maksim [4K]

The equation for carbon-14 emission by Radium-223 nuclei is given below:

$^{223}_{88}Ra \rightarrow\: ^{209}_{82}Pb + \:^{14}_{6}C$

<h3>What is radioactivity?</h3>

Radioactivity is the spontaneous decay of a substance with emission of radiation.

The equation for carbon-14 emission by Radium-223 nuclei is given below:

$^{223}_{88}Ra \rightarrow\: ^{209}_{82}Pb + \:^{14}_{6}C$

In conclusion, the emission of carbon-14 by Radium-223 nuclei produces Lead-209 nuclei.

Learn more about radioactivity at: brainly.com/question/3603596

#SPJ1

5 0

1 year ago

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

<em>B) Radius</em>

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

3 0

3 years ago

Iron filings could be separated from sand using _______.

Deffense [45]

Answer:

A. a magnet

Explanation:

Iron fillings could be separated from sand using a magnet

3 0

3 years ago

Explain in words or using your diagrams how beryllium atoms would react with fluorine atoms.

Marysya12 [62]

Explanation:

To delineate the the nature of the bonds that would be formed between the two elements, let us first write the electronic configuration of the two species;

Be = 2, 2

F = 2, 7

Beryllium is a metal with two valence electrons whereas fluorine is a halogen with seven valence electrons.

When Be loses two electrons it becomes isoelectronic with He;

Be → Be²⁺ + 2e⁻

Also, when fluorine gains an electron, it becomes isoelectronic with Ne;

F + e⁻ → F⁻

This loss and gain of electrons between the two elements creates an electrostatic attraction them and they enter into an electrovalent bond.

Hence;

Be²⁺ + 2F⁻ → BeF₂

8 0

3 years ago