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2 years ago

difference between an Argon atom with a mass number of 38 (Ar-38) and another Argon atom with a mass number of 45 (Ar-45)?

Chemistry

1 answer:

Kay [80]2 years ago

8 0

The difference between Argon-38 and Argon - 45 is the number of neurons in their nuclei.

<h3>What is Isotope?</h3>

Isotopes are two or more types of atoms of the same element that have the same atomic number but differs in mass numbers due to different numbers of neutrons in their nuclei.

Argon atom with a mass number of 38 (Ar-38) - number of neutrons = 20

Argon atom with a mass number of 45 (Ar-45) - number of neutrons = 27

Thus, the difference between Argon-38 and Argon - 45 is the number of neurons in their nuclei.

Learn more about isotopes here: brainly.com/question/364529
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What are the limitations of J.J. Thomson's model of the atom?

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Explanation:

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Is glass melting a physical or chemical change

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3 years ago

A solution of NaCl ( aq ) is added slowly to a solution of lead nitrate, Pb ( NO 3 ) 2 ( aq ) , until no further precipitation o

dezoksy [38]

Answer:

0.218 M of Pb(NO3)2

Explanation:

Equation of the reaction

Pb(NO3)2(aq) + 2NaCl(aq) --> PbCl2(s) + 2NaNO3(aq)

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4 years ago

Which describes the correct procedure when converting a number from scientific notation to standard notation? A If the power of

MissTica

Answer:

I think it's B

Explanation:

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3 years 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 }\$

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

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

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3 years ago