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andreev551 [17]
3 years ago
11

What is Atomic Radius

Chemistry
2 answers:
Sidana [21]3 years ago
6 0

Answer: The atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons. Illustrated Glossary of Organic Chemistry - Atomic radius. Atomic radius: The radius of an atom. This distance between an atom's nucleus and outer electron shell. ... Atomic radius differs with the bonding state of an atom (for example a nonbonded atom of an element versus the same element within a covalent bond). Atomic radius is determined as the distance between the nuclei of two identical atoms bonded together. The atomic radius of atoms generally decreases from left to right across a period. The atomic radius of atoms generally increases from top to bottom within a group.

hope this helps..... Stay safe and have a great weekend!!!! :D

Virty [35]3 years ago
5 0

Answer:

Atomic Radius can be simply defined as the distance between the last shell of electrons and the nucleus of an atom.

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SynthesizeWrite the instructions you would give a classmate to locate the point 27°18'N, 19°2'E on a globe.
SVEN [57.7K]

Answer:

<em>It is known by almost everyone Earth is divided into longitudes and latitudes which are imaginary lines that cut Earth into 2 equal parts.</em>

Explanation:

The main use of dividing Earth into so many imaginary lines is to know no of places perfect location in numbers.  Direction and  tell tales can make the place talked about a bit confusing but the exact longitudes and latitudes known pill and someone in exact  position.

So in the positions given the question  27 ° and 19 ° show the distance of longitude from Prime Meridian While 18 ° and two degrees show the distance from the equator.

6 0
3 years ago
Cual sera el numero de mol de 153,26gr de KMnO4
Fiesta28 [93]

Answer:

0.97014 moles KMnO4

Explanation:

1 g KMnO4 = 0.00633 mol

153.26 g x 0.00633 mol/ 1 g KMnO4 = 0.97014 moles KMnO4

5 0
3 years ago
A sample of 250 g of water are heated from 40°C to 121°C, calculate the amount of heat energy absorbed.
Marysya12 [62]

Answer:

the anwser isn't in the choices

Explanation:

H=MC(change of temp.)

M=mass of water=250g

C=specific heat of water = 4.186 j/g

change in temperature is 121-40= 81

H= 250x4.186x81=84766.5J

7 0
3 years ago
For parts a &amp; b below, derive only the initial value problem set up.
Otrada [13]

Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

\dfrac{dy}{dt}= (rate \ in ) - ( rate \ out )

The rate of the sugar coming into the tank is 0

\text{rate of sugar going out }= \dfrac{y(t) \ pound}{100}\times \dfrac{1 }{min} \\ \\ = \dfrac{y}{100} \ pound/min

So; \\ \\ \\ \dfrac{dy}{dt} = 0 - \dfrac{y}{100}  \\ \\ \implies \dfrac{dy}{y }= -\dfrac{dt}{100 } \\ \\ \implies dx|y| = -\dfrac{t}{100}+C  \\ \\  \implies e*{In|y|}= e^{-\dfrac{t}{100}+C} \\ \\ \implies y = Ce^{-\dfrac{t}{100}}

Initial amount of sugar = 25 Pounds

Now; y(0) = 25

25 = Ce⁰

C = 25

So;  y(t) = 25 e^{-\dfrac{t}{100}

Thus, the amount of sugar at any time t = \mathbf{25 e^{^{-\dfrac{t}{100}}}}

B) For tank B :

\dfrac{dy}{dt}= 1 - \dfrac{y}{50 } \\ \\ \dfrac{dy}{dt}= \dfrac{50-y}{50} \\ \\  \dfrac{dy}{50-y }= \dfrac{dt}{50}

8 0
3 years ago
Think about a large pitcher of ice water and a small cup of ice water with a similar temperature. Which statement about these ob
Bezzdna [24]
B because ice water is water which molocules move fast like air
3 0
3 years ago
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