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Arisa [49]
3 years ago
15

How does atomic radius (size or atom diameter) affect bond distance (distance between atoms)?

Chemistry
2 answers:
inysia [295]3 years ago
8 0
The bond distance is proportional to the atomic radius
yKpoI14uk [10]3 years ago
4 0
When two atoms of the same element are covalently bonded, the radius of each atom will be half the distance between the two nuclei because they equally attract the electrons. The reason for this trend is that the bigger the radii, the further the distance between the two nuclei. Hope this helps:)
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Velocity may be expressed in which I f the following units?
Dmitry_Shevchenko [17]

velocity is speed and direction so it should be expressed in mph f/s (feet per second) with a direction like North, right, left,

ex) 5mph south

3 0
2 years ago
A chemist is using radiation with a frequency of 6.0 x 10^13 Hz
kirza4 [7]

Answer:

λ = 0.5×10⁻⁵ m

E = 39.78×10⁻²¹ J

The given radiation is gamma ray.

Explanation:

Given data:

Frequency of radiation = 6.0 ×10¹³ Hz

Wavelength of radiation = ?

Type of radiation = ?

Energy of radiation = ?

Solution:

Formula:

speed of light = wavelength × frequency

3 ×10⁸ m/s = λ × 6.0 ×10¹³ Hz

Hz = s⁻¹

λ = 3 ×10⁸ m/s /  6.0 ×10¹³  s⁻¹

λ = 0.5×10⁻⁵ m

Energy of radiation:

E = hf

h = planck's constant

f = frequency

E = 6.63×10⁻³⁴ j.s × 6.0 ×10¹³  s⁻¹

E = 39.78×10⁻²¹ J

The given radiation is gamma ray.

4 0
3 years ago
What characteristic do all parts of the electromagnetic spectrum share?
Anastasy [175]

Answer:chicken butt I need some points

Explanation:blm ✊✊✊✊ white lives don't

3 1
2 years ago
Read 2 more answers
HELPP ME !!!! DUE NOW!!!!!!
Deffense [45]

Answer:

the answer is

a

b

d

c

Explanation:

hopeit help

8 0
2 years ago
Read 2 more answers
Metal rings can be coated with a layer of copper using electricity.
Eduardwww [97]

<u>First of all, what is electrolysis?</u>

Electrolysis is the process of breaking down ionic substances using direct current.

<u>Important points about electrolysis </u>

→ Ionic substances contain particles called ions.

→ Electricity is the flow of electrons or ions. For electrolysis to work, the compound must contain ions. The ions must be free to move for electrolysis to occur and it can happen by melting or dissolving an ionic substance in water.

→ Positively charged ions move to the negative electrode. They receive electrons and are <em>reduced</em>. The positive ions move towards the negative electrode because they want to cancel each other out.

→ Negatively charged ions move to the positive electrode.  They lose electrons and are <em>oxidised</em>. The substance that is broken down is called the electrolyte <em>(an electrolyte is just a liquid or solution that can conduct electricity)</em> . The negative ions move towards the positive electrode because they want to cancel each other out.

<h3>Cathode = Negative electrode</h3><h3>Anode = Positive electrode</h3>

Metal ions form at the cathode and non-metal ions form at the anode

How I remember if an element is <em>oxidised</em> or <em>reduced</em> is by remembering OIL RIG

OIL = Oxidation is Loss (of electrons)

RIG = Reduction is Gain (of electrons)

<h2><em><u>The answer to your question</u></em></h2>

1) The first step would be to clean the metal ring and sand it down because when the metal atoms from the electrolyte are deposited onto the ring, they will form a weak bond and they may simply 'fall' off. Also this could affect conductivity and the whole experiment. The more things you do accurately now, the more accurate your result will be.

2) You want to put the solution you are given in to the tank your going to be using.

3) This is basically the main part, you want to set up the circuit, I have attached a diagram at the bottom to show you the circuit. The copper rod will be the anode and the metal ring will be a cathode (ignore the elements).

4) Now turn on the circuit and you will start to see the solution spilt with the the solution now being split some going to the anode and some going the cathode.

5) Then a thin layer should form on the electrode.

Hope this helps :)

<h2><em><u></u></em></h2>

<em><u></u></em>

5 0
3 years ago
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