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joja [24]
3 years ago
11

Cause of iron changing to rust when it reacts with water

Chemistry
1 answer:
stellarik [79]3 years ago
5 0
Iron get OXIDIZED when it reacts with water, and then it get rusted...
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Bond length is the distance between the centers of two bonded atoms. On the potential energy curve, the bond length is the inter
vredina [299]

Answer:

The answer is "152 pm".

Explanation:

The bond length from the values inside the atomic radii is calculated according to the query. This would be the upper limit of a molecule's binding length.

The atomic radius of H= 37.0 \ pm

The atomic radius of Br = 115.0 \ pm

\text{Bond length = Atomic radius of H + Atomic radius of Br}

                    = 37.0\ pm + 115.0 \ pm\\\\= 152\ pm

3 0
3 years ago
Which of these types of radiation has the greatest penetrating power?(1) alpha (3) gamma
Masteriza [31]
(3) Gamma has the greatest penetration power. Alpha has low and both beta and positron have medium. Hope I helped :)
8 0
3 years ago
12. Electrons are very small _<br> located outside the nucleus.
pogonyaev

Answer:

This is not a question but if it is a true or falso statement this is true.

Explanation:

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(5points) Na2[Fe(OH2)2(OH)I3]absorbs photons with energy of 191kJ/mol. Calculate the wavelength (in nm) for a photon absorbed by
AfilCa [17]

Answer:

626.7nm

Explanation:

The energy of a photon is defined as:

E = hc / λ

<em>Where E is the energy of the photon, h is Planck constant (6.626x10⁻³⁴Js), c is speed of light (3x10⁸m/s) and </em>λ is the wavelength of light

The energy of 1 photon is:

(191000 J / mol) ₓ (1 mole / 6.022x10²³) = 3.1717x10⁻¹⁹ J

Replacing:

3.1717x10⁻¹⁹ J = <em>6.626x10⁻³⁴Jsₓ3x10⁸m/s /  </em>λ

λ = 6.267x10⁻⁷m

as 1nm = 1x10⁻⁹m:

6.267x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =

<h3>626.7nm</h3>
6 0
3 years ago
Separate this redox reaction into its balanced component half-reactions. What is the oxidation and reduction half reactions
erma4kov [3.2K]
<span>Separate this redox reaction into its component half-reactions. 
Cl2 + 2Na ----> 2NaCl 

reduction: Cl2 + 2 e- ----> 2Cl-1 
oxidation: 2Na ----> 2Na+ & 2 e- 


2) Write a balanced overall reaction from these unbalanced half-reactions: 

oxidation: Sn ----> Sn^2+ & 2 e- 
reduction: 2Ag^+ & 2e- ----> 2Ag 

giving us 
2Ag^+ & Sn ----> Sn^2+ & 2Ag </span>Steve O <span>· 5 years ago </span><span>
</span>
3 0
3 years ago
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