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igor_vitrenko [27]
3 years ago
6

Explain the function of the valence ring?

Chemistry
1 answer:
kirill115 [55]3 years ago
7 0

Answer:

It provides the electrons that "make" electricity and it contains the electrons that are involved in molecular bond formation.

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What is the difference between the cells in a single-called organism and the cells in a human body
Fed [463]

Answer:

Unicellular organisms are made up of only one cell that carries out all of the functions needed by the organism, while multicellular organisms use many different cells to function. ... Multicellular organisms are composed of more than one cell, with groups of cells differentiating to take on specialized functions.

Explanation:

does that make any sense???

4 0
3 years ago
Pls how do you test for the hydrogen ion concentration of a bottle of water.
iren [92.7K]
Using the pH=-log[ hydrogen ions], assuming the pH of water is 7 then using the concept that kw=[ OH] [H+], then the [H+]= 10 power minus 7,since kw=10 to the power minus 14
4 0
4 years ago
A glass object receives a positive charge by rubbing it with a silk cloth. in the rubbing process, have protons been added to th
MaRussiya [10]
Electrons are the only thing that can freely move from one object to another object. By rubbing the silk cloth on the glass, since the net charge becomes positive, that means that electrons were removed. So answer is : electrons removed.
5 0
3 years ago
At 25 °C, what is the hydroxide ion concentration, [OH-], in an aqueous solution with a hydrogen ion concentration of
Otrada [13]

The hydroxide ion concentration, [OH-] = 2.439 x 10⁻¹⁰

<h3>Further explanation</h3>

Equilibrium of water and the ions :

\tt 2H_2O(l)\rightarrow H_3O^+(aq)+OH^-(aq)

\tt Kc=\dfrac{[H_3O^+][OH^-]}{H_2O}

Kc[H₂O] is the ion equilibrium constant for water ⇒ Kw.

The product of the concentrations of these ions (H₃O⁺ and OH⁻)at equilibrium at 25°C is 1.0 x 10⁻¹⁴ M

[H₃O⁺] [OH⁻] = 1.0 x 10⁻¹⁴ M = Kw

The hydrogen ion concentration of  [H⁺]=4.1 x 10⁻⁵ M, so  the hydroxide ion concentration, [OH-] :

\tt 4.1\times 10^{-5}\times [OH^-]=1.10^{-14}\\\\(OH^-]=\dfrac{1.10^{-14}}{4.1\times 10^{-5}}\\\\(OH^-]=2.439\times 10^{-10}

6 0
3 years ago
___ decrease a sample's boiling point and increase the range of_____
lapo4ka [179]
Sorry i need points✌
3 0
3 years ago
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