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sertanlavr [38]
3 years ago
9

How many valence electrons can atoms of non metals have?

Chemistry
1 answer:
Dmitry_Shevchenko [17]3 years ago
7 0

1 to 8 valence electrons

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Which half-reaction correctly represents oxidation?
Leokris [45]

Answer: Mg\rightarrow Mg^{2+}+2e^- represents oxidation.

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Example: Mg\rightarrow Mg^{2+}+2e^-

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Example: Mg^{2+}+2e^-\rightarrow Mg

Mg\rightarrow Mg^{2+}+2e^- represents oxidation.

4 0
3 years ago
There are four substances, A, B C and D. A and C are pure substances. They cannot be broken down into anything simpler by chemic
yarga [219]

The answer is C Becuz D is a compound and that is the only answer that says it

6 0
3 years ago
After 17.1 thousand years, what percentage of the original carbon-14 would be left in an organism’s remains? After 17.1 thousand
hram777 [196]

Answer:

12.5%

Explanation:

Initial percentage of carbon 14 = 100%

Final percentage = ?

Time passed = 17.1 * 1000 years = 17100 years

Half life of carbon 14 = 5,730 years.

So how many Half lives are in 17100 years?

Number of Half lives = Time passed /  Half life = 3.18 ≈ 3

First Half life;

100% --> 50%

Second Half life;

50% --> 25%

Third Half life ;

25% --> 12.5%

6 0
2 years ago
Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume
DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

<em />

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

<em />

<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>

<em />

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

<h3>pH = 4.56</h3>
4 0
3 years ago
As a sodium atom is oxidized the number of protons in its nucleus
lara31 [8.8K]

Answer:

The proton remains the same.

Explanation:

Oxidation is simply defined as the loss of electron(s) during a chemical reaction either by an atom, molecule or ion.

Oxidation is strictly on the transfer of electron(s) and not proton.

A metal that undergoes oxidation still has its protons intact otherwise it will not be called the ion of the metal since atomic number is called the proton number.

Sodium (Na) undergoes oxidation as follow:

Na —> Na+ + e-

Na is called sodium metal.

Na+ is called sodium ion.

Na has 11 electrons and 11 protons

Na+ has 10 electrons and 11 protons

From the above illustration, we can see that the protons of Na and Na+ are the same why their electrons differ because Na+ indicates that 1 electron has been loss or transferred.

6 0
3 years ago
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