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Nina [5.8K]
3 years ago
13

What is the charge of an aluminum atom that has 13 protons, 13 neutrons, and 10 electrons?

Chemistry
1 answer:
Tomtit [17]3 years ago
5 0
Charge is +3.
The balance of protons and electrons is what affects the charge. The more electrons you have, the more negative the charge will be (this is only if you don't have the same number of protons and electrons ). If you have the same number of protons and electrons, that means the charge is neutral
In this picture you have 3 more protons.
Protons= positive
This means you will have a +3
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What mass of H2 is needed to react with 8.5 g of O2 according to the following equation G plus O2 (G)plus H2 ( G)?
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3 years ago
assume in a different experiment, you prepare a mixture containing 10.0 M FeSCN2+, 1.0 M H+, 0.1 MFe3+ and 0.1 M HSCN. Is the in
Katarina [22]

Answer:

The mixture is not in equilibrium, the reaction will shift to the left.

Explanation:

<em>Based on the equilibrium:</em>

<em>Fe³⁺+ HSCN ⇄ FeSCN²⁺ + H⁺</em>

<em>kc = 30 = [FeSCN²⁺] [H⁺] / [Fe³⁺] [HSCN]</em>

Where [] are concentrations at equilibrium. The reaction is in equilibrium when  the ratio of concentrations = kc

Q is the same expression than kc but with [] that are not in equilibrium

Replacing:

Q = [10.0M] [1.0M] / [0.1M] [0.1M]

Q = 1000

As Q > kc, the reaction will shift to the left in order to produce Fe³⁺ and HSCN untill Q = Kc

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7 0
3 years ago
You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution when ex
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Answer:

a. 1.78x10⁻³ = Ka

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b. It is irrelevant.

Explanation:

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Equilibrium is:

HA ⇄ H⁺ + A⁻

And Ka is defined as:

Ka = [H⁺] [A⁻] / [HA]

The HA reacts with the base, XOH, thus:

HA + XOH → H₂O + A⁻ + X⁺

As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻

That means:

[HA] = [A⁻]

It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:

pH = pKa + log₁₀ [A⁻] / [HA]

Replacing:

2.75 = pKa + log₁₀ [A⁻] / [HA]

As [HA] = [A⁻]

2.75 = pKa + log₁₀ 1

<h3>2.75 = pKa</h3>

Knowing pKa = -log Ka

2.75 = -log Ka

10^-2.75 = Ka

<h3>1.78x10⁻³ = Ka</h3>

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