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Kobotan [32]
3 years ago
10

This exercise uses the radioactive decay model.

Chemistry
1 answer:
Anna11 [10]3 years ago
3 0
For radioactive decay, we can relate current amount, initial amount, decay constant and time using:
N = No x exp(-λt)

Half-life = ln(2)/λ
λ = ln(2) / 5730
N/No = 80% = 0.8
0.8 = exp( -ln(2)/5730 x t)
t = 1844 years
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How might you think it is possible to convert from moles to particles or grams to particles?
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Explanation:

To convert moles to particles or grams to particles, let us have a firm understanding of what a mole is.

A mole is the unit of measuring quantity of particles.

It is the amount of substance that contains the Avogadro's number of particles.

The particle can be atoms, molecules, formula units, electrons, protons, neutrons, etc.

 So, to convert from moles to particles;

    1 mole of a substance  contains 6.02 x 10²³ particles

To convert from grams to particles;

  First convert to moles;

        number of moles  = \frac{mass}{molar mass}  

  So,    1 mole of a substance  contains 6.02 x 10²³ particles

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Scientists in a lab are working on a series of experiments that involve colliding two or more atomic nuclei at very high speeds
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Answer: Scientists are working on nuclear fusion process. Nuclear fusion is the process in which two atomic nuclei combine to form another nuclei. In the process either energy is absorbed or released due to the difference in the mass of reactants and products.

Explanation:

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ruslelena [56]

Answer: C

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When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
ruslelena [56]

Answer:

b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)  + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s)  + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)  + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

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Explanation:

At one end, new crusts are being produced, at other end the crust is being destroyed and this strikes a unique balance.

At the mid-ocean ridges, the lithospheric plates are diverging. This is implies that the earth is pulling apart here. When the earth is pulling apart, new materials from the asthenosphere comes to the surface thereby creating new lithospheric plate.

As new plates are formed, they push back against the old one. New plates are found very close to the margin and it begins to age away from the margin.

On the other end, old plates are taken away from this center to ocean trenches. At oceanic trenches subduction is occurring.

In a subduction, the lithospheric plate plunges deep into the asthenosphere where they are being melted back.

This is a covergent margin.

This process continues in a dynamic manner to cycle matter on earth.

learn more:

Sea floor spreading brainly.com/question/9912731

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