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2 years ago

What is the mass of 2.486 * 10^{24} atoms of neon?

Chemistry

1 answer:

lorasvet [3.4K]2 years ago

4 0

Answer:

The answer is A 83.33 g Ne

Explanation:

Used Avogadros Law

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How many single covalent bonds can halogens form?

Simora [160]

One single covalent bond, hope this helps!

4 0

3 years ago

Free-energy change, ΔG∘, is related to cell potential, E∘, by the equationΔG∘=−nFE∘where n is the number of moles of electrons t

mart [117]

Answer:

a) $\Delta G=372490 J$

b) $\Delta G=-568614 J$

Explanation:

a) The reaction:

$Mg(s) +Fe^{2+}(aq) \longrightarrow Mg^{2+}(aq) + Fe(s)$

The free-energy expression:

$\Delta G=-n*F*E$

$E=E_{red}-E_{ox]$

The element wich is reduced is the Fe and the one that oxidates is the Mg:

$-0.44V=E_{red}-(-2.37V)=1.93V$

The electrons transfered (n) in this reaction are 2, so:

$\Delta G=-2mol*96500 C/mol * 1.93 V$

$\Delta G=-372490 J$

b) If you have values of enthalpy and enthropy you can calculate the free-energy by:

$\Delta G=\Delta H - T* \Delta S$

with T in Kelvin

$\Delta G=-675 kJ*\frac{1000J}{kJ} - 298K*-357 J/K$

$\Delta G=-568614 J$

7 0

3 years ago

If you start with 0.30 m mn2 , at what ph will the free mn2 concentration be equal to 4.6 x 10-11 m?

aksik [14]

If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m

Initial molarity of Mn₂ = 0.30 M

Final molarity of Mn₂ = 4.6 x 10⁻¹¹

pH = ?

Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)

Write the ionic equation

Mn(OH)₂ → Mn⁺² + 2OH⁻

[Mn⁺²] = 4.6 x 10⁻¹¹

We will calculate the concentration of OH⁻ by using Ksp expression

Ksp = [Mn⁺²][OH-]²

[Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴

[OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹

[OH⁻]² = 10⁻³

[OH⁻] = (10⁻³)¹⁽²

[OH⁻] = 0.0316 M

Calculate the pOH

pOH = -log [OH⁻]

pOH = -log [0.0316]

pOH = 1.5

Now calculate pH

pH = 14 - pOH

pH = 14 - 1.5

pH = 12.5

You can also learn about molarity from the following question:

brainly.com/question/14782315

#SPJ4

7 0

1 year ago

What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?

Lubov Fominskaja [6]

Answer:

P = 1/8

Explanation:

The wave function of a particle in a one-dimensional box is given by:

$\psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L})$

Hence, the probability of finding the particle in the one-dimensional box is:

$P = \int_{x_{1}}^{x_{2}} \psi^{2} dx$

$P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx$

$P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx$

Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

$P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]$

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})$

Solving for n=4:

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})$

$P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi})$

$P = \frac{1}{8}$

I hope it helps you!

7 0

3 years ago

Which of the following is true for the theoretical yield of a reaction? (1 point)

just olya [345]

Answer:

It is always twice the value of the actual yield.

Explanation:

3 0

2 years ago