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3 years ago

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A galvanic cell consists of a Al3 /Al half-cell and a standard hydrogen electrode. If the Al3 /Al half-cell standard cell functi

ons as the anode, and the standard cell potential is 1.66 V, what is the standard reduction potential for the Al3 /Al half-cell

1 answer:

11111nata11111 [884]3 years ago

3 0

Answer:

Explanation:

here is your answer:

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Which part of the mantle is still solid but flows like a heavy liquid

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The Earth is composed of four different layers. Many geologists believe that as the Earth cooled the heavier, denser materials sank to the center and the lighter materials rose to the top. Because of this, the crust is made of the lightest materials (rock- basalts and granites) and the core consists of heavy metals (nickel and iron).

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Which of the following is an example of improved functioning as related to the benefits of muscular strength and endurance train

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What are the five main points of kinetic theory of gas?

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2) Gas molecules do not have preferred direction of motion, their motion is completely random. 3) Gas molecules travels in straight line. 4) The time interval of collision between any two gas molecules is very small. 5) The collision between gas molecules and the walls of container is perfectly elastic.

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Calculate the frequency of the n=2 line in the lyman series of hydrogen

Alona [7]

Answer:

Approximately $2.47\times 10^{15}\; \rm Hz$ .

Explanation:

The Lyman Series of a hydrogen atom are due to electron transitions from energy levels $n \ge 2$ to the ground state where $n = 1$ . In this case, the electron responsible for the line started at $n = 2$ and transitioned to

A hydrogen atom contains only one electron. As a result, Bohr Model provides a good estimate of that electron's energy at different levels.

In Bohr's Model, the equation for an electron at energy level $n$ (

$\displaystyle - \frac{k\, Z^2}{n^2}$ (note the negative sign in front of the fraction,)

where

$k = 2.179 \times 10^{-18}\; \rm J$ is a constant.

$Z$ is the atomic number of that atom. $Z = 1$ for hydrogen.
$n$ is the energy level of that electron.

The electron that produced the $n = 2$ line was initially at the

$\begin{aligned} &E_{n = 2} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{2^2} \cr & \approx -5.4475\times 10^{-19}\; \rm J\end{aligned}$ .

The electron would then transit to energy level $n = 1$ . Its energy would become:

$\begin{aligned} &E_{n = 1} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{1^2} \cr & \approx -2.179 \times 10^{-18} \; \rm J\end{aligned}$ .

The energy change would be equal to

$\begin{aligned}&\text{Initial Energy} - \text{Final Energy} \cr &= E_{n = 2} - E_{n = 1} \cr &= -5.4475 \times 10^{-19} - \left(-2.179 \times 10^{-18}\right) \cr & \approx 1.63425\times 10^{-18}\; \rm J \end{aligned}$ .

That would be the energy of a photon in that $n = 2$ spectrum line. Planck constant $h$ relates the frequency of a photon to its energy:

$E = h \cdot f$ , where

$E$ is the energy of the photon.
$h \approx 6.62607015\times 10^{-34}\; \rm J \cdot s$ is the Planck constant.
$f$ is the frequency of that photon.

In this case, $E \approx 1.63425 \times 10^{-18}\; \rm J$ . Hence,

$\begin{aligned} f &= \frac{E}{h} \cr &\approx \frac{1.63425\times 10^{-18}}{6.62607015\times 10^{-34}} \cr & \approx 2.47 \times 10^{15}\; \rm s^{-1}\end{aligned}$ .

Note that $1 \; \rm Hz = 1 \; \rm s^{-1}$ .

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3 years ago

5. Uranium has three common isotopes. If the abundance of 2U which has an atomic mass of 233.878 amu is 0.01%, the

Temka [501]

Uranium has three common isotopes. The average atomic mass of uranium is 23.7 u/atom.

<h3>What are isotopes?</h3>

Isotopes are similar elements that have a different number of protons.

Uranium has three isotopes

The abundance of 2U has an atomic mass of 233.878 is 0.01%

The abundance of U has a mass of 234.892 amu is 0.71%

The **U has an atomic mass of 237.911 amu is 99.28%

Let's assume we have 10 000 atoms of Uranium

Mass of the 0.01 atoms = 0.01 x 233.878 = 2.33878

Mass of the 0.71 atoms = 0.71 x 234.892 = 166.773

Mass of the 99.28 atoms = 99.28 x 237.911 = 23,619.80

2.33878 + 166.773 + 23,619.80 = 23,788.8

23,788.8 / 1000 = 23.7 u/atom

Thus, the average atomic mass of uranium is 23.7 u/atom.

To learn more about average atomic mass, refer to the link:

brainly.com/question/12971445

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1 year ago