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irga5000 [103]
3 years ago
11

Is. Mg(s) --------------> Mg2+(aq) + 2e−oxidation or reduction?​

Chemistry
1 answer:
Kruka [31]3 years ago
5 0

Answer:

Oxidation

Explanation:

Mg loses 2e- for it to gain it's stability hence the reaction above is oxidation

note: Oxidation is the loss of electrons while reduction is the gain of electrons

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Isotopes of elements have same atomic numbers and different mass number(atomic masses).
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Consider the solubilities of a particular solute at two different temperatures.
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How would a collapsing universe affect light emitted from clusters and superclusters? A. Light would acquire a blueshift. B. Lig
Lady_Fox [76]

Answer:

Choice A: Light would acquire a blueshift.

Explanation:

When a universe collapses, clusters of stars start to move towards each other. There are two ways to explain why light from these stars will acquire a blueshift.

Stars move toward each other; Frequency increases due to Doppler's Effect.

The time period t of a beam of light is the same as the time between two consecutive peaks. If \lambda is the wavelength of the beam, and both the source and observer are static, the time period T will be the same as the time it takes for light travel the distance of one \lambda (at the speed of light in vacuum, c).

\displaystyle t = \frac{\lambda}{c}.

Frequency f is the reciprocal of time period. Therefore

\displaystyle f = \frac{1}{t} = \frac{c}{\lambda}.

Light travels in vacuum at a constant speed. However, in a collapsing universe, the star that emit the light keeps moving towards the observer. Let the distance between the star and the observer be d when the star sent the first peak.

  • Distance from the star when the first peak is sent: d.
  • Time taken for the first peak to arrive: \displaystyle t_1 =\frac{d}{c}.

The star will emit its second peak after a time of. Meanwhile, the distance between the star and the observer keeps decreasing. Let v be the speed at which the star approaches the observer. The star will travel a distance of v\cdot t before sending the second peak.

  • Distance from the star when the second peak is sent: d - v\cdot t.
  • Time taken for the second peak to arrive: \displaystyle t_2 =t + \frac{d - v\cdot t}{c}.

The period of the light is t when emitted from the star. However, the period will appear to be shorter than t for the observer. The time period will appear to be:

\begin{aligned}\displaystyle t' &= t_2 - t_1\\ &= t + \frac{d - v\cdot t}{c} - \frac{d}{c}\\&= t + (\frac{d}{c} - \frac{v\cdot t}{c}) -\frac{d}{c}\\&= t - \frac{v\cdot t}{c} \end{aligned}.

The apparent time period t' is smaller than the initial time period, t. Again, the frequency of a beam of light is inversely proportional to its period. A smaller time period means a higher frequency. Colors at the high-frequency end of the visible spectrum are blue and violet. The color of the beam of light will shift towards the blue end of the spectrum when observed than when emitted. In other words, a collapsing universe will cause a blueshift on light from distant stars.

The Space Fabric Shrinks; Wavelength decreases as the space is compressed.

When the universe collapses, one possibility is that clusters of stars move towards each other. Alternatively, the space fabric might shrink, which will also bring the clusters toward each other.

It takes time for light from a distant cluster to reach an observer on the ground. The space fabric keeps shrinking while the beam of light makes its way through the space. The wavelength of the beam will shrink at the same rate. The wavelength of the beam of light will be shorter by the time the beam arrives at its destination.

Colors at the short-wavelength end of the visible spectrum are blue and violet. Again, the color of the light will shift towards the blue end of the spectrum. The conclusion will be the same: a collapsing universe will cause a blueshift on light from distant stars.

8 0
3 years ago
Erika is asked to create a model showing a negatively charged Carbon ion. Which model should she create?
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I am pretty sure u have pictures and this one should be the one that erika should make

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2 years ago
A sample of gas in which [h2s] = 3.50 m is heated to 1400 k in a sealed vessel. after chemical equilibrium has been achieved, wh
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The value of Kc for the thermal decomposition of H₂S is 2.2 x 10⁻⁴ at 1400 K:
                        2 H₂S(g) ↔     2 H₂(g) + S₂(g)
initial                  3.5 M               0           0
at equilibrium     3.5 M - 2x       2x          x
Kc             = [S₂][H₂]² / [H₂S]² 
2.2 X 10⁻⁴ = x(2x)² / (3.5 - 2x)²
2.2 x 10⁻⁴ = 4 x³ / (3.5)² Assuming x <<<<< 3.5
x = 0.088
Thus [H₂S] = 3.324 M
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