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Feliz [49]
3 years ago
5

Ocean water contains 3.5 nacl by mass. what mass of ocean water in grams contains 45.8 g of nacl

Chemistry
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0

Mass percentage of sodium chloride(NaCl) in ocean waters = 3.5 %

That means 3.5 g sodium chloride(NaCl) is present for every 100 g of ocean water.

The given mass of sodium chloride(NaCl) is 45.8 g

Calculating the mass of ocean waters that would contain 45.8 g sodium chloride(NaCl):

45.8 g NaCl *\frac{100g ocean water}{3.5g NaCl}

                     = 1309 g ocean water

Therefore, 45.8 g sodium chloride is present in 1309 g ocean water.

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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.

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3 years ago
Consider the elements: Na, Mg, Al, Si, P.
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Answer:

The elements mentioned in the series are Na, Mg, Al, Si, and P. It can be seen that all these elements are located in the same period. The atomic number of the mentioned elements are Na-11, Mg-12, Al-13, Si-14, and P-15.

a) There will be an increase in the ionization energy with the increase in the element's atomic number across a period. More energy is needed to withdraw an electron from a completely occupied shell in comparison to an incompletely occupied shell.

The atomic number of Na is 11. When one electron is withdrawn from Na, it gets converted into the inert gas configuration of Ne. Thus, it will require more energy to withdraw the second electron from Na. Hence, Na exhibits the lowest second ionization energy.

b) Across the period, with an increase in the element's atomic number, the atomic radii reduces from left to right. Thus, P exhibits the smallest atomic radius.

c) The metallic nature of the elements reduces from left to right across a period in the periodic table. Thus, P is the least metallic element.

d) Diamagnetic signifies towards the element that exhibits pair electrons in its sub-shells. The electronic configuration of Mg is,

1s2 2s2 2p6 3s2

In Mg, no unpaired electrons are present, while all the remaining elements mentioned exhibit unpaired electrons in their valence shell. Thus, Mg is the diamagnetic element.

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What is the effect of adding heat to a gas at constant pressure?
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The pressure will continue to build up eventually causing a release of pressure or an explosion.
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Is condensation of ethanol a physical change or a chemical change
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