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Lemur [1.5K]
2 years ago
15

Density (D) is defined as the ratio of mass (m) to volume (V) and can be determined from the expression D = . Find the density o

f a sample of pure aluminum if a rectangular prism of volume 32.32 cm3 occupies a mass of 87.329 g. (Remember to put a space between your numerical value and the unit.)
Chemistry
1 answer:
Usimov [2.4K]2 years ago
7 0

That is a question about density.

To solve that question, it is important to know how we can obtain the Density of an object. Below, you can see the formula:

\boxed{D=\frac{m}{v} }

D = Density\\m = Mass\\V = Volume

Now, it is easy to answer, right? We just replace the values in the formula. The mass is 87.329g and the volume is 32.32cm³. So,

D=\frac{m}{v} \\\\D = \frac{87.329g}{32.32cm^3} \\\\D \approx 2.70g/cm^3

Therefore, the density of the the aluminium is approximately 2.7 g/cm³.

I hope I've helped. ^^

Enjoy your studies \o/

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A reduced element (which gains electrons) and an oxidized element are required for redox reactions (gives electrons). It is not a redox reaction if we lack both of them (an element can not receive electrons if no element gives electrons and vice versa).

A reduced half and an oxidized half, which always occur together, make up redox processes. While the oxidized half experiences electron loss and an increase in oxidation number, the reduced half obtains electrons and the oxidation number declines. The mnemonic devices OIL RIG, which stand for "oxidation is loss" and "reduction is gain," are simple ways to memorize this. In a redox process, the total number of electrons stays constant. In the reduction half reaction, another species absorbs those that were released in the oxidation half reaction.

In a redox reaction, two species exchange electrons, and they are given unique names:

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  • The ion or molecule that donates electrons is called the reducing agent - by giving electrons it reduces the other species.

Hence, what is oxidized is the reducing agent and what is reduced is the oxidizing agent.

<h3>What is the purpose of oxidizing agents and reducing agents?</h3>

By reducing other compounds and shedding electrons, a reducing agent raises its oxidation state. An oxidizing agent gets electrons by oxidizing other compounds; as a result, its oxidation state lowers.

<h3>What is a redox reaction?</h3>

Oxidation-reduction (or "redox") reactions are chemical processes in which electrons are exchanged between two substances. An oxidation-reduction reaction is any chemical process in which a molecule, atom, or ion alters the number of electrons it has, hence increasing or decreasing its oxidation state.

Learn more about redox reaction:  brainly.com/question/13293425

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(Link; https://www.chemteam.info/SigFigs/SigFigsFable.html)
Anna007 [38]

1. The measurement turned out to be so expensive since the student did not rely on the significant figures to calculate the edge of the cube and approximate it to the next value (2.1 cm possibly) that would allow a simpler construction and therefore its cost was much lower .

2. The student had to take all the significant figures in his calculations, with which he would have:

Volume = \frac{mass}{density}

Volume = \frac{80g}{8.67g/mL}

Volume = 9.2272203 mL

Since the figure to be constructed is a cube, he had to calculate the cube root of the volume to find the value of the edge of the cube:

Edge of the cube = \sqrt[3]{9.2272203 cm^{3} } (Taking into account that cm^{3} is proportional to mL)

Edge of the cube = 2.097443624 cm

Because the cube edge value is so specific, in order to manage his budget, he was able to order a 2.1 cm cube, which would bring the mass up to 80.29287 g, and in the lab reduce one of the faces to the appropriate weight. .

On the other hand, the main thing he had to do was ask how much it would cost to make a cube with those specifications, especially when they mentioned that it would be "expensive" and he only had $50.

The significant figures guarantee the correct operation of a machinery, a gear, a team in general, for which the accuracy will not only be taken to the millimeter, but sometimes microns or much more specific, as in the case of computer components, Therefore, it is very important, if not, and if arbitrary measurements are taken that do not consider significant figures, the components could not function properly, which would cause a loss in time, effort and manpower.

If you want to learn more about exercises with significant figures, you can see the next link: brainly.com/question/11904364?referrer=searchResults

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