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3 years ago

Which activity is most likely involved with the study of electrochemistry?

1 answer:

5 0

The answer is that, from the answer choices, placing zinc and copper in an electrolyte is most likely involved with the study of electrochemistry.

Electrochemistry is the use study of the chemical reactions in relation to electric energy. It comprises the use of electric energy to produce chemical changes and the use of chemical reactions to produce electric energy.

The chemical reactions that produce or use electric energy are carried out in electrolytic cells.

Electrolytic cells consist of a pair of electrodes submerged in an electrolytic solution.

Zinc and copper are good pieces of conductors that work as electrodes in the electrolytic solution. Then, this is the correct answer..

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Why is oxygen an element ? a it is essential to life b it burn c it combines with hydrogen d it cannot be broken down into a sim

MrRa [10]

D.) It cannot be broken down into a simple substance through chemical means...

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2 years ago

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Blank is most if the compounds that make up living things?

Margarita [4]

Answer:

carbon

Explanation:

Carbon forms compounds that make up about 18 percent of all the matter in living things. The processes by which organisms consume carbon and return it to their surroundings constitute the carbon cycle.

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2 years ago

What are 2 interactions between the earth spheres and the carbon cycle?

Fed [463]

Answer:

Explanation:

The spheres interact with each other, and a change in one area can cause a change in another. Humans (biosphere) use farm machinery manufactured from geosphere materials to plow the fields, and the atmosphere brings precipitation (hydrosphere) to water the plants. The biosphere contains all the planet's living things.

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2 years ago

Iodine 131, cesium 137, and strontium 90 are all considered isotopes. What are isotopes?

Nitella [24]

Answer:

Isotopes are variants of an element, where the number of neutrons is different but its number of protons stay the same.

Explanation:

If the number of protons had changed then it would just be a different element.

Electron number changing simply adds charge to the atom, and doesn't change it or anything.

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2 years ago

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The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

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2 years ago