Find the area shaded in yellow. [ ? ] square units
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Answer:
Oxidized and reducing agent: manganese.
Reduced and oxidizing agent: mercury.
Explanation:
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In this case, for the reaction:
We keep in mind that the species that increase the oxidation state is the oxidized one whereas the one that decrease the oxidation state is the reduced one; therefore manganese is the oxidized one as well as the reducing agent because it goes from 0 to +2 and mercury the reduced one as well as the oxidizing agent because it goes from 2+ to 0.
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Considering the ideal gas law, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
- P= 2 atm
- V= ?
- n=
being 2g/mole the molar mass of H2, that is, the amount of mass that a substance contains in one mole.
- R= 0.082
- T= 353 K
Replacing:
2 atm× V = 4.745 moles× 0.082× 353 K
Solving:
V = (4.745 moles× 0.082× 353 K)÷ 2 atm
<u><em>V= 68.67 L</em></u>
Finally, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
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