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Answer:
Base on the question , an acid and a base solution was mixed together in a crucible and all the water in the mixture was evaporated off. The only substance left in the crucible is the salt solution.
Explanation:
Acids are substances that dissolves in water to produce hydrogen ion(H⁺) while base are substances that dissolves in water to produce hydroxide ion (OH⁻).
The chemical reaction between an acid and a base solution is known as a neutralization reaction. The reaction between these 2 compounds will produce salt and water. An acid solution will react with a base solution to form salt and water. An example of this kind of reaction is between HCl (Hydrochloric acid) and NaOH(sodium hydroxide).
HCl + NaOH → NaCl + H₂O . The product form here is sodium chloride(salt) and water
Base on the question , an acid and a base solution was mixed together in a crucible and all the water in the mixture was evaporated off. The only substance left in the crucible is the salt solution. Note that the other product which is water has been evaporated and we are left with only salt solution.
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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