Answer:
In this case, the system doesn't be affected by the pressure change. This means that nothing will happen
Explanation:
We can answer this question applying the Le Chatelier's Principle. It says that changes on pressure, volume or temperature of an equilibrium reaction will change the reaction direction until it returns to the equilibrium condition again.
The results of these changes can define as:
Changes on pressure: the reaction will move depending the quantity of moles on each side of the reaction
Changes on temperature: The reaction will move depending on if it's endothermic or exothermic
Changes on volume: The reaction will move depending the limit reagent and the quantity of moles on each side of the reaction
In the exercise, they mention a change on pressure of the system at constant temperature (that means the temperature doesn't change). As Le Chatelier Principle's says, we must analyze what happens if the pressure increase or decrease. If pressure increase the reaction will move on the side that have less quantity of moles, otherwise, if the pressure decreases the reaction will move to the side that have more quantity of moles. In this case, we can see that both sides of the equation have the same number of moles (2 for the reactants and 2 for the products). So, in this case, we can conclude that, despite the change on pressure (increase or decrease), nothing will happen.
The molecule is stable and can exist even though the number of valence electrons around central atom in the molecule are less than 8.
<h3>Is BF3 molecule stable or not?</h3>
BF3 molecule is a stable molecule because all the electrons present in the outermost shell of boron are covalently bonded with fluorine. Boron in BF3, three bonds is the maximum possible because boron only has 3 electrons to share.
So we can conclude that the molecule is stable and can exist even though the number of valence electrons around central atom in the molecule are less than 8.
Learn more about molecule here: brainly.com/question/26044300
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Answer is: f<span>ormula for the hydrated compound is CuSO</span>₄·3H₂O.
ω(H₂O) = 25,3% = 0,253.
ω(CuSO₄) = 100% - 25,3%.
ω(CuSO₄) = 74,7% = 0,747.
ω(H₂O) : M(H₂O) = ω(CuSO₄) : M(CuSO₄).
0,253 : M(H₂O) = 0,747 : 159,6 g/mol.
M(H₂O) = (0,253 · 159,6 g/mol) ÷ 0,747.
M(H₂O) = 54 g/mol.
N(H₂O) = 54 g/mol ÷ 18 g/mol.
N(H₂O) = 3.
Each 100 cm3 of air, constitutes 78cm3 nitrogen, 21cm3 oxygen and 1cm3 constitutes of other gases like Argon, ozone, carbon dioxide and water vapour in small amounts.
The bladder is responsible.
