The number of moles of H2O is proportional to the number of moles of H2 as much as 4 moles
The calculation reference is the mole of the limiting reagent of the mole of H2O is based on the mol of H2
Stokiometry in Chemistry learns about chemical reactions mainly emphasizing quantitative, such as calculation of volume, mass, number, which is related to the number of ions, molecules, elements, etc.
In chemical calculations, the reaction can be determined, the number of substances that can be expressed in units of mass, volume, mole, or determine a chemical formula, for example, the substance level or molecular formula of the hydrate.
Known variable
4.0 mol of hydrogen and excess oxygen
Asked
the balanced chemical equation
Answer
Reactions that occur:
2H₂ + O₂ ⇒ 2H₂O
The reaction coefficient shows the mole ratio of reactants and reaction products
In determining the mole of the reaction, the calculation reference is the mole of the limiting reagent or the reagent that runs out. Because here it is stated that oxygen (O₂) is reacted under excess conditions, the hydrogen ratio becomes the limiting reagent. So the determination of the mole of H₂O is based on the mol of H₂
In the equivalent equation, the mole of H₂ and H₂O is 2: 2 or can be simplified to 1: 1, so the number of moles of H₂O is proportional to the number of moles of H₂ as much as 4 moles
Excess reactant
The percentage yield
Limiting reactant
Keywords: limiting reactant, excess, reaction coefficient, the balanced chemical equation, H₂O, O₂, H₂
B. decomposition
C. double-replacement reaction
D. synthesis
Answer: Option (C) is the correct answer.
Explanation:
It is given that when an acid and a base form a salt and water.
For example,
Therefore, in this type of reactions there is exchange of ions between the two compounds hence, it is known as double-displacement reaction.
Thus, we can conclude that out of the given options the term an acid and a base form a salt and water describes an acid-base reaction.
Potassium hydroxide KOH is a strong base which can completely ionize to form potassium and hydroxide ions.
The equation representing ionization of KOH,
The mole ratio of hydroxide to KOH from the above equation is
The given concentration of KOH solution=0.30M
Calculating the concentration of hydroxide ions present in 0.30M KOH solution:
Therefore the concentration of hydroxide ion will be equal to 0.30 M in a 0.30 M KOH solution.
The concentration of hydroxide ion in a 0.30 m solution of potassium hydroxide would be equal to 0.30M. Potassium hydroxide is a strong base, it completely dissociates in water producing an equal concentration of hydroxide ions. In a basic solution where the concentration of Hydroxide ions is high, the concentration of hydronium ions will be low.
The concentration of hydroxide ion in a 0.30 m solution of potassium hydroxide (KOH) would be equal to 0.30 M. This is based on the fact that potassium hydroxide is a strong base and it completely dissociates in water, which means one KOH molecule produces one hydroxide ion. Therefore, the concentration of the hydroxide ion [OH-] in the final solution would be equal to the initial concentration of the potassium hydroxide.
This is different from the behavior of weak acids and bases which do not completely dissociate in water, leading to a lower concentration of hydronium (H3O+) or hydroxide (OH-) ions compared to the initial concentration of the acid or base.
However, it's important to note that in solutions with a high concentration of hydronium ions, the concentration of hydroxide ions will decrease according to Le Chatelier's principle, which describes how equilibrium adjusts when it's disturbed. Conversely, in a basic solution—like this 0.30 M KOH solution—where the concentration of hydroxide ions is high, the concentration of hydronium ions will be comparatively low.
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Answer:
Increasing
Explanation: