boyles law states that the volumes of a gas will decrease as pressure increases if the temperature remains constant.
charles law states that the volume of a gas will increase as temp increases if the pressure remains constant.
gay-lussacs law states that the pressure increases as temp increases if the volume remains constant.
Hydrogen gas is produced when dilute hydrochloric acid is added to a reactive metal.
Balanced molecular equation of sodim metal with hydrochloric acid:
2Na(s) + 2HCl(aq) → 2NaCl(aq) + H₂(g).
Ionic equation: 2Na(s) + 2H⁺(aq) + 2Cl⁻(aq) → 2Na⁺ + 2Cl⁻(aq) + H₂(g).
Net ionic equation: 2Na(s) + 2H⁺(aq) → 2Na⁺(aq) + H₂(g).
Sodium is oxidized from oxidation number 0 (Na) to oxidation number +1, hydrogen is reduced from oxidation number +1 to oxidation number 0 (hydrogen gas H₂).
Another example:
Balanced chemical equation: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
Word equation: zinc + hydrochloric acid → zinc chloride + hydrogen gas
More about hydrogen gas:brainly.com/question/24433860
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Answer:
please mark my answer brainliest
Explanation:
its carbon monoxide
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Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,
pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
![\frac{[Deprotonated]}{[Protonated]}=0.01](https://tex.z-dn.net/?f=%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D%3D0.01)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
![\frac{[Protonated]}{[Deprotonated]}=100](https://tex.z-dn.net/?f=%5Cfrac%7B%5BProtonated%5D%7D%7B%5BDeprotonated%5D%7D%3D100)
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
