Answer:
high ph -basic,low pH -acidic
Explanation:
hi I'm from India nice to meet you
Answer:
Explanation:
The given pH = 8.55
Unknown:
[H₃O⁺] = ?
[OH⁻] = ?
In order to find these unknowns we must first establish some relationship.
pH = -log[H₃O⁺]
8.55 = -log[H₃O⁺]
[H₃O⁺] = inverse log₁₀(-8.55) = 2.82 x 10⁻⁹moldm⁻³
To find the [OH⁻],
pH + pOH = 14
pOH = 14 - pH = 14 - 8.55
pOH = 5.45
pOH = -log[OH⁻]
[OH⁻] = inverse log₁₀ (-5.45) = 3.55 x 10⁻⁶moldm⁻³
The solution is basic because it has more concentration of OH⁻ ions compared to H⁺ ions.
Explanation:
When the covalent bonds in a molecule are polarized so that one portion of the molecule experiences a positive charge and the other portion of the molecule experiences a negative charge. This separation of opposite charges creates an electric dipole.
Baking soda and vinegar react because of an acid-base reaction. The product of this creates carbon dioxide.
Answer:
- last option: none of<u> the above.</u>
Explanation:
Describing a solution as<em> concentrated</em> tells that the solution has a relative large concentration, but it is a qualitative description, not a quantitative one, so this does not tell really how concentrated the solution is. This is, the term concentrated is a kind of vague; it just lets you know that the solution is not very diluted, but, as said initially, that there is a relative large amount (concentration) of solute.
One conclusion, of course, is that <u>the solute is soluble</u>: else the solution were not concentrated.
On the other hand, the terms saturated and <em>supersaturated</em> to define a solution are specific.
A saturated solution has all the solute that certain amount of solvent can contain, at a given temperature. A <u>supersaturated solution has more solute dissolved than the saturated solution</u> at the same temperature; superstaturation is a very unstable condition.
From above, there is no way that you can conclude whether a solution is supersaturated or not from the statement that a solution is concentrated, so the answer is<u> none of the above</u>.