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Answer:
<em>The increase in kinetic energy leads to leakage of water from the syringe. When the outside temperature is more than the liquid temperature, say the syringe is out in sunshine, then the liquid becomes slightly warmer.</em>
Energy equals mass times the speed of light squared
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Aluminium oxide is amphoteric. It is easy to see that it is a Bronsted-Lowry base through the following reaction:
Al2O3+6HCl →2AlCl3+3H2O
AlX2OX3+6HCl →2AlClX3+3HX2O
The Alumnium oxide splits and the oxygen accepts a proton, forming water.
But what about the reaction with a base? In my textbook, they say:
Al2O3+NaOH →2NaAlO2+H2O
AlX2OX3+NaOH →2NaAlOX2+HX2O
Now, the textbook claims that Aluminium oxide is an acid merely because it reacts with a base to form a salt and water, as is characteristic of a neutralization reaction.
But I'm not satisfied with this definition. I mean, acids aren't defined as 'things that neutralize bases', we have well-established definitions for them.
I tried to figure out for myself how this could be. Clearly, the Bronsted-Lowry theory cannot be applied here since the compound in question has no protons to donate. Therefore, the only alternative is the Lewis concept. I cannot see how that is applicable in this case.
The most basic definition of "acid" is that it is a proton donor (or one which accepts a lone pair)
All of this stuff is done in an aqueous medium, so we can assume that all aqueous ions and molecules are present. With this assumption (in this case, we are assuming that OH−OHX− is available to react), we get the following equation:
Al2O3+OH−⟶2AlO2−+H+
AlX2OX3+OHX−⟶2AlOX2X−+HX+
Similarly, we get:
Al2O3+6H+⟶2Al3++H2O
AlX2OX3+6HX+⟶2AlX3++HX2O
where it is acting like a proton acceptor (base).
Answer:
A: [H⁺] = 2.4 × 10⁻¹¹ M
B: [H⁺] = 1.2 × 10⁻⁶ M
C: [H⁺] = 1.0 × 10⁻⁸ M
D: A and C are basic and B is acid.
Explanation:
Part A: Calculate [H⁺] for [OH⁻] = 4.1×10⁻⁴ M.
We know the ion-product of water Kw is:
Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]
Then,
Part B: Calculate [H⁺] for [OH⁻] = 8.5×10⁻⁹ M.
Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]
Part C: Calculate [H⁺] for a solution in which [OH⁻] is 100 times greater than [H⁺].
We know that [OH⁻] = 100 . [H⁺]. If we replace this in the ion-product of water:
Kw = 1.0 × 10⁻¹⁴ = [H⁺].[OH⁻]
1.0 × 10⁻¹⁴ = [H⁺]. 100. [H⁺]
1.0 × 10⁻¹⁴ = 100 . [H⁺]²
1.0 × 10⁻¹⁶ = [H⁺]²
[H⁺] = √1.0 × 10⁻¹⁶ = 1.0 × 10⁻⁸ M
Part D: Indicate whether the solution is acidic, basic, or neutral.
A solution is acid when [H⁺] > 10⁻⁷ M, basic when [H⁺] < 10⁻⁷ M and neutral when [H⁺] = 10⁻⁷M. So A and C are basic and B is acid.
The (w/v)% concentration of vinegar contains 5g of acetic in 100mL of solution is approx 0.5 %
<h3>What is Concentration ?</h3>
The concentration of a chemical substance expresses the amount of a substance present in a mixture.
There are many different ways to express concentration.(w/v)% is one of them.
Given ;
- Mass of solute(acetic acid) is 5.0 g
- Volume of solution is 100.0 mL.
We need to convert the grams of acetic acid to moles ;
Formula of acetic acid is CH₃COOH
Thus,
Molar mass of acetic acid = 2(12.01) + 4(1.01) + 2(16)
= 24.02 + 4.04 + 32
= 60.06 grams per mol
Moles of acetic acid = 5.0/60.06
= 0.083 moles
(w/v)% = 0.083 x 60 / 1000 x 100
= 0.498 %
= 0.5 %
Hence, The (w/v)% concentration of vinegar contains 5g of acetic in 100mL of solution is approx 0.5 %
Learn more about weight percentage here ;
brainly.com/question/22575652
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