Answer: a substance that increases the hydroxide ion concentration of a solution
Explanation: According to Arrhenius concept, an acid is a substance which donates hydronium ions ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
ions when dissolved in water and base is a substance which donates hydroxide ions ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
ions in water.
According to the Bronsted Lowry conjugate acid-base pair, an acid is a substance that donates protons and a base is a substance that accepts protons.
According to Lewis concept, an acid is the substance which accepts electron pairs and abase is a substance which donates electron pairs.
Thus a substance that increases the hydroxide ion concentration of a solution is an Arrhenius base.
I believe the correct answer would be the last option. All of the quantities given above contain the same number of particles. We determine this by using the avogadro's number. It represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole.
2 moles of carbon atoms ( 6.022 x 10^23 particles / mole ) = 1.20 x10^24 particles
<span>
2 moles of carbon dioxide molecules </span>( 6.022 x 10^23 particles / mole ) = 1.20 x10^24 particles<span>
2 moles of diatomic oxygen molecules </span>( 6.022 x 10^23 particles / mole ) = 1.20 x10^24 particles
As you can see, no matter what is the gas as long as they have the same number of moles, they would also have same number of particles<span />
<span>what is the atomic mass of an atom of aluminum?
</span>b.13
<span />
Using the Bronsted-Lowry (B/L) Theory tell which is the acid and which the base in .... (c) [H+<span>] = </span>8.1x10-10M [OH-] = 1x10-14/8.1x10-10<span> = 0.12x10</span>-4<span> = 1.2x10</span>-5M ... (c) [OH-] = 4.5x10-7M [H+] = 1x10-14/4.5x10-7<span> = 0.22x10</span>-7<span> = 2.2</span>x10<span>-8</span><span>M</span>
Answer:
The correct answer is 1.33 x 10⁻⁵ M
Explanation:
The concentration of the stock solution is: C= 1.33 M
In the first dilution, the student added 1 ml of stock solution to 9 ml of water. The total volume of the solution is 1 ml + 9 ml = 10 ml. So, the first diluted concentration is:
C₁= 1.33 M x 1 ml/10 ml = 1.33 M x 1/10 = 0.133 M
The second dilution is performed on C₁. The student added 1 ml of 0.133 M solution to 9 ml of water. Again, the total volume is 1 ml + 9 ml = 10 ml. The second diluted concentration is:
C₂= 0.133 M x 1 ml/10 ml = 0.133 M x 1/10= 0.0133 M
Since the student repeated the same dilution process 3 times more (for a total of 5 times), we have to multiply 5 times the initial concentration by the factor 1/10:
Final concentration = initial concentration x 1/10 x 1/10 x 1/10 x 1/10 x 1/10
= initial concentration x (1/10)⁵
= 1.33 M x 1 x 10⁻⁵
= 1.33 x 10⁻⁵ M
