Option (B) is the right answer.
Hydronium ion (
) concentration decreases by the <u>factor of 100</u>, if the pH of a solution increases from 2.0 to 4.0.
<h3>What is pH?</h3>
The hydrogen ion concentration in water is expressed by pH. Specific to aqueous solutions, pH is the <u>negative logarithm</u> of the hydrogen ion (H+) concentration (mol/L) : ![pH = -log_{10}[H_{3}O^{+} ]](https://tex.z-dn.net/?f=pH%20%3D%20-log_%7B10%7D%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D)
Acidic solutions are those with a pH under 7, and basic solutions are those with a pH over 7. At this temperature, solutions with a pH of 7 are neutral (e.g.<u> pure water</u>). The pH neutrality <u>relies on temperature, falling below 7 if the temperature rises above 25 °C</u>.
<h3>Given: </h3>
pH1( initial pH) = 2.0
pH2( initial pH) = 4.0
[H3O+] = initial hydronium concentration
[H3O+]* = final hydronium concentration
<h3>Formula used : </h3>
![pH = -log_{10}[H_{3}O^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-log_%7B10%7D%5BH_%7B3%7DO%5E%7B%2B%7D%5D)
<h3>Solution: </h3>
![pH = - log_{10}[H_{3}O^{+}] \\\\= > 10^{-pH} = [H_{3}O^{+}] \\\\similarly, \\\\10^{-pH} = [H_{3}O^{+}]*\\\\ = > 10^{-4} = [H_{3}O^{+}]*\\\\Now, \frac{[H_{3}O^{+}]*}{[H_{3}O^{+}]} = 10^{-2}](https://tex.z-dn.net/?f=pH%20%3D%20-%20log_%7B10%7D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%5C%5C%5C%5C%3D%20%3E%2010%5E%7B-pH%7D%20%3D%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%5C%5C%5C%5Csimilarly%2C%20%5C%5C%5C%5C10%5E%7B-pH%7D%20%3D%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%2A%5C%5C%5C%5C%20%3D%20%3E%2010%5E%7B-4%7D%20%3D%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%2A%5C%5C%5C%5CNow%2C%20%5Cfrac%7B%5BH_%7B3%7DO%5E%7B%2B%7D%5D%2A%7D%7B%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%20%20%3D%2010%5E%7B-2%7D)
Thus , the concentration of hydronium ion decreases by 100.
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Answer:
-219.99kJ
Explanation:
The acronym '' NADH'' simply stands for what is known as coenzyme 1 with full meaning of Nicotinamide Adenine Dinucleotide Hydride. This substance is useful in the production of energy. The oxidation reaction of NADH causes it to produce NADP⁺ and the oxygen produces water when it is in the reduction process. The balanced equation for the oxidation reaction is given below as:
NADPH ---------------------------------------------------------------------> NADP⁺H⁺ + 2e⁻.
Also, the balanced equation for the reduction reaction is given below as:
O₂ + 2H⁺ + 2e⁻ --------------------------------------------------------------> H₂O.
It can be shown from the above REDOX reaction that the total number of electrons getting transferred is 2.
The Gibbs energy = -nFE. where n = 2, F = faraday's constant = 96485.3329 C and E = overall cell potential.
The overall cell potential = E[ reduction reaction] - E[oxidation reaction] = 0.82 - (- 0.32 ) = 1.14 V.
Hence, the Gibbs energy = - 2 × 96485.3329 × 1.14 = -219.99kJ
Answer:
T = -28.2 °C
Solution:
Data Given;
P = 1820 mmHg = 2.39 atm
V = 5.12 L
m = 19.4 g
R = 0.0821 atm.L.mol⁻¹.K⁻¹
Calculating moles,
n = 19.4 g ÷ 32 g.mol⁻¹
n = 0.606 mol
Assuming that the oxygen gas is acting perfectly, then according to Ideal Gas Equation,
P V = n R T
Solving for T,
T = P V / n R
Putting values,
T = (2.39 atm × 5.12 L) ÷ (0.606 mol × 0.0821 atm.L.mol⁻¹.K⁻¹)
T = 244.7 K
Or,
T = -28.2 °C
Answer: What best describes the reaction is "A single replacement reaction takes place because zinc is more reactive than hydrogen."
To solve for the new water level, let us first calculate
for the volume of the copper cylinder. The formula we can use is:
V = m / ρ
where V is the volume, m is the mass of the copper and ρ
is the density, therefore:
V = 48.85 g / (8.96 g / cm^3)
V = 5.45 cm^3
since cm^3 = mL
V = 5.45 mL
The new water level will be the initial volume of the
graduated cylinder plus the volume of the copper cylinder, therefore:
new water level = 20 mL + 5.45 mL
new water level = 25.45 mL