Hello!
To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:
![pH=pKa+log( \frac{[A^{-}] }{[HA]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%20%7D%7B%5BHA%5D%7D%20%29)

Now, we use the definition of pH and clear [H₃O⁺] from there:
![pH=-log[H_3O^{+}]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%7B%2B%7D%5D%20)
![[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M](https://tex.z-dn.net/?f=%20%5BH_3O%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-pH%7D%20%3D10%5E%7B-3%2C84%7D%3D0%2C00014%20M)
So, the [H₃O⁺] concentration is
0,00014 M
Have a nice day!
Answer:
1.3 L.
Explanation:
- Molarity is the no. of moles of solute per 1.0 L of the solution.
<em>M = (no. of moles of CaSO₄)/(Volume of the solution (L))</em>
<em></em>
M = 0.352 M.
no. of moles of CaSO₄ = mass/molar mass = (62.1 g / 136.14 g/mol) = 0.456 mol,
Volume of the solution = ??? L.
∴ (0.352 M) = (0.456 mol)/(Volume of the solution (L))
<em>∴ (Volume of the solution (L) </em>= (0.456 mol)/(0.352 M) = <em>1.296 L ≅ 1.3 L.</em>
<u>Answer:</u> The above reaction is non-spontaneous.
<u>Explanation:</u>
For the given chemical reaction:

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.
We know that:

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the
of the reaction, we use the equation:


Relationship between standard Gibbs free energy and standard electrode potential follows:

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.
Hence, the above reaction is non-spontaneous.
100°C because all the molecules are moving the fastest past each other
Answer:
24e⁻ are transferred by the reaction of respiration.
Explanation:
C₆H₁₂O₆ + 6O₂ → 6 H₂O + 6CO₂
This is the reaction for the respiration process.
In this redox, oxygen acts with 0 in the oxidation state on the reactant side, and -2 in the product side - REDUCTION
Carbon acts with 0 in the glucose (cause it is neutral), on the reactant side and it has +4, on the product side - OXIDATION
6C → 6C⁴⁺ + 24e⁻
In reactant side we have a neutral carbon, so as in the product side we have a carbon with +4, it had to lose 4e⁻ to get oxidized, but we have 6 carbons, so finally carbon has lost 24 e⁻
6O⁻² + 6O₂ + 24e⁻ → 6O₂²⁻ + 6O⁻²
In reactant side, we have 6 oxygen from the glucose (oxidation state of -2) and the diatomic molecule, with no charge (ground state), so in the product side, we have the oxygen from the dioxide with -2 and the oxygen from the water, also with -2 at the oxidation state. Finally the global charge for the product side is -36, and in reactant side is -12, so it has to win 24 e⁻ (those that were released by the C) to be reduced.