Answer:
Oxidation numbers assign ownership of the electrons to one atom or another in a compound. ... Scientists use these numbers to help name compounds, write formulas and balance chemical equations. They are also useful when studying reactions and they can help you identify when something is oxidized or reduced.
Metallic bond is made up from Metals as the name states it
Answer:
2
Step-by-step explanation:
A. Moles before mixing
<em>Beaker I:
</em>
Moles of H⁺ = 0.100 L × 0.03 mol/1 L
= 3 × 10⁻³ mol
<em>Beaker II:
</em>
Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.
H⁺][OH⁻] = 1 × 10⁻¹⁴ Divide each side by [H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/[H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)
[OH⁻] = 0.01 mol·L⁻¹
Moles of OH⁻ = 0.100 L × 0.01 mol/1 L
= 1 × 10⁻³ mol
B. Moles after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 3 × 10⁻³ 1 × 10⁻³
C/mol: -1 × 10⁻³ -1 × 10⁻³
E/mol: 2 × 10⁻³ 0
You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.
You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.
C. pH
[H⁺] = (2 × 10⁻³ mol)/(0.200 L)
= 1 × 10⁻² mol·L⁻¹
pH = -log[H⁺
]
= -log(1 × 10⁻²)
= 2
Explanation:
For the given reaction 
Now, expression for half-life of a second order reaction is as follows.
![t_{1/2} = \frac{1}{[A_{0}]k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20%3D%20%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5Dk%7D)
....... (1)
Second half life of this reaction will be 
. So, expression for this will be as follows.
= ![\frac{1}{k} [\frac{1}{[A]_{f}} - \frac{1}{[A_{0}]}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bk%7D%20%5B%5Cfrac%7B1%7D%7B%5BA%5D_%7Bf%7D%7D%20-%20%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5D%7D%5D)
...(2)
where ![[A]_{f}](https://tex.z-dn.net/?f=%5BA%5D_%7Bf%7D)
is the final concentration that is, ![\frac{[A]_{0}}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5D_%7B0%7D%7D%7B4%7D)
here and ![[A]_{i}](https://tex.z-dn.net/?f=%5BA%5D_%7Bi%7D)
is the initial concentration.
Hence, putting these values into equation (2) formula as follows.
= ![\frac{1}{k} [\frac{4}{[A]_{0}} - \frac{1}{[A_{0}]}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bk%7D%20%5B%5Cfrac%7B4%7D%7B%5BA%5D_%7B0%7D%7D%20-%20%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5D%7D%5D)
= ![\frac{3}{[A_{0}]k}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B%5BA_%7B0%7D%5Dk%7D)
...... (3)
Now, dividing equation (3) by equation (1) as follows.
= ![\frac{3}{[A_{0}]k} \times [A_{0}]k](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B%5BA_%7B0%7D%5Dk%7D%20%5Ctimes%20%5BA_%7B0%7D%5Dk)
= 3
or, = 3 
Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.
