<span>Reaction: CI2 + H2O ----> HCIO + HCI
Oxidations states:
The oxitation state of Cl2 = 0, because the oxidation state of an atom alone or a molucule with one kind of atom is always 0.
The
oxidation state of Cl in HClO is +1 because the oxidation state of H is
+ 1, the oxidation state of O is - 2, and the molecule is neutral, so
+1 + 1 - 2 = 0
The oxidation state of Cl in HCl is - 1, because the oxidation state of H is +1 and the molecule is neutral, so - 1 + 1 = 0.
Also,
you shall remember that when an atom increases its oxidation state is
is oxidized and when an atoms reduces its oxidations state it is
reduced.
With that you conclude that the right option is the last statement: </span>Cl
has an oxidation number of 0 in Cl2. It is then reduced to CI- with an
oxidation number of –1 in HCl and is oxidized to Cl+ with an oxidation
number +1 in HClO.
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
Answer:
Molality is 0.40 m
Explanation:
Molality is a sort of concentration that indicates the moles of solute dissolved in 1kg of solvent.
To determine molality we need the moles of solute, and the mass of solvent in kg so:
We convert the mass of solvent from g to kg:
740 g . 1kg/1000g = 0.740 kg
We know the moles, so we can determine molality
Molality (mol/kg) = 0.295 mol/ 0.740kg = 0.40 m