Answer:
F₂ (g) + FeI₂ (aq) → FeF₂ (aq) + I₂ (l)
Explanation:
Our reactants are:
F₂ → Fluorine gas, a dyatomic molecule
FeI₂ → Iron (II) iodine
Our products are:
I₂ → Iodine
FeF₂ → Iron (II) fluoride
Then, the reaction is:
F₂ (g) + FeI₂ (aq) → FeF₂ (aq) + I₂ (l)
We see it is completely balanced.
Answer:
k = 0.0306 min-1
Explanation:
The table is given as;
Time, Concentration
0 1.48
5 1.27
10 0.98
15 0.84
The integrated rate law for a first order reaction is given as;
ln [A] = -kt + ln [Ao]
where;
[A] = Final Concentration
[Ao] = Initial Concentration
k = rate constant
t = time
In the table, taking the first two sets of values;
t = 5
k = ?
[Ao] = 1.48
[A] = 1.27
Inserting into the equation;
ln(1.27) = - k (5) + ln(1.48)
ln(1.27) - ln(1.48) = -5k
-0.1530 = -5k
k = -0.1530 / -5
k = 0.0306 min-1
Answer:
See explanation below
Explanation:
In this case, let's see both molecules per separate:
In the case of SeO₂ the central atom would be the Se. The Se has oxidation states of 2+, and 4+. In this molecule it's working with the 4+, while oxygen is working with the 2- state. Now, how do we know that Se is working with that state?, simply, let's do an equation for it. We know that this molecule has a formal charge of 0, so:
Se = x
O = -2
x + (-2)*2 = 0
x - 4 = 0
x = +4.
Therefore, Selenium is working with +4 state, the only way to bond this molecule is with a covalent bond, and in the case of the oxygen will be with double bond. See picture below.
In the case of CO₂ happens something similar. Carbon is working with +4 state, so in order to stabilize the charges, it has to be bonded with double bonds with both oxygens. The picture below shows.
Answer:
Boiling point of the solution is 100.78°C
Explanation:
This is about colligative properties.
First of all, we need to calculate molality from the freezing point depression.
ΔT = Kf . m . i
As the solute is nonelectrolyte, i = 1
0°C - (-2.79°C) = 1.86 °C/m . m . 1
2.79°C / 1.86 m/°C = 1.5 m
Now, we go to the boiling point elevation
ΔT = Kb . m . i
Final T° - 100°C = 0.52 °C/m . 1.5m . 1
Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C
Answer:
Hydrofluoric acid.
Explanation:
To know which of the acid is the strongest, let us determine the pka of each acid. This is illustrated below:
1. Acetic acid
Ka = 1.8x10^-5
pKa =..?
pKa = –logKa
pKa = –Log 1.8x10^-5
pKa = 4.74
2. Benzoic acid
Ka = 6.5x10^-5
pKa =..?
pKa = –logKa
pKa = –Log 6.5x10^-5
pKa = 4.18
3. Hydrofluoric acid.
Ka = 6.8x10^-4
pKa =..?
pKa = –logKa
pKa = –Log 6.8x10^-4
pKa = 3.17
4. Hypochlorous acid
Ka = 3.0x10^-8
pKa =..?
pKa = –logKa
pKa = –Log 3.0x10^-8
pKa = 7.52
Note: the smaller the pKa value, the stronger the acid.
The pka of the various acids as calculated above is given below:
Acid >>>>>>>>>>>>>>>>>> pKa
1. Acetic acid >>>>>>>>>> 4.74
2. Benzoic acid >>>>>>>> 4.18
3. Hydrofluoric acid >>>> 3.17
4. Hypochlorous acid >> 7.52
From the above illustration, we can see that hydrofluoric acid has the lowest pKa value. Therefore, hydrofluoric acid is the strongest among them.
