Answer:
HOFO = (0, 0, +1, -1)
Explanation:
The formal charge (FC) can be calculated using the following equation:
<u>Where:</u>
V: are the valence electrons
N: are the nonbonding electrons
B: are the bonding electrons
The arrange of the atoms in the oxyacid is:
H - O₁ - F - O₂
Hence, the formal charge (FC) on each of the atoms is:
H: FC = 1 - 0 - 1/2*(2) = 0
O₁: FC = 6 - 4 - 1/2*(4) = 0
F: FC = 7 - 4 - 1/2*(4) = +1
O₂: FC = 6 - 6 - 1/2*(2) = -1
We can see that the negative charge is in the oxygen instead of the most electronegative element, which is the F. This oxyacid is atypical.
I hope it helps you!
Answer:
Group 18, also known as the Noble Gasses
Explanation:
Atoms strive for full stability by gaining or losing electrons to get 8 valence electrons in their valence shell, but Group 18 already has 8 electrons in their valence shell, and are therefore already stable in their ground state.
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
Answer: The kilograms of water must evaporate from 8kg of a 25% salt solution to produce 40% salt solution is 3 kg.
Explanation:
According to the ratio and proportion:
where,
= concentration of ist solution = 25%
= mass of ist solution = 8 kg
= concentration of second solution = 40%
= mass of second solution = ? kg
Thus the final solution must have a mass of 5 kg , i.e (8-5)= 3 kg of mass must be evaporated.
Therefore, the mass that must be evaporated from 8kg of a 25% salt solution to produce 40% salt solution is 3 kg.
Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
