Electrolytes are substances that produce ions when they dissolve in water.
What are electrolytes?
When some substances are dissolved in water, they undergo physical or chemical changes, creating ions in solution. These substances form an important class of compounds called electrolytes. Substances that do not release ions when dissolved are called non-electrolytes. A substance is said to be a strong electrolyte if the physical or chemical process that produces ions is inherently 100% efficient (all dissolved compounds produce ions). A solute is said to be a weak electrolyte if only a relatively small portion of the solute undergoes ion production processes.
By measuring the electrical conductivity of aqueous solutions containing substances, substances can be identified as strong, weak, or non-electrolyte. To conduct electricity, a substance must contain free-moving charged species. The best known is the conduction of electricity through metal wires. In this case, the mobile charged unit is the electron.
Therefore, Electrolytes are substances that produce ions when they dissolve in water.
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Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
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If you are provided with Cation and an Anion with different oxidation states, then there ratio in the formula unit is adjusted as such that the oxidation number of one ion is set the coefficient of other ion and vice versa,
Example:
Let suppose you are provided with A⁺² and B⁻¹, so multiply A by 1 and B by 2 as follow,
A(B)₂
In statement we are given with Co⁺³ and SO₄⁻², so multiply Co⁺³ by 2 and SO₄⁻² by 3, hence,
Co₂(SO₄)₃
Result:
Co₂(SO₄)₃ is the correct answer.
Answer:
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Explanation:
Electrolysis is not possible with solid lead (II) bromide. This is because the ions are held in a three-dimensional lattice, unable to move freely to the electrodes. Melting enables the ions to become mobile and to travel to the respective electrodes.
The bulb won't glow when the electrodes are embedded in solid lead bromide. The bulb will glow when the material surrounding the electrodes is molten lead bromide. When an ionic compound is in the molten (liquid) form the positive and negative ions are free to move around.
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