Answer:
An -ite or -ate ending means a polyatomic ion that includes oxygen is in the formula.
ate = one more O than -ite
Explanation:
Example:
NO2 is nitrite
NO3 is nitrate
You can see that the molecules are made of the same elements, however they have more or less of one of the elements (i.e. oxygen).
From the calculation, the concentration of the oxonium ions is 1 * 10^-6 M
<h3>What is concentration?</h3>
The term concentration refers to the amount of substances present in the solution. There are several units that can be used to show the concertation of a substances such as moles/liter, gram per liter, parts per billion, parts per billion, percentage etc.
Now we know that water is composed of the hydrogen and the hydroxide ions and the product of the both is generally known as the ion product of water and have a value of 1 * 10^-14.
If that be the case, we are in order to write the expression;
[H3O+] [OH-] = 1 * 10^-14
[H3O+] = 1 * 10^-14/ [OH-]
[H3O+] = 1 * 10^-14/ 1 x 10^ -8
[H3O+] = 1 * 10^-6 M
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The Nernst equation is:
E = E° - RTlnK/nF
where
E° is the standard potential voltage
R is the universal gas constant = 8.314 J/mol·K
K is the reaction quotient
n is the number of moles electrons transferred
F is Faraday's constant = 96,500 C/mol e⁻
Let's determine K first. The overall reaction is:
Fe(s) + Cd²⁺(aq) --> Fe²⁺(aq) + Cd(s)
Accounting for aqueous phases only,
K = [products]/[reactants] = [Fe²⁺]/[Cd²⁺] = 0.10 M/1.4 M = 1/14
From the reactions written, you can see that 2 electrons were transferred. So, n = 2.
Lastly, the value for E⁰ is the sum of individual E⁰ of the reactions.
E⁰ = E⁰cathode - E⁰anode
Cathode reaction: Fe(s) --> 2e + Fe²⁺ (E⁰cathode = 0.44 V)
Anode reaction: Cd²⁺ + 2e --> Cd(s) (E⁰anode = -0.4 V)
Thus,
E⁰ = 0.44 - -0.4 = 0.84 V
Substituting the values (assume T at room temperature = 298 K),
E = 0.84 - (8.314)(298 K)(ln 1/14)/(2)(96,500)
<em>E = 0.87 V</em>
The thermal energy will flow from the hotter object to the cooler object until their temperatures are equal i.e. until they reach equilibrium.
Answer:
D. equilibrium