HNO3+NaOH ----> H2O
H⁺ +NO3⁻+Na⁺+OH⁻ ---> Na⁺ +NO3⁻ +H2O
H⁺ (aq)+OH⁻(aq)----> H2O(l)
Answer:
Oxygen = 15.999 g/mol
Iron = 3 × 55.845 = 167.535 g/mol
CaCO3 = 20 × 100.0869 = 2001.738 g/mol
The Keq for the reaction N₂ + 3H2 = 2NH3 if the equilibrium concentrations are Keq = 1.5. The correct option is D.
<h3>What is Keq?</h3>
Keq is the ratio of the concentration of reactant to the concentration of the product.
The balanced equation is
N₂ + 3H₂ = 2NH₃
The equilibrium constant is ![\rm \dfrac{[NH_3]^2}{[N_2]\; [H_2]^3}](https://tex.z-dn.net/?f=%5Crm%20%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5C%3B%20%5BH_2%5D%5E3%7D)
The given concentrations of the compounds have been:
Ammonia = 3 M
Nitrogen = 1 M
Hydrogen = 2 M

Thus, the correct option is D. Keq = 1.5.
Learn more about Keq
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False. They don't borrow electrons at all. They already have their respective electron affinities. This is called as electronegativity, and it's an occurence where it already has its own from its actual structure. It never borrows any electrons at all.
Answer:

Explanation:
Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.
The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

The net equation is then:

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:
Actual anode half-equation:

Actual cathode half-equation:

Actual net reaction:
