According to Le Chatelier's principle, when a system at equilibrium is disturbed, the system will react to restore equilibrium. In other words, it will seek to undo the stress. Here, if hydrogen gas is removed, the reaction will shift toward the reactants to re-form it. In the process, more nitrogen will be produced.

Scandium hydroxide dissociates according to this reaction.

\(\displaystyle Sc(OH)_{3}(s) \rightarrow Sc^{3+}(aq) +3(OH)^{-}(aq)\)

As 0.1N HCl is added, it will quantitatively react with hydroxide anions to produce ScCl₃ and water.

\(\displaystyle H^++3Cl^-+Sc^{3+}+OH^-\rightarrow ScCl_3+H_2O\)

Hydroxide will be removed from the above equilibrium, and the system will compensate by shifting the equilibrium to the right, according Le Chatelier's principle. As the equilibrium shifts to the right, more solid scandium hydroxide is hydrolyzed, resulting in increased solubility.

Since the system is originally at equilibrium and a stress is applied, Le Chatelier's principle is to be considered. By increasing the total pressure, the reaction will move in the direction toward which there are less molecules of gas.

\(\displaystyle 2I_{(g)} \rightleftharpoons I_2_{(g)}\)

Looking at the original reaction there are two moles of gas on the left and only one on the right, indicating that the reaction will shift to the right if the pressure is increased.

Us the solubility constant to calculate the answer to this question. We know that:

\(\displaystyle K_{sp} = [Pb^{2+}] \cdot [CrO_{4}^{2-}]\)

and

\(\displaystyle K_{sp} = 1.8 \cdot 10^{-14}\)

Since \(\displaystyle K_{sp}\) is a very small number, we can assume that the concentration of chromate, \(\displaystyle x\), will be very small compared to the concentration of the added lead, 0.019 M. We can use the two known values (\(\displaystyle K_{sp}\) and lead concentration) to solve for the unknown chromate concentration:

\(\displaystyle 1.8 \cdot 10^{-14} = 0.019 M\cdot x\)

\(\displaystyle x = \frac{1.8 \cdot 10^{-14}}{0.019} = 9.5 \cdot 10^{-13}M\)

The activation energy is the minimum amount of energy required to start a chemical reaction. Based on the collision model, increasing the temperature, pressure, and concentration for a chemical reaction will decrease the activation energy. This is because increasing these factors increases the number of collisions of molecules, which is required for a reaction to occur. Therefore, decreasing the temperature would have the opposite effect causing the activation energy to increase, decreasing the likelihood of a reaction taking place.

According to Le Chatelier's principle, if an equilibrium reaction is disturbed by addition of a reactant or product, the equilibrium will shift to counteract the disturbance. Therefore, adding \(\displaystyle H_{2}\) to the system would cause the equilibrium to shift to the right, producing more \(\displaystyle NH_{3}\) in order to reduce the concentration of \(\displaystyle H_{2}\).

The activation energy is the minimum amount of energy required to start a chemical reaction. Based on the collision model, increasing the temperature, pressure, and concentration for a chemical reaction will decrease the activation energy. This is because increasing these factors increases the number of collisions of molecules. Molecules must collide to react.

To determine the solubility product constant, we only need the concentrations and coefficients of the ions. The effective concentration of any pure substance (solid, liquid, or gas) is equal to one by definition, so does not influence the value of \(\displaystyle \small K_{sp}\). The equation for the solubility product constant of this reaction is:

\(\displaystyle K_{sp}=[C]^c[D]^d\)

The units of the solubility product constant will depend on the coefficients of the products. \(\displaystyle \small K_{sp}\) will be a constant for the reaction, and will not change as more solid dissolves or as the reaction progresses.

The longest carbon chain that can be formed is eight carbons. The base molecule is octane.

Using IUPAC rules, substituents should have the lowest possible numbers; thus, we start counting carbons from the right side rather than the left. If you count from the correct side, there are two methyl groups on carbon 3 and one on carbon 5. Thus, the name of the moleculue is 3,3,5-trimethyloctane.

The given molecule is an alkane. The only way to brominate an alkane is with bromine gas and UV light. The energy from the light serves to creat two radical bromines. These radicals are capable of bonding with alkanes. If the given compound were an alkene, either hydrobromic acid or bromine and peroxides would work.