Correct Answer: A. \(\frac{8}{3}\)

Explanation

\(\int_{-1}^{1} (x + 1)^{2}\mathrm {d} x \equiv \int_{-1}^{1} (x^{2} + 2x + 1)\mathrm {d} x\)

= \(\left. \frac{x^{3}}{3} + x^{2} + x \right |_{-1}^{1}\)

= \((\frac{1^{3}}{3} + 1^{2} + 1) - (\frac{(-1)^{3}}{3} + (-1)^{2} + (-1)) = \frac{7}{3} + \frac{1}{3} = \frac{8}{3}\)

Correct Answer: C. periscope

Correct Answer: D. suspense

Correct Answer: B. command

Correct Answer: C. tricky

Correct Answer: B) Increase of concentration of CO

Explanation

In order to understand the factors that affect the equilibrium constant of a reaction, we need to learn about Le Chatelier's principle. Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of the equilibrium will shift to counteract the change.

In the given reaction CO\(_{(g)}\) + H\(_2\)O \(\iff\) CO\(_{2(g)}\) + H\(_{2(g)}\), let's discuss how each option affects the equilibrium:

Option A: Increase of temperature

Changing the temperature of a reaction can affect the position of the equilibrium, but it does not change the equilibrium constant. This is because the equilibrium constant is only dependent on temperature. So, an increase in temperature will shift the equilibrium position, but not change the equilibrium constant.

Option B: Increase of concentration of CO(Correct)

According to Le Chatelier's principle, increasing the concentration of reactants will shift the equilibrium position towards the products, but it will not change the equilibrium constant. The equilibrium constant is a ratio of the concentrations of the products to the reactants, so changing the concentration of one reactant does not change the overall constant value.

Option C: Removal of CO\(_2\) from the mixture

Removing a product from the reaction mixture will shift the equilibrium position towards the products, according to Le Chatelier's principle. However, this also does not change the equilibrium constant, as it is still a ratio of the concentrations of the products to the reactants.

Option D: Decrease of pressure

Decreasing the pressure of a reaction will cause the equilibrium to shift in the direction that has more moles of gas, in order to counteract the decrease in pressure. In this case, since both sides of the reaction have the same number of moles of gas, a decrease in pressure will not affect the equilibrium position or the equilibrium constant.

In conclusion, although some of the options can affect the position of the equilibrium, none of them will change the equilibrium constant of the given reaction. To learn more about this, please read the relevant sections of the recommended textbooks.

Correct Answer: B. public service

Correct Answer: C. iron

Correct Answer: D. N525

Correct Answer: A. N4075