The ideal gas law is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

At constant temperature, Boyle's law states that the product of pressure and volume (PV) is constant, hence P1V1 = P2V2.

According to Raoult's Law, the vapor pressure of the solution is 0.75 * 100 mmHg = 75 mmHg.

Using Gauss's law, the electric field due to an infinite plane sheet of charge is E = σ/2ε₀ on either side of the sheet.

The electric field due to an infinite plane sheet of charge is constant and given by E = σ/2ε₀ on either side of the sheet.

Destructive interference occurs when the path difference is an odd multiple of λ/2 (i.e., (2n+1)λ/2).

f'(1) from left = 2, from right = 2; hence f is differentiable at x = 1.

Mean = 30; Mean Deviation = (|10-30| + |20-30| + |30-30| + |40-30| + |50-30|) / 5 = 10.

The mode is the number that appears most frequently, which is 23.

There is no mode as all values appear only once.

Mean = 6; Mean deviation = (|2-6| + |4-6| + |6-6| + |8-6| + |10-6|)/5 = (4 + 2 + 0 + 2 + 4)/5 = 12/5 = 2.4.

Mean = (4 + 8 + 6 + 5 + 3) / 5 = 26 / 5 = 5.0.

Q1 = 2, Q3 = 4; Interquartile Range = Q3 - Q1 = 4 - 2 = 2.

Mean = 7.5; MAD = (|5-7.5| + |7-7.5| + |8-7.5| + |9-7.5| + |10-7.5|) / 5 = 1.

Mean = 7.5; Variance = [(5-7.5)^2 + (7-7.5)^2 + (8-7.5)^2 + (9-7.5)^2 + (10-7.5)^2] / 5 = 2; Standard Deviation = sqrt(2) = 1.41

The semi-major axis is 4 and the semi-minor axis is 3 after rewriting the equation in standard form.

Substituting x = 1 into the equation gives 1^2 + 2*1 + k = 0 => 1 + 2 + k = 0 => k = -3.

The discriminant must be non-negative: k^2 - 4*1*9 >= 0 => k^2 >= 36 => k <= -6 or k >= 6.

There are 2 moles of gas on the reactant side (2A).

According to Le Chatelier's principle, increasing the concentration of B will shift the equilibrium to the left, resulting in a decrease in the concentration of A.

Decreasing the volume increases the pressure, and the equilibrium will shift towards the side with fewer moles of gas, which is the right side in this case.

Doubling the volume decreases the pressure, which shifts the equilibrium towards the side with more moles of gas, in this case, the right side.

Doubling the volume decreases the pressure, which shifts the equilibrium towards the side with more moles of gas (B and C).

Decreasing the temperature for an endothermic reaction will shift the equilibrium to the left, favoring the formation of HCl.

Decreasing the temperature for an exothermic reaction will shift the equilibrium to the right, favoring the formation of products.

Adding H2 will shift the equilibrium to the right to produce more NH3.

Increasing the pressure will shift the equilibrium to the right, favoring the formation of NH3, which has fewer moles of gas.

Decreasing the volume increases the pressure, which shifts the equilibrium to the side with fewer moles of gas, favoring the production of NH3.

Adding more I2 increases its concentration, which shifts the equilibrium to the right to produce more HI.

Adding more HI increases the concentration of products, which shifts the equilibrium to the left to form more reactants.