The reaction: 2 NO2(g) ↔ N2O4(g) has an equilibrium constant, Kc, of 170 at 298K. Analysis of this system at 298K reveals that 1.02E-1 mol of NO2 and 8.00E-3 mol of N2O4 are present in a 27.0-L flask. Determine the reaction quotient, Q, for this mixture. Enter your answer in scientific notation.

The reaction quotient Q is 20.7

Explanation:

Step 1: Data given

Number of moles NO2 = 1.02 *10^-1 = 0.102 moles

Number of moles N2O4 = 8*10^-3 = 0.008 moles

Volume = 27.0 L

Step 2: The balanced equation

2 NO2(g) ↔ N2O4(g)

Step 3: Calculate concentrations

Concentration = moles / volume

[NO2] = moles NO2 / volume

[NO2] = 0.102 moles / 27.0 L

[NO2] = 0.00378 M

[N2O4] = moles N2O4 / volume

[N2O4] = 0.008 moles / 27.0L

[N2O4] = 2.96 *10^-4 M

Q = [N2O4] / [NO2]²

Q = (2.96 *10^-4) / (0.00378²)

Q = 20.7

The reaction quotient Q is 20.7

When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.

c) should be the answer.