How many moles are 25.0 mL of benzene (density 0.876 g/mL)?

Givens

• V1 = 5.00 L
• V2 =?
• T1 = 298 K
• T2 = 333 K

Formula

V1/T1 = V2/T2

Note: This will be applicable only if the pressure remains constant.

Solution

5.00L / 298 K = x / 333 K. Multiply both sides by 333 K.

5.00 * 333 / 298 = x 333/ 333

V2 = 5.59 L

Answer: The energies of infrared photons are comparable to those linked with various vibrational states of chemical bonds. Molecules can absorb infrared photons of specific wavelengths, highlighting the types and strengths of different chemical bonds present within the molecules.

Explanation:

Infrared spectroscopy evaluates the vibrational energy states found in molecules. When a molecule absorbs infrared photons, the chemical bonds vibrate at distinct frequencies. Scrutinizing the alterations in vibrational energy within a molecule allows for the identification of different bond types and consequently the molecule’s general structure. The vibrational behaviors of a molecule encompass bending, stretching, and scissoring motions.

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and cut (q) down to a third

Explanation:

According to the principle of Le Chatelier, when a system reaches equilibrium and a change is introduced, the system will respond to counteract that change.

Since P and Q are reactants, raising the amount of either one or both without a proportional rise in R (which is a product) will cause the equilibrium to move towards the right. Similarly, if R decreases while P and Q remain constant, this too will push the equilibrium to the right. Thus, Increase(P), Increase(q), and Decrease(R) will lead to a rightward shift in the equilibrium.

Conversely, raising R without increasing P and Q will draw the equilibrium to the left. Likewise, cutting down P and/or Q without a similar reduction in R will shift the equilibrium leftward. Therefore, Increase(R), Decrease(P), Decrease(q), and triple both (Q) and (R) will shift the equilibrium to the left.

If there are equivalent changes in P and Q, with R remaining unchanged, then the equilibrium remains stationary. So, tripling (P) while reducing (q) to one third will not alter the equilibrium.