An oxygen atom has a mass of and a glass of water has a mass of . Use this information to answer the question below. Be sure you

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.

Answer:

The work done in this process will be considered Negative.

Explanation:

The energy transferred by the system to the environment is negative

Therefore, if work is done on the system, it is labeled as positive. Conversely, when work is done by the system, it is regarded as negative.

In this scenario, the argon gas is expanding, and the work is exerted by the system into the surroundings (container), making the sign Negative.

Thus, the result for the work pertaining to this process will carry a Negative sign.

Response: Water molecules migrate from the dilute to the concentrated solution

Clarification:

During osmosis, when a solution is separated by a semipermeable membrane, the solvent (commonly water) moves from the less concentrated solution, regarding solute content, through the semipermeable membrane towards the solution with a higher concentration to balance the concentration levels between the two solutions.

Thus, in this scenario, water molecules flow from the 0.4M sugar solution to the 0.7M sugar solution through the semipermeable membrane.

(c) Cu + S → CuS is classified as a redox reaction

Explanation:

The following reactions are presented:

(a) K₂CrO₄ + BaCl₂ → BaCrO₄ + 2 KCl

(b) Pb²⁺ + 2 Br⁻ → PbBr₂

(c) Cu + S → CuS

Reaction (c) represents a redox reaction, as the oxidation states of the elements are changing. In this case:

Cu + S → CuS

In its elemental form, Cu has an oxidation state of 0, while in CuS (copper sulfide), its oxidation state changes to +2.

Similarly, S in its elemental form has an oxidation state of 0 and is -2 in CuS (copper sulfide).

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redox reactions