A 0.821 gram sample of pure NH F was treated with 25.0 mL of 1.00 M NaOH

Answer:

D

Reasoning:

This is due to the belief that at one time, the galaxy was condensed into a minute spot and the Big Bang event initiated the universe's expansion, propelling galaxies outward into space, distancing them from one another. Dark energy is theorized to exert the force from the Big Bang that drives this expansion.

The resulting temperature is 46.5°C.

Details:

According to Charles's law, the volume of gas, while maintaining constant pressure, correlates directly with temperature in Kelvin.

The formula representing Charles's law is expressed as follows:

We need to determine T2, thus:

V1 = 736 ml = 0.736 L

T1 = 15 ° C

V2 = 2.28 L

Substituting the values gives us:

T2 =

    = 46.5°C

It is evident that as the volume increases, the temperature also rises.

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.

The net ionic equation for the reaction between chromium (III) hydroxide and nitrous acid is:

Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)

Additional Details

An electrolyte dissociates into ions in solution.

Chemical equations can also be represented with ionic species.

Strong electrolytes (fully ionized) are written as separate ions, whereas weak electrolytes (partially ionized) remain as intact molecules.

In ionic equations, spectator ions are those unchanged by the chemical process—they are present both before and after the reaction.

Removing these spectators results in the net ionic equation.

Gases, solids, and water (H₂O) are written as molecules, without ionization.

Therefore, only dissolved compounds are represented by their ions (aq).

The problem involves chromium (III) hydroxide reacting with nitrous acid.

The reaction occurring is:

Cr(OH)₃ (s) + 3HNO₂ (aq) ⇒ Cr(NO₂)₃ (aq) + 3H₂O (l)

Chromium (III) hydroxide is a solid and remains un-ionized, as does water.

Thus, the ionic equation is:

Cr(OH)₃ (s) + 3H⁺ (aq) + 3NO₂⁻ (aq) ⇒ Cr³⁺ (aq) + 3NO₂⁻ (aq) + 3H₂O (l)

The ion 3NO₂⁻ is a spectator ion; removing it yields the net ionic equation:

Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)