The charged particles in the beams that Thomson studied came from atoms. As these particles moved away from their original atoms

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 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.

Response:
             4.5 m³

Resolution:
              The statement indicates the presence of two blocks on a lid of a container with a volume of 9 m³. The lid's weight is equal to that of the two blocks. Thus, there were initially four blocks (or 4 atm pressure) acting on a volume of 9 m³.

After adding four additional blocks on the lid, the pressure rises from 4 atm to 8 atm (2 atm from the lid, 2 atm from the original blocks, and 4 atm from the new blocks).

Hence, The data established is,

                  P₁  =  4 atm

                  V₁  =  9 m³

                  P₂  =  8 atm

                  V₂  =?

Using Boyle's Law,

                               P₁ V₁  =  P₂ V₂

Resolving for V₂,
                               V₂  =  P₁ V₁ / P₂

Substituting values yields:
                               V₂  =  (4 atm × 9 m³) ÷ 8 atm

                               V₂  =  4.5 m³

Ca3(PO4)2 is the correct formula.

Response:

The pKa value is 13.0.

Clarification:

pKa + pKb = 14

For trimethylamine, Kb = 6.3 ×

Calculating pKb: pKb = - log (6.3 × )

= 1.0

Thus, pKa = 14 - pKb = 14 - 1.0

pKa = 13.0

Verification: The typical range for pKa in weak acids is from 2 to 13.