Which equation is derived from the combined gas law? StartFraction V subscript 1 over T subscript 1 EndFraction equals StartFrac

Answer:

Explanation:

The relationship between the new temperature scale and the absolute temperature scale is defined as follows

Aw = 2 K

for K = 273.15 (the freezing point of water on the absolute scale)

Aw = 2 x 273.15 = 546.3 K

Each division of the new scale is equivalent to half that of each division on the absolute scale

each division of the new scale is minimal.

The value of R = 8.314 J per mole per K

Here, per K corresponds to 2Aw

Hence, the value of R in the new scale = 8.314/2 J per mole per Aw

= 4.157 J per mole per Aw

k = R / N

= 4.157 / 6.02 x 10²³

= .69 x 10⁻²³

= 6.9 x 10⁻²⁴ J per molecule per Aw .

Answer:

pH = 8.0

Explanation:

Initially, we need to determine the moles of NaOH.

Consider the balanced reaction.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio among C₂H₄O₃, NaOH, and C₂H₃O₃Na is 1: 1: 1. Therefore, if 7.2 × 10⁻⁴ moles of NaOH completely react with 7.2 × 10⁻⁴ moles of C₂H₄O₃, they will yield 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

To find the concentration of C₂H₃O₃Na, we note that:

C₂H₃O₃Na dissociates as follows:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

The anion C₂H₃O₃⁻ originates from a weak acid, which thus goes through basic hydrolysis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

Given that the pKa for C₂H₄O₃ is 3.9, we can derive pKb for C₂H₃O₃⁻ using this relation:

pKa + pKb = 14

pKb = 14 - 3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can compute [OH⁻] through the equation:

[OH⁻] = √(Kb.Cb)               where Cb indicates the base's initial concentration

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

We can then find pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0