Answer please!!!!!!!!!!!!!!!!!!!!!!!!✌✌✌

Which statements regarding the Henderson-Hasselbalch equation are true? If the pH of the solution is known as is the pKa for the

Sonbull [250]

Answer:

1, 2, and 3 are true.

Explanation:

The Henderson-Hasselbalch equation is:

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. TRUE

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

If you know pH and pka:

10^(pH-pka) = $\frac{[A^-]}{[HA]}$

The ratio will be: 10^(pH-pka)

At pH = pKa for an acid, [conjugate base] = [acid] in solution. TRUE

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

0 = log₁₀ $\frac{[A^-]}{[HA]}$

10^0 = $\frac{[A^-]}{[HA]}$

1 = $\frac{[A^-]}{[HA]}$

As ratio is 1, [conjugate base] = [acid] in solution.

At pH >> pKa for an acid, the acid will be mostly ionized. TRUE

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

If pH >> pKa, 10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]

At pH << pKa for an acid, the acid will be mostly ionized. FALSE

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

If pH << pKa, 10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]

I hope it helps!

6 0

3 years ago

Which horizon would have the highest amount of organic material?

alexandr402 [8]

O horizon is the answer

5 0

3 years ago

3. During an experiment where 50.0 mL of a 1.0 M acid solution was mixed with 50.0 mL of a 1.0 M base solution, the temperature

Aleksandr-060686 [28]

Answer:

$\large \boxed{\text{-61 kJ$\cdot$mol$^{-1}$}}$

Explanation:

Data:

H₃O⁺ + OH⁻ ⟶ 2H₂O

V/mL: 50.0 50.0

c/mol·L⁻¹: 1.0 1.0

ΔT = 6.5 °C

ρ = 1.210 g/mL

C = 4.18 J·°C⁻¹g⁻¹

C_cal = 12.0 J·°C⁻¹

Calculations:

(a) Moles of acid

$\text{Moles of acid} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}\\\\\text{Moles of base} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}$

(b) Volume of solution

V = 50.0 mL + 50.0 mL = 100.0 mL

(c) Mass of solution

$\text{Mass of solution} = \text{100.0 mL} \times \dfrac{\text{1.10 g}}{\text{1 mL}} = \text{110.0 g}$

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

nΔH + mCΔT + C_calΔT = 0

0.0500ΔH + 1.10×4.18×6.5 + 12.0×6.5 = 0

0.0500ΔH + 2989 + 78.0 = 0

0.0500ΔH + 3067 = 0

0.0500ΔH = -3067

ΔH = -3067/0.0500

= -61 000 J/mol

= -61 kJ/mol

$\text{The enthalpy of reaction is $\large \boxed{\textbf{-61 kJ$\cdot$mol$^{\mathbf{-1}}$}}$}$

Note: The answer can have only two significant figures because that is all you gave for the change in temperature.

7 0

3 years ago

Which of the following reasons correctly explains the color changes that take place when ethylenediamine (C2N2H8) is added to th

pickupchik [31]

Answer:

The water ligands surrounding the cobalt metal center are being replaced by ethylenediamine and chloride ligands which results in a different crystal field splitting. Thus, the energy associated with electron transitions between the do-orbitals will differ for the two compounds showing a color change.

Explanation:

The five d-orbitals are usually degenerate. Upon approach of a ligand, the d-orbitals split into two sets of orbitals depending in the nature of the crystal field.

The magnitude of crystal field splitting is affected by the nature of the ligand. Ligands having filled p-π orbitals such as ethylenediamine lead to greater crystal field splitting.

The change in the colour that takes place when ethylenediamine is added to the solution of cobalt(II) chloride occurs due to a different crystal field splitting pattern. Thus, the energy associated with electron transitions between the d-orbitals now differ for the two compounds showing a color change.

7 0

3 years ago

What is the scale factor?

professor190 [17]

Answer:

Scale Factor: 1 : 2

X = 11

Explanation:

3 / 3 = 1

6 / 3 = 2

The scale factor should be 1:2

So 'X' should be 11 since 5.5 * 2 = 11

4 0

3 years ago

Answer please!!!!!!!!!!!!!!!!!!!!!!!!✌✌✌​

Answer please!!!!!!!!!!!!!!!!!!!!!!!!✌✌✌