Using mathematical expressions drive the compression factor

What assumptions do we make in order to use the Henderson-Hasselbalch equation? a. Both the weak acid and its conjugate base are

zepelin [54]

Answer:

The final and initial concentration of the acid and it's conjugate base are approximately equal, that is we use the weak acid approximation.

Explanation:

The Henderson-Hasselbalch is used to calculate the pH of a buffer solution. It depends on the weak acid approximation.

Since the weak acid ionizes only to a small extent, then we can say that [HA] ≈ [HA]i

Where [HA] = final concentration of the acid and [HA]i = initial concentration of the acid.

It also follows that [A^-] ≈ [A^-]i where [A^-] and[A^-]i refer to final and initial concentrations of the conjugate base hence the answer above.

7 0

3 years ago

What are the different elements that make up a molecule of glucose?

Yanka [14]

Answer:

Glucose has a chemical formula of: C6H12O6 That means glucose is made of 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms. You will be building one type of sugar called glucose.

Explanation:

4 0

3 years ago

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A chemist mixes 1.00 g CuCl2 with an excess of (NH4)2HPO4 in dilute aqueous solution . He measures the evolution of 670 J of hea

kaheart [24]

Answer:

$\Delta H$ will be 90054 J

Explanation:

Number of moles = (mass)/(molar mass)

Molar mass of $CuCl_{2}$ = 134.45 g/mol

So, 1.00 g of $CuCl_{2}$ = $\frac{1.00}{134.45}mol$ of $CuCl_{2}$ = 0.00744 mol of $CuCl_{2}$

0.00744 mol of $CuCl_{2}$ produces 670 J of heat

So, 1 mol of $CuCl_{2}$ produces $\frac{670}{0.00744}J$ of heat or 90054 J of heat

4 0

4 years ago

Consider the decomposition of red mercury(II) oxide under standard state conditions. )H0 T SFE ڮ( H M 0 H (a) Is the decompositi

sertanlavr [38]

The question is incomplete, complete question is :

Consider the decomposition of red mercury(II) oxide under standard state conditions.

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

Given :

$\Delta S_{HgO}^o=70.29 J/mol K$

$\Delta S_{Hg}^o=75.9 J/mol K$

$\Delta S_{O_2}^o=205.2 J/mol K$

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer:

a) The decomposition is spontaneous under standard state conditions.

b)The reaction will spontaneous above -419.69 Kelvins.

Explanation:

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

Given :

$\Delta S_{HgO}^o=70.29 J/mol K$

$\Delta S_{Hg}^o=75.9 J/mol K$

$\Delta S_{O_2}^o=205.2 J/mol K$

Entropy change of the reaction ; ΔS

$\Delta S=[2\times \Delta S_{Hg}^o+1\times \Delta S_{O_2}^o]-[2\times \Delta S_{HgO}^o]$

$=[2\times 75.9 J/mol K+1\times 205.2 J/molK]-[2\times 70.29 J/molK]=216.42 J/mol K$

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K

1 kJ = 1000 J

At standard condition the value of temperature = T = 298 K

ΔG = ΔH - TΔS

ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol

ΔG < 0 ( spontaneous)

The decomposition is spontaneous under standard state conditions.

b) Above what temperature does the reaction become spontaneous

Let the ΔG = 0

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K

ΔG = ΔH - TΔS

0 = -90830 J/mol K - T × 216.42 J/mol K

T = -419.69 K

The reaction will spontaneous above -419.69 Kelvins.

6 0

3 years ago

Then, explain how natural resources are identified and why natural resources are unevenly distributed. Be sure to use informatio

Levart [38]

D s w a s. srvevshaagahBahWhaba scavenger

8 0

3 years ago

Using mathematical expressions drive the compression factor​

Using mathematical expressions drive the compression factor