What type of solid is candle wax ? Amorphous or Crystalline?

A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc

Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

7 0

2 years ago

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$