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3 years ago

All of the following are true concerning enzymes, except that they:

Chemistry

1 answer:

suter [353]3 years ago

8 0

Answer:

D (or E If properly listed to include the active site option)

Explanation:

A. Is true

Enzymes are organically biochemical catalyst and thus they can speed up the rate of chemical reaction in the body

B is true

They are catalysts as said earlier

C is true

They have active sites. An enzyme does not act on all substrates. They have particular group on which they can act. For example, we have carbohydrates enzymes that act on carbohydrates substrate only. This enzymes have no business acting on a protein substrate.

D. Enzymes are proteins

One of the important characteristics of enzymes is that they are protenious in nature

E. This is wrong. Enzymes like any over catalyst are not consumed in the course of the biochemical reaction

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What is the ph of a 0.0055 m ha (weak acid) solution that is 8.2% ionized?

Trava [24]

Answer is: pH value of weak is 3.35.
Chemical reaction (dissociation): HA(aq) → H⁺(aq) + A⁻(aq).
c(HA) = 0.0055 M.
α = 8.2% ÷ 100% = 0.082.
[H⁺] = c(HA) · α.
[H⁺] = 0.0055 M · 0.082.
[H⁺] = 0.000451 M.
pH = -log[H⁺].
pH = -log(0.000451 M).
pH = 3.35.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity an aqueous solution.

3 0

3 years ago

How long does it take the sun to melt a block of ice at 0∘c with a flat horizontal area 1.0 m2 and thickness 1.8 cm ? assume tha

VLD [36.1K]

This is a problem involving heat transfer through radiation. The solution to this problem would be to use the formula for heat flux.

ΔQ/Δt = (1000 W/m²)∈Acosθ

A is the total surface area:
A = (1 m²) + 4(1.8 cm)(1m/100 cm)(√(1 m²))
A = 1.072 m²

ΔQ is the heat of melting ice.
ΔQ = mΔHfus
Let's find its mass knowing that the density of ice is 916.7 kg/m³.
ΔQ = (916.7 kg/m³)(1 m²)(1.8 cm)(1m/100 cm)(333,550 J/kg)
ΔQ = 5,503,780 J

5,503,780 J/Δt = (1000 W/m²)(0.05)(1.072 m²)(cos 33°)
Δt = 122,434.691 s or 34 hours

4 0

3 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$