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Ede4ka [16]
3 years ago
11

A biologist working in a lab adds a compound to a solution that contains an enzyme and substrate. This particular compound binds

reversibly to the enzyme at the active site. Once the compound is bound to the enzyme, the rate of catalysis of substrate to product is greatly reduced. Which of the statements are true of the compound?
Chemistry
1 answer:
ad-work [718]3 years ago
6 0

Answer:

Adding more substrate would overcome  the effect of the compound

Explanation:

  • Enzymes are biochemical catalysts that speed up chemical reactions. They act on specific substrate to convert them to products.
  • Compounds known as inhibitors slow down the rate of enzyme activity.
  • Inhibitors are classified as competitive and non-competitive inhibitors.
  • Competitive inhibitors will compete with the substrate to bind the active sites on the enzyme. The effect of competitive inhibitors may be reduced by increasing the concentration of the substrate.
  • The compound added by the biologist was a competitive inhibitor and therefore adding more substrate would overcome its effect on enzyme catalysis
  • Non-competitive inhibitors binds the active site of the enzyme permanently and prevents the substrate from accessing the active sites.

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Which best represents a homogeneous mixture of an element and a compound.
Allushta [10]

Answer:

The composition of air this is because it vmade up of oxygen, nitrogen, Nobel gages and Carbon dioxide

3 0
3 years ago
The moon is 250,000 miles away. How many inches is it from earth?
Lapatulllka [165]

Answer:

1.584e10 = 15,840,000,000,000

Explanation:

250,000 miles

multiply the length value by 63360

25e4 x 63360

= 1.584e10

3 0
3 years ago
Read 2 more answers
Which ice cream rather have Chocolate- Mint or Cookie IN' Cream?
Effectus [21]

Answer:

Cookies N Cream

Explanation:

Its The

5 0
4 years ago
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The heat of vaporization for benzaldehyde is 48.8 kj/mol, and its normal boiling point is 451.0 k. use this information to deter
user100 [1]

Answer:

The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.

Explanation:

  • To solve this problem, we use Clausius Clapeyron equation: ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂).
  • The first case: P₁ = 1 atm = 760 torr and T₁ = 451.0 K.
  • The second case: P₂ = <em>??? needed to be calculated</em> and T₂ = 61.5 °C = 334.5 K.
  • ΔHvap = 48.8 KJ/mole = 48.8 x 10³ J/mole and R = 8.314 J/mole.K.
  • Now, ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂)
  • ln(760 torr /P₂) = (48.8 x 10³ J/mole / 8.314 J/mole.K) (1/451 K - 1/334.5 K)
  • ln(760 torr /P₂) = (5869.62) (-7.722 x 10⁻⁴) = -4.53.
  • (760 torr /P₂) = 0.01075
  • Then, P₂ = (760 torr) / (0.01075) = 70691.73 torr.

So, The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.

7 0
3 years ago
Equal volumes of two solutions, one containing a strong acid at pH 2 and the other containing a strong base at pH 12, are mixed.
mash [69]

Answer:

7  

Explanation:

Assume we have 1 L of each solution.

Solution 1

\text{[H$^{+}$]}= 10^\text{-pH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of H}^{+} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol H}^{+}}{\text{1 L solution}} = 10^{-2}\text{ mol H}^{+}

Solution 2

pH = 12

pOH = 14.00 - pOH = 14.00 - 12 = 2.0

\text{[OH$^{-}$]}= 10^\text{-pOH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of OH}^{-} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol OH}^{-}}{\text{1 L solution}} = 10^{-2}\text{ mol OH}^{-}

3. pH after mixing

               H⁺  +  OH⁻ ⟶ H₂O

I/mol:     10⁻²    10⁻²  

C/mol:   -10⁻²   -10⁻²

E/mol:      0        0

The H⁺ and OH⁻ have neutralized each other. The pH will be that of pure water.

pH = 7

8 0
4 years ago
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