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irina [24]
3 years ago
11

You are working with a specific enzyme-catalyzed reaction in the lab. You are a very careful experimentalist, and as a result, a

t the beginning of each of your experiments, you measure the temperature in the lab. On days 1 through 5, the temperature in the lab was 20oC. Today is day 6 of your experiment, and the temperature in the lab is 30oC. How do you predict that this will alter the rate of your enzyme-catalyzed reaction?
A. It will decrease the rateB. It will increase the rateC. It could possibly increase or decrease the rateD. it will not affect the rate
Chemistry
1 answer:
lord [1]3 years ago
7 0

B. It will increase the rate

Explanation:

In this experiment, the rate of the reaction will increase with increase in temperature. Since enzymes are chemical catalysts that speeds up the rate of chemical reactions, one must know that a high temperature favors the rate of chemical reaction.

  • At a high temperature, the catalyst action of the enzyme will increase rapidly.
  • Caution must be taken because at extremely high temperatures, the enzymes can become denatured.

Learn more:

proteins as enzymes: brainly.com/question/13022851

#learnwithBrainly

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Which type of chemical reaction takes place in an electrolytic cell
Firlakuza [10]

Answer:

The reaction that takes place in an electro-chemical cell is the Redox reaction, a type of reversible reaction combining both oxidation and reduction reactions.

8 0
3 years ago
If a solution initially contains 0.260 M HC2H3O2, what is the equilibrium concentration of H3O+ at 25 ∘C? Express your answer in
arlik [135]

Answer:

2.16 × 10⁻³

Explanation:

Step 1: Given data

Concentration of the acid (Ca): 0.260 M

Acid dissociation constant (Ka): 1.80 × 10⁻⁵

Step 2: Write the acid dissociation equation

HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)

Step 3: Calculate the concentration of H₃O⁺ at equilibrium

We will use the following expression.

[H_3O^{+} ]= \sqrt{Ka \times Ca } = \sqrt{1.80 \times 10^{-5} \times 0.260 } = 2.16 \times 10^{-3}

8 0
3 years ago
Acetone has a boiling point of 56.5 celcius. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57
vodomira [7]

Answer:

0.515 g

Explanation:

<em>Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?</em>

<em />

Step 1: Given data

Temperature (T): 57°C

Pressure (P): 730 mmHg

Volume (V): 250 mL

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = 57°C + 273.15 = 330 K

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 760 mmHg.

730 mmHg × (1 atm/760 mmHg) = 0.961 atm

Step 4: Convert "V" to L

We will use the conversion factor 1 L = 1,000 mL.

250 mL × (1 L/1,000 mL) = 0.250 L

Step 5: Calculate the moles (n) of acetone

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K

n = 8.87 × 10⁻³ mol

Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone

The molar mass of acetone is 58.08 g/mol.

8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g

8 0
3 years ago
A liquid boils when its vapor pressure equals the pressure of the atmosphere.
JulijaS [17]
The answer is true. A liquid boils when its vapor pressure equals the pressure of the atmosphere. At this point, vapor can now readily escape from the liquid phase because it has an equal pressure now with its surrounding which, basically, is in the vapor phase.
3 0
3 years ago
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What is similar about igneous rock and metamorphic rock formation?
Harman [31]

Answer:

I think they both need heat.

Explanation:

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