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vodka [1.7K]
3 years ago
7

What is the biological significance of the temperature at which the amylase catalyzed reaction

Chemistry
1 answer:
ipn [44]3 years ago
7 0

Answer:

The biological significance is that it is the normal human body temperature and also the optimum temperature of the enzyme.

Explanation:

  • Enzymes are biological catalysts that speed up the rate of chemical reactions.
  • Enzymes catalyze specific reactions by working on a specific substrate to convert it into a product.
  • The rate of enzyme activity depends on several factors which include pH, temperature, substrate concentration, and enzyme concentration among others.
  • Enzymes work best at a specific pH and temperature known as optimum pH and optimum temperature respectively.
  • In this case, enzyme amylase works best at a temperature of 37° C which is equivalent to the normal human body temperature.
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What is the relative rate of diffusion of hydrogen and nitrogen ​
vredina [299]

Answer:

2.645

Explanation:

Rate of diffusion formula:

Sqrt(mass2/mass1)

>>sqrt(14/2)

(Note:Hydrogen must exist in dwiatomic, [H2])

3 0
2 years ago
Is thiamine mononitrate an ionic bond or covalent bond?
timofeeve [1]
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3 years ago
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g Write a molecular equation for the precipitation reaction that occurs (if any) when the following solutions are mixed. If no r
Mrac [35]

Answer:

CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓

Explanation:

We identify the reactants:

CuBr₂ and Pb(CH₃COO)₂

The products will be: Cu(CH₃COO)₂ and PbBr₂

You may know these information:

Salts from acetate are soluble.

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PbBr₂ is formed, so this will be our precipitate

The equation is:

CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓

8 0
3 years ago
What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a
Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})

Where:

Q: is the energy transferred = 5.0 MJ

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T_{i}: is the initial temperature = 1000 K

T_{f}: is the final temperature = 500 K

Hence, the entropy change is:

\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}                                    

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!          

7 0
3 years ago
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