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uysha [10]
3 years ago
7

Which component is required for the activation of some enzymes?

Chemistry
2 answers:
shepuryov [24]3 years ago
5 0
The best answer is letter A. 
The main factors that change the speed of enzymatic reactions are temperature, pH and substrate concentration (quantity). <span>There are enzymes that need other associated molecules to work. These molecules are called enzyme co-factors. They can be organic ions like mineral salts.</span>
Alexus [3.1K]3 years ago
5 0

The correct answer is option A.

Presence of a co-factor is required for the activation of some enzymes. Co-factors are generally metal ions or small molecules (non-protein).Co-factors helps enzymes bind substrate tightly.  For example, NAD⁺/NADH is a co-factor which is required for the activation of the enzyme dehydrogenase. NAD⁺ is the oxidized form and NADH is the reduced form.

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Balance the following Chemical Equation:<br> NaBr +CaCl2-&gt; NaCl+ CaBr2
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First write all of the compounds/atoms in either side then fill in existing values and balance


Na- 1
Br- 1
Ca- 1
Cl- 2

Na- 1
Cl- 1
Ca-1
Br-2

Balance to get

2NaBr+CaCl2=2NaCl+CaBr2
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Define adhesion and give an example?
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3 0
2 years ago
in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?
dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of C_3H_5N_3O_9

\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol

Now we have to calculate the moles of CO_2,O_2,N_2\text{ and }H_2O

The balanced chemical reaction is:

4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)

From the balanced chemical reaction we conclude that,

As, 4 moles of C_3H_5N_3O_9 react to give 12 moles of CO_2

So, 1 moles of C_3H_5N_3O_9 react to give \frac{12}{4}=3 moles of CO_2

and,

As, 4 moles of C_3H_5N_3O_9 react to give 1 moles of O_2

So, 1 moles of C_3H_5N_3O_9 react to give \frac{1}{4}=0.25 moles of O_2

and,

As, 4 moles of C_3H_5N_3O_9 react to give 6 moles of N_2

So, 1 moles of C_3H_5N_3O_9 react to give \frac{6}{4}=1.5 moles of N_2

and,

As, 4 moles of C_3H_5N_3O_9 react to give 10 moles of H_2O

So, 1 moles of C_3H_5N_3O_9 react to give \frac{10}{4}=2.5 moles of H_2O

Now we have to calculate the mole fraction of water.

\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}

\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

p_{H_2O}=X_{H_2O}\times p_T

where,

p_{H_2O} = partial pressure of water vapor gas  = ?

p_T = total pressure of gas  = 58 atm

X_{H_2O} = mole fraction of water vapor gas  = 0.345

Now put all the given values in the above formula, we get:

p_{H_2O}=X_{H_2O}\times p_T

p_{H_2O}=0.345\times 58atm=20.01atm

Therefore, the partial pressure of the water vapor is, 20.01 atm

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4 0
3 years ago
Read 2 more answers
How many grams of o2 are required to produce 358.5 grams of zno? 2zn + o2 ® 2zno?
Murrr4er [49]
The balanced equation for the reaction is ;
2Zn + O2 —> 2ZnO
The stoichiometry of O2 to ZnO is 1:2
The mass of ZnO formed - 358.5 g
The number of moles formed - 358.5 g / 81.4 g/mol = 4.4 moles
Therefore number of O2 moles reacted = 4.4 moles /2 = 2.2 mol
Mass of O2 reacted = 2.2 mol x 32 g/mol = 70.4 g
6 0
3 years ago
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