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yaroslaw [1]
3 years ago
9

Match the following words.

Chemistry
1 answer:
azamat3 years ago
7 0

Answer:

1. active site

2. substrate.

3. activity

4. cofactor

5. complex

6. coenzyme.

Explanation:

1. A substrate binds to an enzyme at the active site where the reaction occurs.

2. In a catalyzed reaction, a reactant is often called a  substrate.

3  activity is a measure of how fast an enzyme can convert the reagent to the  product.

4. An inorganic substance necessary for the function of some enzymes is called a cofactor.

5. When properly aligned, the enzyme and substrate form an enzyme-substrate (ES) complex.

6. A small organic molecule necessary for the function of some enzymes is  called a  coenzyme.

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If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
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Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

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Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

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n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

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2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

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