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rewona [7]
3 years ago
13

Cofactors are important because A. they are the building blocks of proteins. B. they signal cells to release enzymes. C. they pr

event spikes in blood sugar. D. they serve to inactivate an enzyme, so as to better regulate chemical reactions. E. they enable enzymes to bind to their substrates.
Chemistry
2 answers:
olasank [31]3 years ago
7 0

Answer: E. they enable enzymes to bind to their substrates.

Explanation:

The cofactors are the molecules that help the enzyme catalyze the reactions. The Cofactor can be defined as a non-protein chemical compound that either loosely or tightly binds to the enzyme. This facilitates in performing the reactions that enzyme cannot perform alone. They can be  divided into prosthetic group and coenzymes. The cofactors enhances the binding affinity of the enzymes with their substrates so that products can be formed.

katrin2010 [14]3 years ago
7 0

Answer:

The answer is E: they enable enzymes to bind to their substrates.

Explanation:

Cofactors are inorganic substrates that are needed to: 1) produce a chemical reaction between the enzyme and the substrate, 2) increase the rate of catalysis, and 3) attach to the enzyme just like a prosthetic hand, to effectively allow enzymes carry out the catalysis of the reaction.

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The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?
Aloiza [94]

Answer:

0.0468 g.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>

8 0
3 years ago
HELPPP!! Which of the following describes light? (Select all that apply.)
Mnenie [13.5K]

Answer: A.Light travels in a straight line., B.Light behaves in a predictable way. , D.Light curves around corners or obstructions

-Hope this helps<3

5 0
2 years ago
Read 2 more answers
A class of substances known as __________ are used for their medicinal properties.
Misha Larkins [42]

Answer: Alkaloids

A class of substances known as alkaloids are used for their medicinal properties.

Explanation:

Alkaloids are small but complex organic substances with at least one nitrogen atom in its ring structure. Hence, they have strong basic properties, and are produced in naturally by some plants.

Examples of alkaloids and their medicinal properties are as follows:

- caffeine, used in certain drug and drinks to stimulate the nervous system

- morphine, used to reduce pain and induce sleep in humans.

- cocaine, nicotine etc

6 0
3 years ago
Read 2 more answers
1. Calculate the molarity when 403 g MgSO4 is dissolved to make a 5.25 L solution. Round to two significant digits.
lisov135 [29]

Answer:

1. Molarity of MgSO₄ = 0.6 M

2. Molarity of AgNO₃ = 0.06 M

3. volume of acetic acid = 250 mL

4. Molarity of NaCl= 2.3 M

5. Mass of C₁₂H₂₂O₁₁ = 4.1 g

6. Mass of C₁₀H₈ (naphthalene) = 77 g

7. mass of ethanol = 11 g

8. Mass of DDT = 0.011 mg

Explanation:

Ans 1.

Data given

mass of MgSO₄ = 403

Volume of solution = 5.25 L

Solution:

First we will find mole of solute

                 no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of MgSO₄ = 24 + 32 + 4(16)

Molar mass of MgSO₄ = 120 g/mol

Put values in equation 1

             no. of moles = 403 g / 120 g/mol

             no. of moles = 3.36 mol

Now to find molarity

Formula used

            Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

           M = 3.36 mol / 5.25 L

           M = 0.64

So, round the figure to two significant digits.

Molarity of MgSO₄ = 0.64 M

_____________________

Ans 2.

Data given

mass of AgNO₃ = 1.24 g

Volume of solution = 125 mL

convert mL to L

1000 mL = 1 L

125 mL = 125/1000 = 0.125

Solution:

First we will find mole of solute

                 no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of AgNO₃ = 108 + 14 + 3(16)

Molar mass of AgNO₃ = 170 g/mol

Put values in equation 1

             no. of moles = 1.24 g / 170 g/mol

             no. of moles = 0.0073 mol

Now to find molarity

Formula used

            Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

           M = 0.0073 mol / 0.125 L

           M = 0.06

So, round the figure to two significant digits.

Molarity of AgNO₃ = 0.06 M

______________________

Ans 3.

Data Given:

volume of acetic acid = 500 mL

% solution of acetic acid (M)= 50%

volume of acetic acid needed = ?

Solution:

formula used

    percent of solution = volume of solute/ volume of solution x 100

Put Values in above formula

                        50 % = volume of solute / 500 mL x 100

Rearrange the above equation

                   volume of solute =   50 x 500 mL /100

                   volume of solute =   250 mL

So, volume of acetic acid = 250 mL

______________________

Ans 4

Data Given:

Molarity of NaCl (M1) = 6 M

Volume of NaCl (V1) = 750 mL

convert mL to L

1000 ml = 1 L

750 ml = 750/1000 = 0.75 L

Volume of dilute solution (V2)= 2 L

Molarity of dilute solution (M2) = ?

Solution:

Dilution Formula will be used

                M1V1 = M2V2 . . . . . . (1)

Put values in equation 1

               6  M x 0.75 L = M2 x 2 L

Rearrange the above equation

               M2 = 6 M x 0.75 L / 2 L

               M2 = 2.25 M

So, round the figure to two significant digits.

Molarity of NaCl= 2.3 M

_________________________

Ans 5.

Data Given:

Molarity of C₁₂H₂₂O₁₁ = 0.16 M

Mass of C₁₂H₂₂O₁₁ (m) = ?

Volume of dilute solution = 75 mL

convert mL to L

1000 ml = 1 L

75 ml = 75/1000 = 0.075 L

Solution:

As we know

            Molarity = no.of moles/ liter of solution . . . . . . .(1)

we also know that

           no.of moles = mass in grams / molar mass . . . . . (2)

Combine both equation 1 and 2

            Molarity = (mass in grams / molar mass) / liter of solution . . . . (3)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 12(12) +22(1) +11(16)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 144 +22 + 176

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 342 g/mol

Put values in equation in equation 3

              0.16 M = (mass in grams / 342 g/mol) / 0.075 L

Rearrange the above equation

         mass in grams = 0.16 mol/L x 0.075 L x 342 g/mol

         mass in grams = 4.104 g

So, round the figure to two significant digits.

Mass of C₁₂H₂₂O₁₁ = 4.1 g

____________________

The remaing portion is in attachment

8 0
4 years ago
Doubling the mass and object has, will increase its kinetic energy more than doubling its speed.
Bess [88]

<u><em>Answer: The answer is False</em></u>

I did it today and got a 100%

Sorry if i'm late.

7 0
3 years ago
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