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NNADVOKAT [17]
2 years ago
13

A 4.0g Glass was heated from 5°C to 45°C after absorbing 32 J of heat. What is the specific heat of the glass? 

Chemistry
1 answer:
IRINA_888 [86]2 years ago
3 0

Answer:

c=0.2\ J/g^{\circ} C

Explanation:

Given that,

Mass of a glass, m = 4 g

Initial temperature, T_i=5^{\circ} C

Final temperature, T_f=45^{\circ} C

Heat absorbed, Q = 32 J

We need to find the specific heat of the glass. The formula for the heat absorbed is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{32\ J}{4\ g\times (45-5)^{\circ} C}\\\\=0.2\ J/g^{\circ} C

So, the required specific heat of the glass is 0.2\ J/g^{\circ} C.

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and the specific Heat of gold=  0.13 J/g°

and ΔT = (Tf- Ti) = 10°C - 50°C = -40 °C

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∴Heat H = 40 g * 0.13 J/g° * -40
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Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM             |   [S]=KM                  |  [S]>>KM                     | Not true

____________  |   Half of the active  | Reaction rate is         | Increasing

[E_{free}] is about   |    sites are filled of  |    independent of      |  [E_{Total}] will                                            

 equal to [E_{total}]. |                                 |   [S]                             | lower KM

_____________________________________________|____________

[ES] is much       |                                 | Almost all active

 lower than         |                                 | sites are filled

[E_{free}]                  |                                 |

Explanation:

Generally the combined enzyme[ES] is mathematically represented as

                   [ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)

for Michaelis-Menten equation

Where [S] is the substrate concentration and K_M is the Michaelis constant

Considering the statement [S] < < K_M

  Looking at the equation [S] is denominator so it can be ignored(it is far too small compared to K_M)  hence the above equation becomes

               [ES] = \frac{[E_{total}][S]}{K_M}

Since [S] is less than K_M it means that \frac{[S]}{K_M}  < < 1

so it means that [ES] < < [E_{total}]

  What this means is that the  number of combined enzymes[ES] i.e the number of occupied site is very small compared to the the total sites [E_{total}]  i.e the total enzymes concentration which means that the free sites [E_{free}]  i.e the concentration of free enzymes is almost equal to [E_{total}]

Considering the second statement

      [S] = K_M

So  this means that equation one would now become

           [ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

      [S] >>K_M

In this case the K_M in the denominator of equation 1 would be neglected and the equation becomes

       [ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]

This means that almost all the sites are occupied with substrate

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In this case also the K_M at the denominator would be neglected as a result of the statement hence the equation becomes

                v = \frac{V_{max}[S]}{[S]} = V_{max}

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing [E_{total}] will lower K_M

This is because K_M does not depend on enzyme concentration it is a property of a enzyme

             

       

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