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Musya8 [376]
2 years ago
12

The ΔG for a particular enzyme-catalyzed reaction is -20 kcal/mol. If the amount of enzyme in the reaction is doubled, what will

be the ΔG for the new reaction?
Chemistry
1 answer:
LUCKY_DIMON [66]2 years ago
3 0

Answer:

The value of the of ΔG for the new reaction will be same as the given value of -20kcal/mol.

Explanation:

In an enzyme-catalyzed reaction, the concentration or amount of enzyme will not affect the equilibrium constant of the reaction due to which ΔG for the reaction will remain unaffected. Here enzymes are acting as a catalyst that cannot alter law thermodynamics and equilibrium of the reaction.

\Delta G=\Delta G^o+RT\ln K_{eq}

Since the enzyme amount will not affect the equilibrium of the reaction, the value of ΔG will remain the same as given -20 kcal/mol.

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Consider a process in which the entropy of a system increases by 125 J K−1 and the entropy of the surroundings decreases by 125
expeople1 [14]

Answer : The process is not spontaneous.

Explanation :

As, we know that:

Change in entropy = Change in entropy of system + Change in entropy of surrounding

As we are given in question, the entropy of surroundings decrease by the same amount as the entropy of the system increases.

For the given reaction to be spontaneous, the total change in entropy should be positive.

Given :

Entropy change of system = +125J/K

Entropy change of surroundings = -125J/K

Total change in entropy = Entropy change of system + Entropy change of surroundings

Total change in entropy = 125 J/K + (-125 J/K)

Total change in entropy = 0

The process is at equilibrium because the entropy change is equal to zero. So, the process is not spontaneous.

4 0
2 years ago
How many capsules containing 75mg of Tamiflu could be produced from 155g of star anise.?
jeyben [28]
From the question you will find that:
one capsule of tamiflu is obtained from 2.6 g of star anise.
1 capsule   = 2.6 g tamiflu
? capsules = 155 g tamiflu
by cross multiplication = \frac{(1 x 155)}{2.6} = 59 capsules
7 0
2 years ago
What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c
brilliants [131]

Answer:

Molecular formula → PbSO₄ → Lead sulfate

Option c.

Explanation:

The % percent composition indicates that in 100 g of compound we have:

68.3 g of Pb, 10.6 g of S and  (100 - 68.3 - 10.6) = 21.1 g of O

We divide each element by the molar mass:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

We divide each mol by the lowest value to determine, the molecular formula

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Molecular formula → PbSO₄ → Lead sulfate

8 0
2 years ago
Read 2 more answers
The student is now told that the four solids, in no particular order, are calcium bromide (CaBr2), sugar (C6H12O6), benzoic acid
lara [203]

MgBr2, 3 ions per mole=best conductor

KBr, 2 ions per mole, 2nd best conductor

Benzoic acid, weak acid, slightly ionized, weak conductor

Sugar, molecular, non ionized, non conductor

8 0
2 years ago
A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contain
krek1111 [17]

Answer:

- 0.0249% Sb/cm

-1.2465 * 10^9 \frac{atoms}{cm^3.cm}

Explanation:

Given that:

One surface contains 1 Sb atom per  10⁸  Si atoms and the other surface contains 500 Sb atoms per  10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb  per cm can be calculated as follows:

The difference in concentration = \delta_c

The distance \delta_x = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing  1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per   10⁸ silicon atoms can be calculated as follows:

\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%

= - 0.0249% Sb/cm

b) The concentration (c_1) of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}

Lattice parameter = 5.4307 Å;  To cm ; we have

= 5.4307A^0* \frac{10^{-8}cm}{ A^0}

c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

= 0.00499*10^{17}atoms/cm^3

The concentration (c_2) of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

   =  \frac{4*10^{-3}}{1.601*10^{-22}}

   = 2.4938*10^{17}atoms/cm^3

Finally, to calculate the concentration gradient

(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}

(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}

= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}

8 0
2 years ago
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