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aev [14]
4 years ago
12

Calculate the ΔG°' for the reaction with 3 significant figures with no label for the dimension (just the number). fructose-6-pho

sphate → glucose-6-phosphate. Given the equilibrium constant is 1.97 and the physiological relevant temperature is 37° C.
Chemistry
1 answer:
seraphim [82]4 years ago
6 0

Answer:

ΔG° = 1747.523

Explanation:

The parameters mentioned are;

Gibbs Free energy ΔG°

Equilibrium constant Kc

Temperature T = 37 + 273 = 310 (upon conversion to kelvin temperature)

The formular relating all three parameters is given as;

ΔG° =  -RTlnKc

Where; R = rate constant = 8.314 J⋅K−1⋅mol−1

Upon solving;

ΔG° = - 8.314 * 310 * ln(1.97)

ΔG° = 1747.523

You might be interested in
How many moles of C are in a sample of C5H12 that also contains 22.5g of H?
pishuonlain [190]

The number of moles of C present in C₅H₁₂ that contains 22.5 g of H is 9.375 moles

<h3>How to determine the mass of C₅H₁₂ that contains 22.5 g of H</h3>

1 mole of C₅H₁₂ = (12×5) + (1×12) = 72 g

Mass of H in 1 mole of C₅H₁₂ = 12 × 1 = 12 g

Thus,

12 g of H is present in 72 g of C₅H₁₂

Therefore,

22.5 g of H will be present in = (22.5 × 72) / 12 = 135 g of C₅H₁₂

<h3>How to determine the mole of C present in 135 g of C₅H₁₂</h3>

72 g of C₅H₁₂ contains 5 moles of C

Therefore,

135 g of C₅H₁₂ will contain = (135 × 5) / 72 = 9.375 moles of C

Thus, 9.375 moles of C is present in C₅H₁₂ that contains 22.5 g of H

Learn more about mole:

brainly.com/question/13314627

#SPJ1

8 0
2 years ago
HelpUsing a 2-cm-thick piece of cardboard over a radiation source would be mosteffective for protecting against which type of ra
Lisa [10]
It would protect best against C) Alpha radiation, as Beta radiation is stopped by lighter metals such as aluminium, and Gamma radiation can only be stopped by heavier metals such as lead.
3 0
4 years ago
Read 2 more answers
How do I do an energy profile diagram
Marina CMI [18]

Answer:

first u have to like draw like a number line uk how dey do in math but instead of num u add wht eva da question askin u

Explanation:

5 0
3 years ago
The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests tha
lakkis [162]

Answer:

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

Explanation:

Volume of the radiator = 3.6 L

Percentage of antifreeze = 60%

Total volume of anti freeze in radiator = 60% of 3.6 L :

\frac{60}{100}\times 3.6 = 2.16 L

Percentage of water= 40%

Given,optimal cooling of the engine is obtained with only 50% antifreeze.

So, now we want to reduce the percentage of antifreeze from 60% to 50 %

Volume of coolant removed = x

Volume of water added = x

Volume of anti freeze removed = 60% of(x) = 0.6x

Volume of antifreeze left in radiator  =50% of 3.6 L = \frac{50}{100}\times 3.6=1.8 L

1.8 Liter is the desired volume of the antifreeze.

Total volume - Removed volume = desired  volume

2.16 L - 60% of( x) = 1.8 L

2.16 L-0.6x=1.8 L

x=\frac{2.16 l- 1.8 L}{0.6}=0.6 L

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

4 0
4 years ago
Part A
Roman55 [17]

These are two questions and two answers

Question 1.

Answer:

  • <u>7.33 × 10 ⁻³ c</u>

Explanation:

<u>1) Data:</u>

a) m = 9.11 × 10⁻³¹ kg

b) λ =  3.31 × 10⁻¹⁰ m

c) c = 3.00 10⁸ m/s

d) s = ?

<u>2) Formula:</u>

The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

Solve for s:

  • s = h / (m.λ)

Substitute:

  • s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg ×  3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s

To express the speed relative to the speed of light, divide by c =  3.00 10⁸ m/s

  • s =  2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³

Answer: s = 7.33 × 10 ⁻³ c

Question 2.

Answer:

  • 2.06 × 10 ⁻³⁴ m.

Explanation:

<u>1) Data:</u>

a) m = 45.9 g (0.0459 kg)

b) s = 70.0 m/s

b) λ =  ?

<u>2) Formula:</u>

Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

  • λ = h / (m.s)

Substitute:

  • λ =  6.626×10⁻³⁴J.s / ( 0.0459 kg ×  70.0 m/s) = 2.06 × 10 ⁻³⁴ m

As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.

Answer: 2.06 × 10 ⁻³⁴ m.  

5 0
3 years ago
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