How does the rate of reaction proceed with time

Glycolysis is the process by which energy is harvested from glucose by living things. Several of the reactions of glycolysis are

mote1985 [20]

Answer:

Unfavorable reactions: Reaction A and Reaction B.

Coupled reactions favorable: Reactions B and C and Reactions A and C.

Net change:

Reactions B and C : -14.2kJ/mol

Reactions A and B : 30.1kJ/mol

Reactions A and C: -16.7kJ/mol

Explanation:

A reaction is thermodynamically favorable (spontaneous) if ΔG < 0. Thus, the unfavorable reactions -ΔG > 0- are:

Reaction A and reaction B.

When reaction C is coupled with reaction B and reaction A the chemical equation is:

ATP + fructose-1,6-phosphate ⟶ ADP + fructose-1,6-bisphosphate

ΔG = 16.3 - 30.5 = -14.2 kJ/mol

ATP + glucose ⟶ ADP + glucose-6-phosphate

ΔG = 13.8 - 30.5 = -16.7 kJ/mol

The coupled reaction of A and B has a change in free energy of:

ΔG = 13.8 + 16.3 = 30.1 kJ/mol

4 0

4 years ago

Regarding acids, bases, and pH, which of these statements is true? Choose one: A. Substances that release protons when they diss

Mamont248 [21]

Answer: A

Explanation:

An acid is a substance that dissolves in water to produce proton (H +), which then combines with Water molecule (H2O) to give Hydroxonium ion (H3O +).

Acid also have a pH value below 7 on the pH scale.

8 0

3 years ago

Read 2 more answers

A sample of damp air in a 1.00 L container exerts a total pressure of 741.0 torr at 20 oC; but when it is cooled to -10 oC, the

AlexFokin [52]

From the information presented in the question, the number of molecules present of water present is obtained 2.41 × 10^21 molecules.

From the information we have;

Volume of the damp air = 1 L

Pressure of the damp air = 741.0 torr or 0.975 atm

Temperature of the gas = 20 oC + 273 = 293 K

R = 0.082 atm LK-1mol-1

Number of moles = ?

n =PV/RT

n = 0.975 × 1/0.082 × 293

n = 0.041 moles

Volume of water vapor = 1 L

Temperature of water = -10 oC + 273 = 263 K

Pressure of the gas = 607.1 torr or 0.799 atm

R = 0.082 atm LK-1mol-1

n= PV/RT

n = 0.799 × 1/ 0.082 × 263

n = 0.037 moles

Number of moles of water = 0.041 moles - 0.037 moles = 0.004 moles

If 1 mole = 6.02 × 10^23 molecules

0.004 moles = 0.004 moles × 6.02 × 10^23 molecules/1 mole

= 2.41 × 10^21 molecules

Learn more: brainly.com/question/2510654

5 0

3 years ago

15 Points. Pls give a real answer this is due in an hour.

bonufazy [111]

Answer:

This tells us that for every 2 moles of potassium chlorate that are decomposed, then 3 moles of oxygen gas is produced.

7 0

3 years ago

(2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) ) If 165 mL of oxygen is produced at 30.0 °C and 90.0 kPa, what mass of KClO3 was decomposed

soldier1979 [14.2K]

Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 KClO₃ → 2 KCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

KClO₃: 2 moles
KCl: 2 moles
O₂: 3 moles

The molar mass of the compounds is:

KClO₃: 122.45 g/mole
KCl: 74.45 g/mole
O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

KClO₃: 2 moles ×122.45 g/mole= 244.8 grams
KCl: 2 moles ×74.45 g/mole= 148.9 grams
O₂: 3 moles ×32 g/mole= 96 grams

<h3>Ideal gas law</h3>

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that the gas occupies.
T is the temperature of the gas.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

<h3>Number of O₂ produced.</h3>

165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:

P= 90 kPa= 0.888231 atm (being 101.325 kPa= 1 atm)
V= 165 mL= 0.165 L (being 1000 mL= 1 L)
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 30 C= 303 K (being 0 C= 273 K)

Replacing in the ideal gas law:

0.888231 atm× 0.165 L = n× 0.082 $\frac{atmL}{molK}$ × 303 K

Solving:

n= (0.888231 atm× 0.165 L)÷ (0.082 $\frac{atmL}{molK}$ × 303 K)

<u><em>n= 0.0059 moles</em></u>

Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.

<h3>Mass of KClO₃ required</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ are produced by 244.8 grams of KClO₃, 0.0059 moles of O₂ are produced by how much mass of KClO₃?

$mass of KClO_{3}= \frac{0.0059 moles of O_{2}x 244.8 grams of KClO_{3}}{3 moles of O_{2}}$

<u><em>mass of KClO₃= 0.48144 grams</em></u>

Finally, 0.48144 grams of KClO₃ was decomposed.

Learn more about

the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

4 0

2 years ago