All categories

3 years ago

The half-reactions involved in the lactate dehydrogenase (ldh) reaction and their standard reduction potentials are

1 answer:

5 0

Enzyme lactate dehydrogenase catalyses the conversion of lactate to pyruvate. In this reaction lactate loses two electrons (becomes oxidized) and is converted to pyruvate. NAD+ gains two electrons (is reduced) and is converted to NADH. The half reactions involved are;

NAD+ + H+ + 2e- = NADH -0.320
Pyruvate + 2H+ + 2e- = Lactate -0.185
Giving an overall reaction of;
Lactate + NAD+ = Pyruvate + NADH + H+

You might be interested in

PLEASE HELPP Tell me everything you know about "balancing the equation" for science

Free_Kalibri [48]

Watch melissa maribel explains it amazingly on her yt channel

3 0

3 years ago

How many moles of water are produced from 40.0 g of C2H6?

inn [45]

Answer:

4 moles of water

Explanation:

this is a combustion reaction, so the balanced equation is: 2C2H6 + 7O2 → 4CO2 + 6H2O.

the molar mass of C2H6 is 30.07g, so 40.0 g of C2H6 is 1.33 moles of C2H6.

mole ratio of H2O to C2H6 is 6/2, or 3.

1.33 moles C2H6 * 3 moles H2O/1 mole C2H6 = 4 moles H2O

4 0

2 years ago

Which term best defines the feeding behavior of most gulls?

White raven [17]

Answer:

Gulls, or colloquially seagulls, are seabirds of the family Laridae in the suborder Lari. They are most closely related to the terns and skimmers and only distantly related to auks, and even more distantly to waders. Until the 21st century, most gulls were placed in the genus

6 0

1 year ago

A solution is made by dissolving 0.656 mol of nonelectrolyte solute in 869 g of benzene. calculate the freezing point, tf, and b

solmaris [256]

Answer is: the freezing point is 1.63°C and boiling point is 82.01°C..

1) n(nonelectrolyte solute) = 0.656 mol.
m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.
b(solution) = n(nonelectrolyte solute) ÷ m(C₆H₆).
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.

2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.

3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.

4 0

3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago