Which is NOT a difference between cytochrome C and ubiquinone (U)?A: U accepts e- from NADH, cytochrome C accepts e- from FADH2
B: U carries 2 e- and 2 H+, cytochrome C carries 1 e- and no H+
C: U pumps 2 H+ into IMM from matrix, cytochrome C pumps 1 H+

Answers

Answer 1

Answer:

The correct Statement is B: U carries 2 e- and 2 H+, cytochrome C carries 1 e- and no H+. It claims that ubiquinone carries 2 electrons and 2 protons, whereas cytochrome C carries 1 electron and no protons, is not a difference between the two but a similarity.

The statement that is NOT a difference between cytochrome C and ubiquinone (U) is Option B: U carries 2 e- and 2 H+, cytochrome C carries 1 e- and no H+. This is because both molecules act as electron carriers, with ubiquinone carrying two electrons and two protons (H+), while cytochrome C can only carry one electron without carrying a proton.

Moreover, in Statement A and Statement C, ubiquinone and cytochrome c play different roles in the electron transport chain, therefore those statements describe actual differences between ubiquinone and cytochrome C.

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Please help! Multiple choice question! Which of the following gradients allow oxygen to leave the blood and enter body tissues and carbon dioxide to leave body tissues and enter the blood?

A. High concentration of oxygen in the blood; high concentration of carbon dioxide in the body tissues
B. Low concentration of oxygen in the blood; high concentration of carbon dioxide in the body tissues
C. High concentration of oxygen in the blood; low concentration of carbon dioxide in the body tissues
D. Low concentration of oxygen in the blood; low concentration of carbon dioxide in the body tissues

Answers

Answer:

A. High concentration of oxygen in the blood; high concentration of carbon dioxide in the body tissues

Explanation:

Diffusion occurs from a region of higher concentration to a region of lower concentration. Diffusion occurs against a concentration gradient.

High concentration of oxygen in the blood causes oxygen to leave the blood and enter body tissues. High concentration of carbon dioxide in the body tissues causes carbon dioxide to leave body tissues and enter the blood.

According to my mind and my biology test, it is B) Low concentration of oxygen IN THE BLOOD; high concentration co2 in the tissue

How does the skin help regulate body temperature?

HELP FAST

Answers

Answer:

1. The body's immense blood supply (when blood vessels dilate or become larger heat loss occurs) When the blood vessels constrict it holds in heat

2. The skins blood supply also helps maintain body temperature (helps maintain homeostasis which is the "normal" state of our body)

3. Humidity in the air effects thermoregulation (heat regulation in the body) by limiting sweat evaporation (usually sweat evaporates into the environment when the humidity is low because the air has low "water levels" per say while you have high "water levels" because of your sweat

A cylinder with a movable piston contains a sample of gas having a volume of 6.0 liters at 293 K and 1.0 atmosphere. What is the volume of the sample after the gas is heated to 303 K, while the pressure is held at 1.0 atmosphere?(1) 9.0 L (3) 5.8 L
(2) 6.2 L (4) 4.0 L

Answers

The volume of sample when the gas is heated to 303 K is $\boxed{{\text{(2)6}}{\text{.2 L}}}$

Further explanation:

Charles’s law:

Charles’s work showed that at constant pressure, the volume-temperature relationship for a fixed amount of gas is linear. In other words, Charles’s law can be stated that at constant pressure, the volume occupied by a fixed amount of a gas is directly proportional to its absolute temperature (Kelvin). This relationship is known as Charles’s law.

The mathematical representation of Charles’s law is,

${\mathbf{V}}\propto{\mathbf{T}}$ [P and n are constant]

Where,

⦁ V is volume occupied by the fixed quantity of gas.

⦁ T is the temperature of a gas.

⦁ P is the pressure of a gas.

⦁ n denotes the number of moles of gas.

The relationship can also be expressed as,

$\frac{{\text{V}}}{{\text{T}}} = {\text{constant}}$ [P and n are constant]

Or it can also be expressed as follows:

$\frac{{{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}} = \frac{{{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}$ …… (1)

Here,

${{\text{V}}_1}$ is the initial volume of gas.

${{\text{V}}_2}$ is the final volume of gas.

${{\text{T}}_1}$ is the initial temperature of the gas.

${{\text{T}}_2}$ is the final temperature of the gas.

Rearrange equation (1) to calculate ${{\text{V}}_2}$ .

${{\text{V}}_{\text{2}}} = \frac{{{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}$ …… (2)

The value of ${{\text{V}}_1}$ is 6 L.

The value of ${{\text{T}}_1}$ is 293 K.

The value of ${{\text{T}}_2}$ is 303 K.

Substitute these values in equation (2).

$\begin{aligned}{{\text{V}}_{\text{2}}}&=\frac{{\left({{\text{6 L}}}\right)\left( {{\text{303 K}}}\right)}}{{\left( {{\text{293 K}}}\right)}}\n&={\text{6}}{\text{.20477 L}}\n&\approx{\mathbf{6}}{\mathbf{.2 L}}\n\end{aligned}$

So option (2) is the correct answer.

Learn more:

1. Law of conservation of matter states: brainly.com/question/2190120

2. Calculation of volume of gas: brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: Charles’s law, volume, temperature, pressure, volume temperature relationship, absolute temperature, constant pressure, relationship, V directly proportional to T, ideal gas, ideal gas equation number of moles, moles, P, n, V, T, volume of gas, 6.2 L, 303 K, 293 K, 6 L.

The volume of the sample after the gas is heated to 303 K is 6.2 L

The correct answer to the question is Option (2) 6.2 L

From the question given above, the following data were obtained:

Initial volume (V₁) = 6 L

Initial temperature (T₁) = 293 K

Final temperature (T₂) = 303 K

Pressure = constant

Final volume (V₂) =?

Using the Charles' law equation, we can obtain the final volume of the gas as follow:

$(V_(1))/(T_(1)) = (V_(2))/(T_(2))\n\n(6)/(293) = (V_(2))/(303)\n\n$

Cross multiply

293 × V₂ = 6 × 303

293 × V₂ = 1818

Divide both side by 293

$V_(2) = (1818)/(293)\n\n$

V₂ = 6.2 L

Therefore, the volume of the sample after the gas is heated to 303 K is 6.2 L

The correct answer to the question is Option (2) 6.2 L

Learn more: brainly.com/question/16927784

Differentiate between atoms, elements, molecules and compounds

*URGENT

Answers

Atoms are the simpliest form of matter. But an atom can be classified as an element and an element can be presented various substance. An element can be oxygen, hydrogen and so on. The combination of two or more elements is a compound. A compound can be a molecule, same as the atom to element.

Simple chem question - please help!Given the equation: N2 +2O2 → 2NO2, what mass of oxygen is required to react completely with 7g of nitrogen?

A) 8 g
B) 16 g
C) 32 g
D) 64 g

Answers

Molar mass O₂ = 32.0 g/mol and N₂ = 28.0 g/mol

N₂ +2 O₂ → 2 NO₂

28.0 g -------> 32.0 * 2
7 g ----------> ?

Mass of O₂ = 7 * 32.0 * 2 / 28.0

Mass of O₂ = 448 / 28.0

= 16 g of O₂

Answer B

hope this helps!

Answer:

guy above is right

Explanation:

How does a warm front form