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QveST [7]
3 years ago
14

More ATP per carbon is produced from oxidation of fatty acids as compared to glucose. However, fatty acids are used primarily to

provide fuel for ATP synthesis during long periods of exercise. Why can't fatty acids be used to fuel short, intense periods of exercise?
A. Short, intense periods of exercise preferentially activate ATP synthesis from glucose.
B. Short, intense periods of exercise do not signal release of fatty acids from adipose tissue.
C. Short, intense periods of exercise do not require ATP synthesis.
D. Short, intense periods of exercise do not provide sufficient time for oxidative phosphorylation to occur.
Chemistry
1 answer:
Luda [366]3 years ago
8 0

Answer:

The correct option is D

Explanation:

Although, option A appears correct but the reason for option A is actually option D. What this means is that, during short intense exercise (anaerobic exercise), the body preferentially activates ATP synthesis from glucose (glycolysis) even though the beta oxidation of fatty acid produces more energy (ATP) - this is because <u>glycolysis occurs many times faster (about a 100 times faster) than the beta oxidation of fatty acid/oxidative phosphorylation</u>. Thus, it can be said that short, intense periods of exercise do not provide sufficient time for oxidative phosphorylation to occur.

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Which civilization formed a confederation thay may have impacted the us government
lana66690 [7]

Answer:

Anasazi

Explanation:

3 0
3 years ago
Tetrachloromethane, CCl4 is produced from the substitution reaction between methane and chlorine gas. If the rate of formation o
Korolek [52]

The rate of disappearance of chlorine gas : 0.2 mol/dm³

<h3>Further explanation</h3>

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

For reaction :

\tt aA+bB\rightarrow cC+dD

The rate reaction :

\tt -\dfrac{1}{a}\dfrac{d[-A]}{dt}= -\dfrac{1}{b}\dfrac{d[-B]}{dt}=\dfrac{1}{c}\dfrac{d[C]}{dt}=\dfrac{1}{d}\dfrac{d[D]}{dt}

Reaction for formation CCl₄ :

<em>CH₄+4Cl₂⇒CCl₄+4HCl</em>

<em />

From equation, rate of reaction = rate of formation CCl₄ = 0.05 mol/dm³

Rate of formation of  CCl₄  = reaction rate x coefficient of  CCCl₄

0.05 mol/dm³ = reaction rate x 1⇒reaction rate = 0.05 mol/dm³

The rate of disappearance of chlorine gas (Cl₂) :

Rate of disappearance of  Cl₂  = reaction rate x coefficient of  Cl₂

Rate of disappearance of  Cl₂ = 0.05 x 4 = 0.2 mol/dm³

4 0
3 years ago
Darin skateboarded to the north at 4 m/s. In 5 seconds, he sped up to 7 m/s. What was his acceleration?
Julli [10]
To solve this problem we can use following equation.

v =u + at

Where v is the final velocity (m/s), u is the initial velocity (m/s), a is the acceleration (m/s²) and t is the time taken (s).

v = 7 m/s
u = 4 m/s
a = ?
t = 5 s

By applying the equation, we can get
    7 m/s = 4 m/s + a x 5 s
    3 m/s = a x 5 s
          a = 0.6 m/s²

Hence, the acceleration is 0.6 m/s² towards north.

Answer is "C".
4 0
3 years ago
How many atoms are in a sample of 175 grams of sodium (Na)? The answer needs to be <br> a x 10^b
ValentinkaMS [17]
Atomic mass Sodium ( Na ) = 22.98 u.m.a

22.98 g ----------------- 6.02x10²³ atoms
175 g ------------------- ?? atoms

175 x ( 6.02x10²³) / 22.98 =

4.58x10²⁴ atoms of Na

hope this helps!

4 0
3 years ago
WILL MARK BRAINLIEST!!!!
Lyrx [107]

Answer:

5.702 mol K₂SO₄

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Compounds
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 993.6 g K₂SO₄

[Solve] moles K₂SO₄

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of K: 39.10 g/mol

[PT] Molar Mass of S: 32.07 g/mol

[PT] Molar mass of O: 16.00 g/mol

Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 993.6 \ g \ K_2SO_4(\frac{1 \ mol \ K_2SO_4}{174.27 \ g \ K_2SO_4})
  2. [DA] Divide [Cancel out units]:                                                                         \displaystyle 5.7015 \ mol \ K_2SO_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄

7 0
3 years ago
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