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AnnZ [28]
4 years ago
10

Carbohydrate loading Group of answer choices involves a reduction in the intensity of workouts with a corresponding increase in

the percentage of carbohydrate intake over several days before a competition. does not increase glycogen stores to any significant degree. involves little exercise and a high-carbohydrate diet the first 3 days, followed by heavy exercise and a low-carbohydrate diet right before competition. involves loading up on carbohydrate-laden foods on the mor
Chemistry
1 answer:
SOVA2 [1]4 years ago
7 0

Answer:

the correct option would be:

The group of response options implies a reduction in the intensity of the workouts with a corresponding increase in the percentage of carbohydrate intake for several days before a competition.

Since the carbohydrate load is an increase in glycogen reserves as an energy source accompanied by a decrease in muscle demand. This is often used in high-performance activities, where strict competencies are required.

Although today some professionals do not support that, but rather support a diet with carbohydrates and proteins.

Explanation:

Carbohydrate loading increases glycogen reserves, it is accompanied by a muscle rest plan, without fatigue of muscle fibers.

The purpose of this is to exhaust the muscle fibers in maximum demands such as the competencies, ensuring a necessary energy source that supplies this reaction, for which glycogen reserves are needed.

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Does anyone know the properties of group 1 on the periodic table?
bija089 [108]
The alkali metals make up Group 1 of the periodic table. This family consists of the elements lithium, sodium, potassium, rubidium, cesium, and francium (Li, Na, K, Rb, Cs, and Fr, respectively). Group one elements share common characteristics. They are all soft, silver metals. Due to their low ionization energy, these metals have low melting points and are highly reactive. The reactivity of this family increases as you move down the table. Alkali metals are noted for how vigorously they react with water. Due to this, they are often stored in mineral oil and are not found in their elemental forms in nature. These characteristics can be explained by examining the electronic structure of each element in this group. Alkali metals have one valence electron. They readily give up this electron to assume the noble gas configuration as a cation. This makes the elements in this group highly reactive.hope this helps you ok.
3 0
3 years ago
How can one determine if a bond between two atoms is ionic covalent or metallic?
Wewaii [24]

Answer:

By the Pauling rule, of EN

Explanation:

EN means electronegativity, the ability of atoms to attract electrons.

In the periodic table each atom, has a value of EN.

When there are two atoms bonded, you must substract the EN (the high - the low) → ΔEN

It depends on the ΔEN, that you can define a bond as ionic, covalent or metallic.

Ionic ΔEN → > 1.7

Covalent polar 0.4 < ΔEN > 1.7

Covalent non polar 0.4 < ΔEN

Two metals have always a metallic bond

5 0
3 years ago
HCl reacts with Barium Hydroxide to produce barium chloride and water. how many ml of a 3.00 M hydrochloric acid solution would
ollegr [7]
THINK OF GUMMY BEARS IN A TEST TUBE AND IN THAT TEST TUBE THEIRS A SUBSTANCE ACID IT EX PLODS LOOK IT UP ON YOU TUBE
3 0
4 years ago
You are counting red blood cells, which objective will be best to count the number of cells, high power or low power objective a
BARSIC [14]

Answer:

Low power

Explanation:

Low power would allow for the full image of the red blood cells and would appear as small circles.

4 0
3 years ago
Your calculated density of aluminum is d = 2.69 g/cm3. Aluminum’s accepted density is 2.70 g/cm3. Without writing the "%" sign,
stepan [7]

Answer:

0.37 %

Explanation:

Given that:

Calculated density of aluminum = 2.69 g/cm³

Accepted density of aluminum = 2.70 g/cm³

Error\ percentage=\frac {|Accepted\ value-Calculated\ value|}{Accepted\ value}\times 100

Thus, applying values as:

Error\ percentage=\frac {|2.70-2.69|}{2.70}\times 100

<u>Percent error = 0.37 %</u>

7 0
4 years ago
Read 2 more answers
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