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.
<span>AX(aq)+BY(aq)→no precipitate
AX(aq)+BZ(aq)→precipitate
this two equations imply
</span>
AX(aq) is soluble and <span>BY(aq) is insoluble
the answer is
</span><span>E. BY</span>
Answer:
the concentration of PCl5 in the equilibrium mixture = 296.20M
Explanation:
The concept of equilibrium constant was applied where the equilibrium constant is the ration of the concentration of the product over the concentration of the reactants raised to the power of their coefficients. it can be in terms of concentration in M or in terms of Pressure in atm.
The detaied steps is as shown in the attached file.
Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:
