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andrezito [222]
3 years ago
15

How can I find the solution needed for amount of reaction?

Chemistry
1 answer:
liq [111]3 years ago
7 0

Answer:

0.48L of NaOH

Explanation:

Data obtained from the question include:

Number of mole of NaOH = 0.725mol

Molarity of NaOH = 1.50M

Volume of NaOH =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.725/1.5

Volume = 0.48L

Therefore, 0.48L of NaOH is needed for the reaction

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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.

7 0
3 years ago
Imagine that A and B are cations and X, Y, and Z are anions, and that the following reactions occur:
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3 years ago
The equilibrium constant, Kc, for the following reaction is 3.61×10-4 at 426 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently lar
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Answer:

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2 years ago
How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl
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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:

12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}

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:

4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}

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:

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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:

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