Based on recommended amount of carbohydrate, a basketball player should consume about 17 - 34 ounces of gatorade g series during the hour-long game.
<h3>How many ounces of endurance formula gatorade g series, endurance formula should a basketball player consume during an hour-long game if it contains 14 grams of carbohydrate per 8 ounces?</h3>
Carbohydrates are food substances metabolized easily by the body to produce energy.
Given that the recommended amount of carbohydrate to consume to maintain performance is 30–60 g/h.
Also 14 grams of carbohydrate found in 8 ounces of the drink.
30 g of carbohydrate will be present in 30 × 8/14 = 17.1 ounces of gatorade g series
60 g of carbohydrate will be present in 60 × 8/14 =34.3 ounces of gatorade g series.
Therefore, a basketball player should consume about 17 - 34 ounces of gatorade g series during the hour-long game.
Learn more about carbohydrates at: brainly.com/question/797978
PH=-lg[H⁺]
[H⁺]=2.88*10⁻² mol/L
pH=-lg(2.88*10⁻²)=1.54
pH=1.54
4 this is a covalent compound it is named
Answer:
- The last option: <u><em>Decrease the volume to increase pressure and to increase concentration.</em></u>
Explanation:
You can support your choice on basis of the collision theory.
According to the collision theory, the chemical reactions happen when the molecules collide with each other, in the correct orientation and with enough kinetic energy to overcome the activation energy.
As consequence, the larger the number of collisions the larger the rate of a reaction.
In the case of a reaction that involves gases, decreasing the volume, will increase the pressure and the concentration (volume is inversely related to both the pressure and the concentration) causing the molecules to be closer to each other and to collide with higher frequency, this is you will expect more collisions, and so an increase on the rate of the reaction. That is expressed by the last choice: decrease the volumen to increase pressure and to increase concentration.
Answer:
C₇H₁₄O₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1.152 g
Mass of CO₂ = 2.726 g
Mass of H₂O = 1.116 g
Empirical formula =?
Next, we shall determine the mass of carbon, hydrogen and oxygen present in the compound. This can be obtained as follow:
For carbon (C):
Mass of CO₂ = 2.726 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g/mol
Mass of C = 12/44 × 2.726
Mass of C = 0.743 g
For hydrogen (H):
Mass of H₂O = 1.116 g
Molar mass of H₂O = (2×1) + 16
= 2 + 16
= 18 g/mol
Mass of H = 2/18 × 1.116
Mass of H = 0.124 g
For oxygen (O):
Mass of compound = 1.152 g
Mass of C = 0.743 g
Mass of H = 0.124 g
Mass of O =?
Mass of O = (Mass of compound) – (Mass of C) + (Mass of H)
Mass of O = 1.152 – (0.743 + 0.124)
Mass of O = 1.152 – 0.867
Mass of O = 0.285 g
Finally, we shall determine the empirical formula for the compound as follow:
C = 0.743 g
H = 0.124 g
O = 0.285 g
Divide by their molar mass
C = 0.743 / 12 = 0.062
H = 0.124 / 1 = 0.124
O = 0.285 / 16 = 0.018
Divide by the smallest
C = 0.062 / 0.018 = 3.44
H = 0.124 / 0.018 = 7
O = 0.018 / 0.018 = 1
Multiply by 2 to express in whole number.
C = 3.44 × 2 = 7
H = 7 × 2 = 14
O = 1 × 2 = 2
Empirical formula => C₇H₁₄O₂