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.
They should identify the confounding variable.
Some condition that is not examined by the scientist might alter the experiment result. That condition is called confounding variable. If the method of the experiment same but result is very different, there should be unidentified confounding variable. It could be air humidity, temperature, ventilation, light, time of the year or anything that might not be seen by naked eye.
Try to redo the experiment with controlling variable as much as possible.
The drug has a concentration of 6 mg per 1 ml. Therefore, to know that number of ml containing 25 grams, we will simply do cross multiplication as follows:
amount of drug = (25 x 1) / 6 = 4.1667 ml
Therefore, for the patient to receive 25 mg of methimazole, he/she should take 4.1667 ml of the drug solution.
✨Tysm for the points kurama :)✨
Answer:
T° freezing solution → -11.3°C
T° boiling solution → 103.1 °C
Explanation:
Assuming 100 % dissociation, we must find the i, Van't Hoff factor which means "the ions that are dissolved in solution"
This salt dissociates as this:
SnCl₄ (aq) → 1Sn⁴⁺ (aq) + 4Cl⁻ (aq) (so i =5)
The formula for the colligative property of freezing point depression and boiling point elevation are:
ΔT = Kf . m . i
where ΔT = T° freezing pure solvent - T° freezing solution
ΔT = Kb . m . i
where ΔT = T° boiling solution - T° boiling pure solvent
Freezing point depression:
0° - T° freezing solution = 1.86°C/m . 1.22 m . 5
T° freezing solution = - (1.86°C/m . 1.22 m . 5) → -11.3°C
Boiling point elevation:
T° boiling solution - 100°C = 0.512 °C/m . 1.22 m . 5
T° boiling solution = (0.512 °C/m . 1.22 m . 5) + 100°C → 103.1 °C
