Answer:
Carbohydrates
Explanation:
Increased exercise intensity means the overall need for energy increases. As we increase exercise intensity we increase our glucose uptake and oxidation which far exceeds uptake, indicating that muscle stores of glycogen are being used. At moderate intensities (65%) there is an increased need for muscle glycogen and muscle triglycerides which is fat. At higher levels of intensities (85%) there is an even greater need for energy, and this is met almost solely by an increased uptake of glucose from the blood and from muscle glycogen.
In the case of fats as an energy fuel source at high intensities, increasing levels of intensity increases fat oxidation but once we get into higher levels of intensity, we return to levels of fat oxidation similar to very low intensities.
Answer:
See explanation
Explanation:
The central atom in the perbromate ion is bromine. The chemical symbol of bromine is Br. There are no lone pairs around the central bromine atom. The ion is tetrahedral in shape hence we expect a bond angle of 109°. 27 which is the ideal tetrahedral bond angle. The actual bond angle of the prebromate ion is 109.5°. The perbromate ion is BrO4^-
The observed bond angle is very close to the ideal value because of the absence of lone pairs of electrons from the central atom in the ion.
<u>Answer:</u> The molarity of HCl solution is 0.262 M
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

We are given:
Given mass of HCl = 0.3366 g
Molar mass of HCl = 36.5 g/mol
Volume of the solution = 35.23 mL
Putting values in above equation, we get:

Hence, the molarity of HCl solution is 0.262 M.
Combined gas law is
PV/T = K (constant)
P = Pressure
V = Volume
T = Temperature in Kelvin
For two situations, the combined gas law can be applied as,
P₁V₁ / T₁ = P₂V₂ / T₂
P₁ = 3.00 atm P₂ = standard pressure = 1 atm
V₁ = 720.0 mL T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K
By substituting,
3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
V₂ = 2012.6 mL
hence the volume of gas at stp is 2012.6 mL
Answer:
The answer should be 1000 kg / m3