Answer:
The correct answer is -all of the above.
Explanation:
Muscle fatigue is a reduced ability in work capacity caused by work itself. It is known that altering oxygen is contracting skeletal muscle affects performance. Reduced O2 supply increases the rate of muscle fatigue.
The lactic acid is accumulated as it forms rapidly but the breaking of the lactic acid is slow down, which causes muscle fatigue. Less ATP and glycogen in muscle results in fatigue as the muscle is not able to generate energy to power contractions and therefore contributes to muscle fatigue.
They move fast enough to overcome the forces of attraction that hold them together, becoming a gas.
ITS THAT :)
The pH = 2.41
<h3>Further explanation</h3>
Given
5.0% by mass solution of acetic acid
the density of white vinegar is 1.007 g/cm3
Required
pH
Solution
Molarity of solution :
Ka for acetic acid = 1.8 x 10⁻⁵
[H⁺] for weak acid :
![\tt [H^+]=\sqrt{Ka.M}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D)
Input the value :
![\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.839}\\\\(H^+]=0.00388=3.88\times 10^{-3}\\\\pH=3-log~3.88=2.41](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.839%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D0.00388%3D3.88%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5CpH%3D3-log~3.88%3D2.41)
Answer:
2.5 × 10² ppm
Explanation:
Step 1: Given data
- Mass of the sample: 200. g
Step 2: Convert 0.050 g to μg
We will use the conversion factor 1 g = 10⁶ μg.
0.050 g × 10⁶ μg/1 g = 5.0 × 10⁴ μg
Step 3: Calculate the concentration of NaCl in ppm
The concentration of NaCl in ppm is equal to the micrograms of NaCl per gram of the sample.
5.0 × 10⁴ μg NaCl/200. g = 2.5 × 10² ppm
The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1
number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂
we can use ideal gas law equation to find the volume of H₂
PV = nRT
where
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation
101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL
therefore answer is
c. 448 mL
