Answer:
c. one must have a higher kinetic energy than the other system
Answer:
Volume of chlorine = 61.943 mL
Explanation:
Given:
Volume of the water in the Pool = 18,000 gal
also,
1 gal = 3785.412 mL
thus,
Volume of water in pool = 18,000 × 3785.412 = 68,137,470 mL
Density of water = 1.00 g/mL
Therefore,
The mass of water in the pool = Volume × Density
or
The mass of water in the pool = 68,137,470 mL × 1.00 g/mL = 68,137,470 g
in terms of million = 
or
= 68.13747 g
also,
1 g of chlorine is present per million grams of water
thus,
chlorine present is 68.13747 g
Now,
volume = 
or
Volume of chlorine = 
or
Volume of chlorine = 61.943 mL
Add 7 water atom to the right hand side to adjust the quantity of oxygen. Increase Cr(+3) by two to adjust the quantity of Cr. Duplicate Cl-by two to adjust the quantity of chlorine molecules.
Cr2O7[2-](aq) +2 Cl[-](aq) < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
Presently adjust that charges.
you have - 4 charges on the left hand side, while +18 charges on the right hand side, there for include 14H+ the left hand side to adjust the charges
Cr2O7[2-](aq) +2 Cl[-](aq)+14H+ < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
take note of that the oxidation number of hydrogen in water is +1
C₆H₁₂O₆, or glucose, is oxidized in the presence of oxygen to form carbon dioxide and water. The reaction equation for this is:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
Therefore, if 6 moles of oxygen are consumed, we can see from the equation that one mole of C₆H₁₂O₆ will be consumed.
