Answer:
The appropriate answer is "9.225 g".
Explanation:
Given:
Required level,
= 63 ppm
Initial concentration,
= 22 ppm
Now,
The amount of free SO₂ will be:
= 
= 
= 
The amount of free SO₂ to be added will be:
= 
= 
∵ 1000 mg = 1 g
So,
= 
= 
Thus,
"9.225 g" should be added.
1-PRIMARY ALKANOL 2-SECONDARY ALKANOL 3-TERTIARY ALKANOL
Answer:
So 1 mole
Explanation:
PV = nRT
P = Pressure atm
V = Volume L
n = Moles
R = 0.08206 L·atm·mol−1·K−1.
T = Temperature K
standard temperature = 273K
standard pressure = 1 atm
22.4 liters of oxygen
Ok so we have
V = 22.4
P = 1 atm
PV = nRT
n = PV/RT
n = 22.4/(0.08206 x 273)
n = 22.4/22.40
n = 1 mole
Answer:
The atomic weight of an element represents the ratio of the average mass of atoms of a chemical element.
