H20-h&o
Gold-au
Nitric acid-h,n,o
Cobalt-co
Oxygen-o
MAKE SURE THEY ARE CAPITALIZED
Answer:
47.5 g of water can be formed
Explanation:
This is the reaction:
CH₄ + 2O₂ → CO₂ + 2H₂O
Methane combustion.
In this process 1 mol of methane react with 2 moles of oxygen to produce 2 moles of water and 1 mol of carbon dioxide.
As ratio is 1:2, I will produce the double of moles of water, with the moles of methane I have.
1.320 mol .2 = 2.64 moles
Now, we can convert the moles to mass (mol . molar mass)
2.64 mol . 18g/mol = 47.5 g
“Flow is the volume of fluid that passes in a unit of time. In water resources, flow is often measured in units of cubic feet per second (cfs), cubic meters per second (cms), gallons per minute (gpm), or other various units.”
A) in pure water :
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0
change +X +X
Equ X X
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X* X
∴X = √2.1 x 10*-10
∴X = 1.4 x 10^-5
∴ the solubility = X = 1.4 X 10^-5
B) In 1.6 x 10^-3 m Na2CrO4
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0.0016
Change +X +X
Equ X X+0.0016
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X*(X+0.0016) by solving for X
∴ X = 1.3 x 10^-7
∴ solubility =X = 1.3 x 10^-7