This is an incomplete question, here is a complete question.
The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.
Answer : The molar concentration of oxygen is, 
Explanation :
As we know that,
where,
= molar solubility of 
= ?
= partial pressure of = 0.2 atm = 1.97×10⁻⁶ Pa
= Henry's law constant = 4.34 × 10⁹ g/L.Pa
Now put all the given values in the above formula, we get:
Now we have to molar concentration of oxygen.
Molar concentration of oxygen = 
Therefore, the molar concentration of oxygen is,
Using the exponential decay model; we calculate "k"
We know that "A" is half of A0
A = A0 e^(k× 5050)
A/A0 = e^(5050k)
0.5 = e^(5055k)
In (0.5) = 5055k
-0.69315 = 5055k
k = -0.0001371
To calculate how long it will take to decay to 86% of the original mass
0.86 = e^(-0.0001371t)
In (0.86) = -0.0001371t
-0.150823 = -0.0001371 t
t = 1100 hours
Answer:
The answer to your question is V = 0.108 L or 108 ml
Explanation:
Data
Volume = ?
mass = 0.405 g
Temperature = 273°K
Pressure = 1 atm
Process
1.- Convert mass of Kr to moles
83.8 g of Kr -------------------- 1 mol
0.405 g ------------------- x
x = (0.405 x 1) / 83.8
x = 0.0048 moles
2.- Use the Ideal gas law to solve this problem
PV = nRT
- Solve for V
V = nRT / P
- Substitution
V = (0.0048)(0.082)(273) / 1
- Simplification
V = 0.108 / 1
- Result
V = 0.108 L
Answer:
chyba się zniszczy
Explanation:
ponieważ sie film prześwietli
