Answer:
Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)
Explanation:
Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P
Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm
=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.
∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>
The rate constant for 1st order reaction is
K = (2.303 /t) log (A0 /A)
Where, k is rate constant
t is time in sec
A0 is initial concentration
(6.82 * 10-3) * 240 = log (0.02 /A)
1.63 = log (0.02 /A)
-1.69 – log A = 1.63
Log A = - 0.069
A = 0.82
Hence, 0.82 mol of A remain after 4 minutes.
The correct answer is option A. Three, dumbbell.
The p sublevel has __three_orbitals that are _dunmbbell_-shaped.
The three orbitals of the p sublevel are oriented in three directions along the x, y and Z axis. The orbitals are called px, py and pz. Each p orbital can have a maximum of 2 electrons. Since there are three p orbitals, the p sublevel can have a maximum of 6 electrons.
Answer : The correct option is, 
Explanation :
To calculate the pressure of gas we are using ideal gas equation as:
where,
P = pressure of gas = ?
V = volume of gas = 0.046 L
n = number of moles of gas = 3.4
R = gas constant = 8.314 L.kPa/mol.K
T = temperature of gas = 298 K
Now put all the given values in the above formula, we get:
Therefore, the pressure of gas is,
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