Response:
9.9 ml of 0.200M NH₄OH(aq)
Reasoning:
3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)
What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?
1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution
1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)
=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters
<span>128 g/mol
Applying Graham's law of effusion, we can utilize the formula:
r1/r2 = sqrt(m2/m1)
where
r1 = effusion rate of gas 1
r2 = effusion rate of gas 2
m1 = molar mass of gas 1
m2 = molar mass of gas 2
Given that the atomic weight of oxygen is 15.999, the molar mass of O2 = 2 * 15.999 = 31.998.
We can now insert the known values into Graham's equation to find m2.
r1/r2 = sqrt(m2/m1)
2/1 = sqrt(m2/31.998)
4/1 = m2/31.998
Thus, we find m2 to be 127.992.
Rounding to three significant figures yields 128 g/mol</span>
Response:
The conclusion to your inquiry is Pressure 1 = 1.73 atm
Clarification:
Data provided
Volume 1 = 5 l
Pressure 1 =?
Volume 2 = 12 l
Pressure 2 = 0.72 atm
Procedure
To resolve this issue, Boyle's law is applied
Pressure 1 x Volume 1 = Pressure 2 x Volume 2
-Finding Pressure 1
Pressure 1 = Pressure 2 x Volume 2 / Volume 1
-Replace values
Pressure 1 = 0.72 x 12 / 5
-Calculating
Pressure 1 = 8.64/5
-Final outcome
Pressure 1 = 1.73 atm
Response:
Clarification:
refer to the solution provided below
