Answer:
The correct order will be
a. Transfer the measured amount of NaCl to the volumetric flask.
e. Dissolve the NaCl in less than 250 mL of water and mix well.
b. Dilute the solution with water to the 250.0 mL mark.
Explanation:
Preparation of NaCl solution in 250.0 ml volumetric flask:
Add the weighed NaCl directly to volumetric flask and add small amount of water to it and mix it will until all NaCl gets dissolved( if not add small water amount of water more)
After dissolving NaCl add the water upto the mark.
The correct order will be
a. Transfer the measured amount of NaCl to the volumetric flask.
e. Dissolve the NaCl in less than 250 mL of water and mix well.
b. Dilute the solution with water to the 250.0 mL mark.
Answer:
The partial pressure of SO₃ is 82.0 atm
Explanation:
The equilibrium constant Kp is equal to <em>the equilibrium pressure of the gaseous products raised to the power of their stoichiometric coefficients divided by the equilibrium pressure of the gaseous reactants raised to the power of their stoichiometric coefficients</em>.
For the reaction,
2 SO₂(g) + O₂(g) → 2 SO₃(g)
![Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm](https://tex.z-dn.net/?f=Kp%20%3D%200.345%20%3D%20%5Cfrac%7B%28pSO_%7B3%7D%29%5E%7B2%7D%20%7D%7B%28pSO_%7B2%7D%29%5E%7B2%7D%20%5Ctimes%20pO_%7B2%7D%20%7D%5C%5CpSO_%7B3%7D%20%3D%20%5Csqrt%5B%5D%7B0.345%20%5Ctimes%20%28pSO_%7B2%7D%29%5E%7B2%7D%20%5Ctimes%20pO_%7B2%7D%20%7D%20%5C%5CpSO_%7B3%7D%20%3D%20%5Csqrt%5B%5D%7B0.345%20%5Ctimes%20%2835.0%29%5E%7B2%7D%20%5Ctimes%2015.9%20%7D%20%5C%5CpSO_%7B3%7D%20%3D%2082.0%20atm)
Answer:
- 0.0249% Sb/cm
Explanation:
Given that:
One surface contains 1 Sb atom per 10⁸ Si atoms and the other surface contains 500 Sb atoms per 10⁸ Si atoms.
The concentration gradient in atomic percent (%) Sb per cm can be calculated as follows:
The difference in concentration = 
The distance 
= 0.2-mm = 0.02 cm
Now, the concentration of silicon at one surface containing 1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per 10⁸ silicon atoms can be calculated as follows:
= - 0.0249% Sb/cm
b) The concentration 
of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:
Lattice parameter = 5.4307 Å; To cm ; we have
= 
= 
The concentration 
of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:
= 
= 
Finally, to calculate the concentration gradient
Answer: acetone molecule ( CH₃-CO-CH₃)
Explanation:
1) Acetone is CH₃-CO-CH₃
2) That is a molecule (build up of covalent bonds).
3) When dissolved, covalent bonded compounds remain as separate molecules, then it is said that the major species present in the solution is the molecule. The molecules of acetone are surrounded (sovated) by the molecules of water.
This as opposed to the case of ionic compounds that ionize. When a compound as NaCl dissolves in water, it ionizes completely, so the major speceies are not NaCl formulas, but the ions Na⁺ and Cl⁻, not molecules.
That leads to the answer: the major species present when acetone is dissolved in water is the molecules of acetone (you do not need to state the fact that the molecules of water are part of the solution, because that is not the target of the question).
Dissociation=k×no of moles
percentage of dissociation=9.0×10^-4×1×100
knowing that x%=x/100,we then say;
x/100=9.0×10^-4×1×100
therefore, x=100×100×9×10^-4×1
x=9
x percentage of dissociation=9%
