During a titration the following data were collected. A 10. mL portion of an unknown monoprotic acid solution was titrated with
1 answer:
Answer:
8.0 moles
Explanation:
Since the acid is monoprotic, 1 mole of the acid will be required to stochiometrically react with 1 mole of NaOH.
Using the formula:
Concentration of acid = ?
Volume of acid = 10 mL
Concentration of base = 1.0 M
Volume of base = 40 mL
mole of acid = 1
mole of base = 1
Substitute into the equation:
Concentration of acid = 40/10 = 4.0 M
To determine the number of moles of acid present in 2.0 liters of the unknown solution:
Number of moles = Molarity x volume
molarity = 4.0 M
Volume = 2.0 Liters
Hence,
Number of moles = 4.0 x 2.0 = 8 moles
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Answer:
59.2 grams
Explanation:
We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:
There are 59.2 grams of water in this solution.
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