The mass of nitric acid required to make the given solution is 0.0627 g.
The given parameters:
- <em>Volume of the acid, V = 250 mL</em>
- <em>pH of the acid, = 2.4</em>
The hydrogen ion (H⁺) concentration of the nitric acid is calculated as follows;

The molarity of the nitric acid is calculated as follows;

The number of moles of the nitric acid is calculated as follows;

The molar mass of nitric acid is calculated as;

The mass of the nitric acid contained in the calculated number of moles is calculated as;

Thus, the mass of nitric acid required to make the given solution is 0.0627 g.
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Answer:
569K
Explanation:
Q = 3.5kJ = 3500J
mass = 28.2g
∅1 = 20°C = 20 + 273 = 293K
∅2 = x
c = 0.449
Q = mc∆∅
3500 = 28.2×0.449×∆∅
3500 = 12.6618×∆∅
∆∅ = 3500/12.6618
∆∅ = 276.4220
∅2 - ∅1 = 276.4220
∅2 = 276.4220 + ∅1
∅2 = 276.4220 + 293
∅2 = 569.4220K
∅2 = 569K
Explanation:
Identify the areas of the following atomic model, Explain how these are related to the formatio