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:
Neutralization
Explanation:
When an acid react with base it form the salt and water. The reaction is also called neutralization reaction because both neutralize each other.
In neutralization reaction equal amount of acid and base react to neutralize each other and equal amount of water and salt are formed. When pH does not reach to 7 its means there is less amount of one of reactant which is not fully neutralize.
Neutralization reactions are also used as first aid. For example when someone is dealing with HCl for cleaning purpose of toilet and get touched. It is advised to neutralize it with soap, milk or egg white.
Example:
Hydrochloric acid when react with the sodium hydroxide, a salt sodium chloride and water are formed.
Chemical equation:
HCl + NaOH → NaCl + H₂O
Titration:
Neutralization reactions are also used to determine the concentration of solution. Titration is a quantitative technique in which acid or base is gradually added into the solution whose concentration is to be determine until the neutral point is reached.
Answer:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓
Explanation:
We identify the reactants:
CuBr₂ and Pb(CH₃COO)₂
The products will be: Cu(CH₃COO)₂ and PbBr₂
You may know these information:
Salts from acetate are soluble.
Bromide can make solid salts with these cations: Ag⁺, Pb²⁺, Hg₂²⁺, Cu⁺
PbBr₂ is formed, so this will be our precipitate
The equation is:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓
<h3>
Answer:</h3>
5.55 mol C₂H₅OH
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Tables
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
[Given] 500. g C₆H₁₂O₆ (Glucose)
[Solve] moles C₂H₅OH (Ethanol)
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH
[PT] Molar mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:

- [DA} Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH