An intensive property does not change when you take away
some of the sample. The procedures that a student could use to examine the
intensive property of a rectangular block of wood are the hardness, color,
density and molecular weight.
Answer: its the first one buster
Answer:
Ionic equation:
Na⁺(aq) + OH⁻(aq) + H⁺(aq) + Cl⁻(aq) → H₂O(l) + Na⁺(aq) + Cl⁻ (aq)
Explanation:
Chemical equation:
NaOH(aq) + HCl(aq) → H₂O(l) + NaCl (aq)
Balanced chemical equation:
NaOH(aq) + HCl(aq) → H₂O(l) + NaCl (aq)
Ionic equation:
Na⁺(aq) + OH⁻(aq) + H⁺(aq) + Cl⁻(aq) → H₂O(l) + Na⁺(aq) + Cl⁻ (aq)
Net ionic equation:
OH⁻(aq) + H⁺(aq) → H₂O(l)
The Cl⁻(aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
<h3>
Answer:</h3>
0.000538 mol Pb
<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>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.24 × 10²⁰ particles Pb (lead)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
Our final answer is in 3 sig figs, no need to round.
Answer:
The molar concentration of HCl in the aqueous solution is 0.0131 mol/dm3
Explanation:
To get the molar concentration of a solution we will use the formula:
<em>Molar concentration = mass of HCl/ molar mass of HCl</em>
<em></em>
Mass of HCl in the aqueous solution will be 40% of the total mass of the solution.
We can extract the mass of the solution from its density which is 1.2g/mL
We will further perform our analysis by considering only 1 ml of this aqueous solution.
The mass of the substance present in this solution is 1.2g.
<em>The mass of HCl Present is 40% of 1.2 = 0.48 g.</em>
The molar mass of HCl can be obtained from standard tables or by adding the masses of Hydrogen (1 g) and Chlorine (35.46 g) = 36.46g/mol
Therefore, the molar concentration of HCl in the aqueous solution is 0.48/36.46 = 0.0131 mol/dm3
