The dilution formula can be used to find the volume needed
c1v1 = c2v2
Where c1 is concentration and v1 is volume of the concentrated solution
And c2 is concentration and v2 is volume of the diluted solution to be prepared
c1 - 0.33 M
c2 - 0.025 M
v2 - 25 mL
Substituting these values in the equation
0.33 M x v1 = 0.025 M x 25 mL
v1 = 1.89 mL
Therefore 1.89 mL of the 0.33 M solution needs to be diluted up to 25 mL to make a 0.025 M solution
<h3>
Answer:</h3>
128 g HCl
<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>
<u>Stoichiometry</u>
- Reaction Mole Ratios
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl (aq) + H₂ (g)
↓
[RxN - Balanced] 2Mg (s) + 2HCl (aq) → 2MgCl (aq) + H₂ (g)
[Given] 3.25 mol Mg
[Solve] x g HCl
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Mg → 2 mol HCl
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of Cl - 35.45 g/mol
Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol
<u>Step 3: Stoich</u>
- [S - DA] Set up:

- [S - 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>
127.61 g HCl ≈ 128 g HCl
Answer:
less than 5 centimeters because it is a nonvascular plant
Explanation:
Mosses are a group of plants under the division Bryophyta. They are said to be the most primitive plant life in existence as they lack true roots, stems and leaves. They also lack vascular system, hence, they are regarded as non-vascular plants. They usually grow in very small sizes (about 0.2 - 10cm).
According to this question, Penny bought a club moss plant for her water garden and needs to know how tall the plant will grow so she know how much space it will need. Since it is a miss plant that lacks vascular tissues i.e. nonvascular, it will likely grow less than 5 centimeters in height.
That would be phosphorus. It’s electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^3