Given 35.0 grams of aluminum at 22.0 °C. What will be the final temperature of the Al if 300.0 joules of heat energy is added to
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<h3>Answer:</h3>
2 M
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Unit 0</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Aqueous Solutions</u>
- Molarity = moles of solute / liters of solution
<u>Step 1: Define</u>
36.7 g CaF₂
300 mL H₂O
<u>Step 2: Identify Conversions</u>
Molar Mass of Ca - 40.08 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of CaF₂ - 40.08 + 2(19.00) = 78.08 g/mol
1000 mL = 1 L
<u>Step 3: Convert</u>
<em>Solute</em>
- Set up:
- Multiply:
<em>Solution</em>
- Set up:
- Multiply:
<u>Step 4: Find Molarity</u>
- Substitute [M]:
- Divide:
<u>Step 5: Check</u>
<em>Follow sig fig rules and round.</em> <em>We are given 1 sig fig as our lowest.</em>
1.56677 M ≈ 2 M
Answer:
A. 0.143 M
B. 0.0523 M
Explanation:
A.
Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).
KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄
The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:
1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol
The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.
5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:
M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M
B.
Let's consider the neutralization of potassium hydroxide and perchloric acid.
KOH + HClO₄ → KClO₄ + H₂O
When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.