<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:
2 moles
Explanation:
In one mole of O2 there are 16 grams. So in 2 moles there are 32 grams
Answer:
The answer to your question is all the formulas in bold has the same empirical formula
Explanation:
Data
Empirical formula CH₂O
Process
To solve this problem factor the subscripts of each formula and compare the result with the empirical formula given.
a) C₂H₄O₂ factor 2 2(CH₂O)
b) C₃H₆O₃ factor 3 3(CH₂O)
c) CH₂O₂ this formula can not be simplified
d) C₅H₁₀O₅ factor 5 5(CH₂O)
e) C₆H₁₂O₆ factor 6 6(CH₂O)
Answer:
5x10⁻⁶ = [HTeH₄O₆⁺]
Explanation:
The first dissociation equilibrium of the telluric acid in water is:
H₂TeH₄O₆ + H₂O ⇄ HTeH₄O₆⁺ + H₃O⁺
Using H-H equation for telluric acid:
<em>pH = pKa + log₁₀ [HTeH₄O₆⁺] / [H₂TeH₄O₆]</em>
pKa of telluric acid is -logKa1
pKa = -log 2.0x10⁻⁸
pKa = 7.699
As concentration of [H₂TeH₄O₆] is 0.25M, replacing in H-H equation:
3.00 = 7.699+ log₁₀ [HTeH₄O₆⁺] / [0.25M]
-4.699 = log₁₀ [HTeH₄O₆⁺] / [0.25M]
2x10⁻⁵ = [HTeH₄O₆⁺] / [0.25M]
<h3>5x10⁻⁶ = [HTeH₄O₆⁺]</h3>