How many milligrams remain of a 15.0 mg sample of radium-226 after 6396 years? The half life of radium-226 is 1599 years.
1 answer:
You might be interested in
When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = =74.8 %.
Answer:
1.33 × 10²⁴ molecules CO₂
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Reading a Periodic Table
- Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Explanation:
<u>Step 1: Define</u>
97.3 g CO₂
<u>Step 2: Define conversions</u>
Avogadro's Number
Molar Mass of C - 12.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol
<u>Step 3: Convert</u>
= 1.33138 × 10²⁴ molecules CO₂
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules.</em>
1.33138 × 10²⁴ molecules CO₂ ≈ 1.33 × 10²⁴ molecules CO₂