Answer:
2K (s) + Cl₂ (g) ⇒ 2KCl (s)
Explanation:
Potassium and chlorine gas combine to form potassium chloride which is an ionic compound. The reaction is a type of combination reaction in which chlorine is being added to the metal, potassium.
Potassium reacts violently with the chlorine which is yellowish green in color to produce white solid of potassium chloride.
The balanced reaction is shown below as:
2K (s) + Cl₂ (g) ⇒ 2KCl (s)
Answer: Photosynthesis and cellular respiration are related because the reactants for photosynthesis is the products (plus ATP) is cellular respiration. Also the products for cellular respiration is the reactants for cellular respiration.
Explanation:
The fraction of the original amount remaining is closest to 1/128
<h3>Determination of the number of half-lives</h3>
- Half-life (t½) = 4 days
- Time (t) = 4 weeks = 4 × 7 = 28 days
- Number of half-lives (n) =?
n = t / t½
n = 28 / 4
n = 7
<h3>How to determine the amount remaining </h3>
- Original amount (N₀) = 100 g
- Number of half-lives (n) = 7
- Amount remaining (N)=?
N = N₀ / 2ⁿ
N = 100 / 2⁷
N = 0.78125 g
<h3>How to determine the fraction remaining </h3>
- Original amount (N₀) = 100 g
- Amount remaining (N)= 0.78125 g
Fraction remaining = N / N₀
Fraction remaining = 0.78125 / 100
Fraction remaining = 1/128
Learn more about half life:
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1. The empirical formula of the hydrocarbon is CH₃
2. The molecular formula of the hydrocarbon is C₂H₆
<h3>How to determine the mass of Carbon </h3>
- Mass of CO₂ = 1.47 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 1.47
Mass of C = 0.4 g
<h3>How to determine the mass of H</h3>
- Mass of compound = 0.5 g
- Mass of C = 0.4 g
- Mass of H = ?
Mass of H = (mass of compound) – (mass of C)
Mass of H = 0.5 – 0.4
Mass of H =0.1 g
<h3>1. How to determine the empirical formula </h3>
- C = 0.4 g
- H = 0.1 g
- Empirical formula =?
Divide by their molar mass
C = 0.4 / 12 = 0.03
H = 0.1 / 1 = 0.1
Divide by the smallest
C = 0.03 / 0.03 = 1
H = 0.1 / 0.03 = 3
Thus, the empirical formula of the compound is CH₃
<h3>2. How to determine the molecular formula</h3>
- Empirical formula = CH₃
- Molar mass = 30 g/mol
- Molecular formula =?
Molecular formula = empirical × n = mass number
[CH₃]n = 30
[12 + (3×1)]n = 30
15n = 30
Divide both side by 15
n = 30 / 15
n = 2
Molecular formula = [CH₃]n
Molecular formula = [CH₃]₂
Molecular formula = C₂H₆
Learn more about empirical formula:
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Answer:
≅ 16.81 kJ
Explanation:
Given that;
mass of acetone = 31.5 g
molar mass of acetone = 58.08 g/mol
heat of vaporization for acetone = 31.0 kJ/molkJ/mol.
Number of moles = 
Number of moles of acetone = 
Number of moles of acetone = 0.5424 mole
The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;
Hence;
The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol
The heat required to vaporize 31.5 g of acetone = 16.8144 kJ
≅ 16.81 kJ
