Answer:
a would be the correct answer
<h3>Answer:</h3>
Molar Mass = 56 g.mol⁻¹
<h3>Explanation:</h3>
Data Given:
Mass = 5.00 μg = 5.0 × 10⁻⁶ g
Number of Molecules = 5.38 × 10¹⁶ Molecules
Step 1: Calculate Moles of 1-Butene:
As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.
The relation between Moles, Number of Particles and Avogadro's Number is given as,
Number of Moles = Number of Particles ÷ 6.022 × 10²³
Putting values,
Number of Moles = 5.38 × 10¹⁶ Molecules ÷ 6.022 × 10²³
Number of Moles = 8.93 × 10⁻⁸ Moles
Step 2: Calculate Molar Mass of 1-Butene:
As,
Mole = Mass ÷ M.Mass
Solving for M.Mass,
M.Mass = Mass ÷ Mole
Putting values,
M.Mass = 5.0 × 10⁻⁶ g ÷ 8.93 × 10⁻⁸ mol
M.Mass = 55.99 g.mol⁻¹ ≈ 56 g.mol⁻¹
This is a strong acid but the actual value of the pH will depend on its strength.
Acids have a pH of less than 7.
Answer:
1.) 13 g C₄H₁₀
2.) 41 g CO₂
Explanation:
To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O
48 g O₂ 1 mole 2 moles C₄H₁₀ 58.124 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 13 g C₄H₁₀
48 g O₂ 1 mole 8 moles CO₂ 44.007 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 41 g CO₂
84.24 g of water (H₂O)
Explanation:
We have the following chemical reaction:
2 H₂O → 2 H₂ + O₂
Now we calculate the number of moles of products.
number of moles = mass / molar weight
number of moles of H₂ = 50 / 2 = 25 moles
number of moles of O₂ = 75 / 32 = 2.34 moles
We see from the chemical reaction that for every 2 moles of H₂ produced there are 1 mole of O₂ produces for every 25 moles of H₂ produced there are 12.5 moles of O₂ but we only have 2.35 moles of O₂ available. The O₂ will be the limiting quantity from which we devise the following reasoning:
if 2 moles of H₂O produces 1 mole of O₂
then X moles of H₂O produces 2.34 mole of O₂
X = (2 × 2.34) / 1 = 4.68 moles of H₂O
mass = number of moles × molar weight
mass of H₂O = 4.68 × 18 = 84.24 g
Learn more about:
limiting reactant
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