Answer:
Oxygen is the limiting reactant.
Explanation:
Based on the reaction:
C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O
<em>1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of CO₂ and 11 moles of H₂O.</em>
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10.0g of sucrose (Molar mass: 342.3g /mol) are:
10.0g C₁₂H₂₂O₁₁ × (1mole / 342.3g) = 0.0292 moles of C₁₂H₂₂O₁₁
And moles of 10.0g of oxygen (Molar mass: 32g/mol) are:
10.0g O₂ × (1mole / 32g) = 0.3125 moles of O₂
For a complete reaction of 0.0292 moles of C₁₂H₂₂O₁₁ you need (knowing 12 moles of oxygen react per mole of sucrose):
0.0292 moles of C₁₂H₂₂O₁₁ × (12 moles O₂ / 1 mole C₁₂H₂₂O₁₁) = <em>0.3504 moles of O₂</em>
As you have just 0.3125 moles of O₂, <em>oxygen is the limiting reactant.</em>
Second ionization energy is defined by the equation: it is the energy needed to remove second electron from each job in 1 mole of gaseous 1+ ions to give gaseous 2+ ions. More ionization energies. You can then have as many successive ionization energies as there are electrons in the original atom.
Answer:
Berryllium
Explanation:
its the most reactive (mark brainliest plz)
Enthalpy of reaction: -41.15 kJ
Entropy change: -42.36 kJ
Answer:
The answer to the question is
The star’s approximate radial velocity is 68.52 km/s
Explanation:
To solve the
The formula is
where
= velocity of the star
λ = Star's spectrum wavelength = 656.45 nm
= Rest wavelength = 656.30 nm
c = Speed of light = 299 792 458 m / s
Therefore we have
or = 68518.7699 m/s or 68.52 km/s