Answer:
depends on how many you have...
Explanation:
The correct answer is approximately 11.73 grams of sulfuric acid.
The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.
The balanced equation is:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g
40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.
Answer:
Na has the most similar configuration.
Explanation:
Na electron configuration: 1s²2s²2p⁶3s¹ or [Ne] 3s₁
Mg electron configuration: 1s²2s²2p⁶3s² or [Ne] 3s²
Be electron configuration: 1s²2s² or [He] 2s²
This is because Na and Mg are right next to each other in the same period (horizontal).
<span>The answer is curie.
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Answer:

Explanation:
The Rydberg equation gives the wavelength λ for the transitions:

where
R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

Data:

λ = 657 nm
Calculation:
