Number of atoms in 1 mole = 6.023 * 10²³
Number of atoms in 0.46 mole = 0.46 * 6.023 * 10²³
It is = 2.77 * 10²³ atoms (Approx.)
In short, Your Answer would be Option B
Hope this helps!
Answer: 287.8 cm3
Explanation:
Given that:
Initial volume of gas V1 = 350 cm3
Initial pressure of gas P1 = 740 mmHg
New volume V2 = ?
New pressure P2 = 900 mmHg
Since, pressure and volume are involved while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
740 mmHg x 350 cm3 = 900mmHg x V2
V2 = (740 mmHg x 350 cm3) /900mmHg
V2 = 259000 mmHg cm3 / 900mmHg
V2 = 287.8 cm3
Thus, the gas will occupy 287.8 cubic centimeters at the new pressure.
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.
