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Answer:
the pressure in the pipe in the case when there is no net force on the car is 81,726 N/m^2
Explanation:
a. The computation of the pressure in the pipe in the case when there is no net force on the car is shown below
As we know that
Pressure = F ÷ area
Also
F = mg
Now
= (1500 × 9.8) ÷π (0.24)^2
= 81,726 N/m^2
Hence, the pressure in the pipe in the case when there is no net force on the car is 81,726 N/m^2
I think it’s Object 3
Hopefully this helps
Ba²⁺ + 2Cl⁻ + 2H⁺ + SO₄²⁻ = BaSO₄ (precipitate) + 2H⁺ + 2Cl⁻
Ba²⁺ + SO₄²⁻ = BaSO₄
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.
