The effort force is the force applied to a machine. Work input is the work done on a machine. The work input of a machine is equal to the effort force times the distance over which the effort force is exerted.
Answer:
The ionization of 0.250 moles of H₂SO₄ will produce 0.5 moles of H⁺ (hydrogen ion)
Explanation:
From the ionization of H₂SO₄, we have
H₂SO₄ → 2H⁺ + SO₄²⁻
Hence, at 100% yield, one mole of H₂SO₄ produces two moles of H⁺ (hydrogen ion) and one mole of SO₄²⁻ (sulphate ion), therefore, 0.250 moles of H₂SO₄ will produce 2×0.250 moles of H⁺ (hydrogen ion) or 0.5 moles of H⁺ (hydrogen ion) and 0.25 moles of SO₄²⁻ (sulphate ion).
That is; 0.250·H₂SO₄ → 0.5·H⁺ + 0.250·SO₄²⁻.
The vascular cambium also known as the main cambium the wood cambium plural cambium is a plant tissue located between the xylem and the phloem
Answer:
i = 2.483
Explanation:
The vapour pressure lowering formula is:
Pₐ = Xₐ×P⁰ₐ <em>(1)</em>
For electrolytes:
Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ
Where:
Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)
4.5701g of MgCl₂ are:
4.5701g ₓ (1mol / 95.211g) = 0.048000 moles
43.238g of water are:
43.238g ₓ (1mol / 18.015g) = 2.400 moles
Replacing in (1):
0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm
0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)
2.4mol + i*0.048mol = 2.4mol / 0.9527
2.4mol + i*0.048mol = 2.5192mol
i*0.048mol = 2.5192mol - 2.4mol
i = 0.1192mol / 0.048mol
<em>i = 2.483</em>
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I hope it helps!
Answer:
the youngest is G and the oldest is K
