Answer:
the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C
Explanation:
Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.
The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(
)
Substituting the values into the equation, it becomes
E = σ/ε₀() = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(
) = 7.12 × 10⁵(
) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C
Therefore, the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C
<span>polar, Hadley, Ferrel, polar, Hadley, Ferrel. I hope this helps you.</span>
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<span>A- solute is the answer you were looking for
</span>
Answer:
14.3 g SO₃
Explanation:
2S + 3O₂ → 2SO₃
First, find the limiting reactant. To do that, calculate the mass of oxygen needed to react with all the sulfur.
5.71 g S × (1 mol S / 32 g S) = 0.178 mol S
0.178 mol S × (3 mol O₂ / 2 mol S) = 0.268 mol O₂
0.268 mol O₂ × (32 g O₂ / mol O₂) = 8.57 g O₂
There are 10.0 g of O₂, so there's enough oxygen. The limiting reactant is therefore sulfur.
Use the mass of sulfur to calculate the mass of sulfur trioxide.
5.71 g S × (1 mol S / 32 g S) = 0.178 mol S
0.178 mol S × (2 mol SO₃ / 2 mol S) = 0.178 mol SO₃
0.178 mol SO₃ × (80 g SO₃ / mol SO₃) = 14.3 g SO₃
