<span>mass percent of carbon (c) in CCl</span>₄ = mass of C/mass of CCl₄
= 12 x 100/153.8 = 7.8%
If you collected 7.0 grams of something the percentage recovery would be 70% (7/10 x 100).
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A percentage recovery is something like that
Answer:
Dipole moment of H₂O: 2.26 Debye
Bond angle in H₂S: 89.7°
Explanation:
The total dipole moment(μr) for H₂O can be calculated by the sum of the dipole moments of each bond of O-H. Because the dipole moment is a vector, the sum of these vectors can be calculated by the cosine law.
μr² = (μO-H)² + (μO-H)² + 2*(μO-H)*(μO-H)*cos(104.5°)
μr² = 1.85² + 1.85² + 2*1.85*1.85*cos(104.5°)
μr² = 3.4225 + 3.4225 + 6.845*(-0.2504)
μr² = 5.13115
μr = √5.13115
μr = 2.26 Debye.
For H₂S:
0.95² = 0.67² + 0.67² + 2*0.67*0.67*cosθ
0.9025 = 0.4489 + 0.4489 + 0.8978cosθ
0.8978cosθ = 0.0047
cosθ = 0.00523
θ = arcos0.00523
θ = 89.7°
