Answer: D) 36.0 km/hr, downstream
Explanation:
For downstream motion of the boat, the actual velocity of the boat is the sum of velocity of the water current and the velocity of the boat due to pushed by wind.
Velocity of water current, v = 15 km/h
Velocity of the boat going downstream, u = 21 km/h
Actual velocity of the boat = v'
v' = v + u
⇒v' = 15 km/h + 21 km/h
⇒u = 21 km/h +15 km/h = 36.0 km/h downstream
Thus, the correct answer is option D.
The answer would be B. Brimone. I had the same question before, but let me know if it is not right. Cause certain schools have the same questions but different answers for them.
Answer:
molecular weight (Mb) = 0.42 g/mol
Explanation:
mass sample (solute) (wb) = 58.125 g
mass sln = 750.0 g = mass solute + mass solvent
∴ solute (b) unknown nonelectrolyte compound
∴ solvent (a): water
⇒ mb = mol solute/Kg solvent (nb/wa)
boiling point:
- ΔT = K*mb = 100.220°C ≅ 373.22 K
∴ K water = 1.86 K.Kg/mol
⇒ Mb = ? (molecular weight) (wb/nb)
⇒ mb = ΔT / K
⇒ mb = (373.22 K) / (1.86 K.Kg/mol)
⇒ mb = 200.656 mol/Kg
∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg
moles solute:
⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute
molecular weight:
⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol
When an element losses its electron its called a cation. When an element accepted that electron it called anion. This is called an ionic bond.
Answer:
Shifts the equilibrium to the left. reduces solubility.
Explanation:
- MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
S S 2S
∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²
⇒ 4S² * S = 6.4 E-9
⇒ 4S³ = 6.4 E-9
⇒ S³ = 1.6 E-9
⇒ S = 1.1696 E-3 M
- NaF(s) → Na+(aq) + F-(aq)
0.10M 0.10M 0.10M
- MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
S' S' 2S' + 0.10
⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²
If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:
⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'
⇒ S' = 6.4 E-7 M
∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption
We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).