Answer: Option (B) is the correct answer.
Explanation:
Surface tension is defined as the attractive forces experienced by the surface molecules of a liquid by the molecules present beneath the surface layer of the liquid.
And, viscosity is defined as the ability of a liquid to resist its flow. When a substance has high viscosity then it is known as a viscous substance.
Since, it is given that viscosity of liquid B is more than liquid A. Therefore, liquid B has more resistive force on its surface as compared to liquid A. As a result, time taken by liquid B is more than time taken by liquid A.
Also, Surface tension = ![\frac{Force}{Length}](https://tex.z-dn.net/?f=%5Cfrac%7BForce%7D%7BLength%7D)
Surface tension of liquid B is more than liquid A. Therefore,
.
Thus, we can conclude that tA will be less than tB.
Answer:
The answer to your question is None of your answers is correct, maybe the data are wrong.
Explanation:
Data
Concentration 1 = C1 = 1 M
Volume 2 = 5 ml
Concentration 2 = 0.05 M
Volume 1 = x
To solve this problem use the dilution formula
Concentration 1 x Volume 1 = Concentration 2 x Volume 2
Solve for Volume 1
Volume 1 = (Concentration 2 x Volume 2)/ Concentration 1
Substitution
Volume 1 = (0.05 x 5) / 1
Simplification
Volume 1 = 0.25/1
Result
Volume 1 = 0.25 ml
2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Mno4- (aq) + Cl- (aq) → Mn2+ + Cl2 (g)
Required
Half reaction
Solution
1. Add coefficient(equalizing atoms in reaction)
2. Adding H₂O on the O-deficient side.
3. Adding H⁺ on the H-deficient side.
4. Adding e⁻
5. Equalizing the number of electrons and sum the all the half-reaction
1.MnO₄⁻(aq) = Mn²⁺(aq) reduction
2.MnO4(aq) = Mn2+(aq) + 4H2O
3. MnO4-(aq) + 8H+ = Mn2+(aq) + 4H2O
4. MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O
Cl⁻(aq) = Cl₂(g) oxidation
1. 2Cl-(aq) = Cl2 (g)
2-3 none
4. 2Cl-(aq) = Cl2 (g) + 2e-
5.
MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O x2
2Cl-(aq) = Cl2 (g) + 2e- x5
2MnO4-(aq) + 16H+ + 10e- = 2Mn2+(aq) + 8H2O
10Cl-(aq) = 5Cl2 (g) + 10e-
<em>2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)</em>