I think it was sue to she has all the details to seem like it was her
Given that <span>sample a has a higher melting point than sample
b. Therefore, sample a is a longer chain of a </span><span>fatlike solid substance. It could also be that the bonds present in sample a is much stronger which will require more energy to break. Hope this answers the question.</span>
Answer:
V ∝ abc
Explanation:
This task is a joint variation task involving only direct proportionality:
Direct variation is one in which two variables are in direct proportionality to each other. This means that as one increases, the other variable also increases and vice - versa.
Joint variation is one in which one variable is dependent on two or more variables and varies directly as each of them.
In this exercise:
If a ∝ b and a ∝ c, then a ∝ bc
Taking the above three proportionalities,
V ∝ a ∝ b ∝ c
V ∝ a ∝ bc
V ∝ abc
Answer:
Static Friction.
Explanation:
Friction is the force that resists the relative motion between the surfaces sliding against each other.
Static friction is friction between objects that are not in relative motion with each other.
The coefficient of static friction, typically denoted as μs,
Static friction arises due to surface roughness( relative term)
The static friction force can be overcome by an applied maximum force
F max = μs x N
N= normal force
Any force smaller than F max attempting to slide one surface over the other is opposed by a frictional force of equal magnitude and opposite direction.
Any force larger than F max overcomes the force of static friction and causes sliding to occur.
This maximum force is sometimes called the limiting value also. Here that value is 75 N.
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
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