Ofcoure the elephante as the elephant has more weight
Answer:
As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.
Explanation:
Depression in freezing point :

where,
=depression in freezing point =
= freezing point constant
m = molality ( moles per kg of solvent) of the solution
As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:
- If molality of the solution in high the depression in freezing point of the solution will be more.
- If molality of the solution in low the depression in freezing point of teh solution will be lower .
Relative lowering in vapor pressure of the solution is given by :

= Vapor pressure of pure solvent
= Vapor pressure of solution
= Mole fraction of solute

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.
- Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
- lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.
Answer:
The change in temperature, 
Explanation:
Given that,
The temperature in Spearfish, South Dakota, rose from
in just 2 minutes. We need to find the temperature change in Celsius degrees. Change in temperature is given by final temperature minus initial temperature such that,

The relation between degrees Celsius and degrees Fahrenheit is given by :

Here, F = 49 degrees

So, the change in temperature is 9.45 degree Celsius. Hence, this is the required solution.
Answer:

Explanation:
The force of gravity acting on the satellite is given by:

where
G is the gravitational constant
is the Earth's mass
m is the mass of the satellite
r is the distance of the satellite from the Earth's centre
Here we have
m = 700 kg

Substituting into the equation, we find:

<em>Note that the distance mentioned in the problem (2.4 x 10^6 meters) is not realistic, since it is less than the radius of the Earth (6.37 x 10^6 meters).</em>