That's two different things it depends on:
-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.
Here's what I have in mind for an experiment to show those two dependencies:
-- a closed box with a wall down the middle, separating it into two closed sections;
-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.
-- a tiny fan that blows air through a tube into the hole in one outer wall.
<u>Experiment A:</u>
-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
<span><em>Show that the 1 ounce of water evaporated faster </em>
<em>when it had more surface area.</em></span>
============================================
============================================
<u>Experiment B:</u>
-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the <em>first section </em>of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section. Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
<em>Show that it took longer to evaporate when the air </em>
<em>blowing over it was already loaded with vapor.</em>
==========================================
Answer:
C.Supersaturated
Explanation:
There are three types of solution:
<u>SATURATED SOLUTION</u>:
It is the solution that contains maximum amount of solute dissolved in a solution in normal conditions.
<u>UNSATURATED SOLUTION</u>:
It is the solution that contains less than the maximum amount of solute dissolved in a solution in normal conditions. It has space for more solute to be dissolved in it.
<u>SUPERSATURATED SOLUTION:</u>
It contains more than the maximum amount of solute dissolved in it. Such a solution has no capacity to dissolve any more solute under any conditions.
Since the sugar is no more dissolving in the tea and has settled down. Therefore, the solution is:
<u>C.Supersaturated</u>
Answer:
Loss, 
Explanation:
Given that,
Mass of particle 1, 
Mass of particle 2, 
Speed of particle 1, 
Speed of particle 2, 
To find,
The magnitude of the loss in kinetic energy after the collision.
Solve,
Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.
Applying the conservation of momentum to find the speed of two particles after the collision.
V = 6.71 m/s
Initial kinetic energy before the collision,
Final kinetic energy after the collision,
Lost in kinetic energy,
Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.
Answer:
Elements in Group 14 could lose four, or gain four electrons to achieve a noble gas structure. In fact, if they are going to form ions, Group 14 elements form positive ions. Carbon and silicon form covalent bonds. Carbon's millions of organic compounds are all based on shared electrons in covalent bonds.
Explanation:
Answer:
The equations of kinematics is applied for the motion with constant acceleration (including zero), but the condition is that the acceleration should be in the direction of the motion (positive or negative).
In circular motion, the acceleration is radial (centripetal), which means that the acceleration is always perpendicular to the motion of the object, therefore the equations of kinematics cannot be applied.
