On a hot day, a student places a glass of cold lemonade on a table outdoors. After a few minutes, water droplets have formed on the outside of the glass.

Is energy absorbed or released by the cold lemonade? Explain your answer.
Compare average kinetic energy for the air molecules and lemonade molecules when the student first places the lemonade outdoors. Explain your answer.
Explain how and why the water droplets form on the outside of the glass.

Answer:

Hydroelectric pumped storage can be used to eliminate the conditional nature of wind and solar energy. At times of peak production of energy from either solar or wind (during summer and windy days), the excess electrical energy produced can be stored using hydroelectric pumped storage methods. When conditions no longer favour energy production from either wind or solar sources, these stored energy can then be regenerated for use.

Explanation:

Pumped-storage hydroelectricity is a type of hydroelectric energy storage used by electric power systems to store excess electrical power during periods of low demand for later release at periods when demand for energy rises again. It stores energy in the form of gravitational potential energy of water, pumped from a lower level reservoir to a higher level reservoir.

At times of low electrical demand, excess generated energy is used to pump water into the upper reservoir. When there is higher demand, water is released back into the lower reservoir through a turbine, generating electricity.

This form of energy storage is useful in circumventing the conditional nature of renewable energy sources such as wind and solar energy. At times of peak production of energy from either solar or wind (during summer and windy days), the excess electrical energy produced can be stored using hydroelectric pumped storage methods. When conditions no longer favour energy production from either wind or solar sources, these stored energy can then be regenerated for use.

Answer:

0.208333 M

Explanation:

M2=M1V1/V2=(0.25 M)(125mL)/(125mL+25 mL)= 0.208333 M

Answer:

It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams ...

Explanation:

It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams ...

Answer: The value of acid ionization constant [tex]K_a[/tex] for acetic acid is [tex]1.87\times 10^{-5}[/tex]

Explanation:

[tex]CH_3COOH\rightarrow H^+CH_3COO^-[/tex]

 cM              0             0

[tex]c-c\alpha[/tex]        [tex]c\alpha[/tex]          [tex]c\alpha[/tex]  

So dissociation constant will be:

[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]

Give c= 0.85 M and [tex]pH[/tex] = 2.4

[tex]pH=-log[H^+][/tex]

[tex][H^=}=c\times \alpha=10^{-2.4}=3.98\times 10^{-3}[/tex]

[tex]K_a=?[/tex]

Putting in the values we get:

[tex]K_a=\frac{(3.98\times 10^{-3})^2}{(0.85-3.98\times 10^{-3})}=1.87\times 10^{-5}[/tex]

Thus the value of acid ionization constant [tex]K_a[/tex] for acetic acid is [tex]1.87\times 10^{-5}[/tex]

Answer:

they help to calm the bees when the person keeping the bees inspects there hives.