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Naily [24]
3 years ago
15

Demonstrate your knowledge of waves.Describe the difference between mechanical and electromagnetic waves. Give an example of eac

h kind of wave related to telecommunications. Answer using complete sentences and key vocabulary
PLEASE HELP
Chemistry
1 answer:
4vir4ik [10]3 years ago
8 0

There are some methods that differentiate waves; one thing is by its medium. Electromagnetic waves did not require a medium for transmission such as in a vacuum while mechanical waves require a medium to travel such as air, water or anything that can serve as a transmission aid.  

The most common sample of EM waves in telecommunication is radio, light and infra-red signals. An example of a mechanical wave is a sound wave, which requires air to travel. Oscillating molecules made the sound waves. 

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You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acetate plus acetate concentration
vovangra [49]

<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>

Given:

Acetic Acid/Sodium Acetate buffer of pH = 5.0

Let HA = acetic acid

A- = sodium acetate

Total concentration [HA] + [A-] = 250 mM ------(1)

pKa(acetic acid) = 4.75

Based on Henderson-Hasselbalch equation

pH = pKa + log[A-]/[HA]

[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77

[A-] = 1.77[HA] -----(2)

From (1) and (2)

[HA] + 1.77[HA] = 250 mM

[HA] = 250/2.77 = 90.25 mM

[A-] = 1.77(90.25) = 159.74 mM



7 0
3 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
The volume of 1.20 lol of gas at 61.3 kPa and 25.0 degree Celsius is
aniked [119]

Answer:

V = 48.5 L

Explanation:

Converting °C to K and kPa to atm

T = 25.0°C + 273.15 = 298.15 K

P = 61.3 kPa × (1 atm / 101.325 kPa) = 0.60498 atm

Calculating the volume of gas

V = nRT / P

V = (1.20 mol)(0.082057 L•atm/mol•K)(298.15 K) / 0.60498 atm

V = 48.5 L

4 0
2 years ago
A pure gold bar is made up of only particles. <br><br>A.nickel<br>B.copper <br>C.god<br>D.gold​
Minchanka [31]

Answer: gold

Explanation:

4 0
3 years ago
We have two surfaces A and B where same amount of Force is acting. Area of Surface A is greater than Area of Surface B. which of
Phoenix [80]

Answer:

Pressure on Surface A > Pressure on Surface B

Explanation:

Pressu

re on Surface B > Pressure on Surface A

7 0
3 years ago
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