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Scrat [10]
3 years ago
15

5. What is the difference between a reflected and a diffracted sound wave?

Chemistry
1 answer:
d1i1m1o1n [39]3 years ago
3 0

Answer:

Reflection involves a change in direction of waves when they bounce off a barrier; refraction of waves involves a change in the direction of waves as they pass from one medium to another; and diffraction involves a change in direction of waves as they pass through an opening or around a barrier in their path.

Explanation:

https://www.physicsclassroom.com/class/waves/Lesson-3/Reflection,-Refraction,-and-Diffraction#:~:text=Diffraction%20of%20Waves-,Reflection%20involves%20a%20change%20in%20direction%20of%20waves%20when%20they,a%20barrier%20in%20their%20path.

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What is the volume of 1.56 kg of a compound whose molar mass is 81.86 g/mole and whose density is 41.2 g/ml?
hjlf

Answer:

v = 37.9 ml

Explanation:

Given data:

Mass of compound = 1.56 kg

Density = 41.2 g/ml

Volume of compound = ?

Solution:

First of all we will convert the mass into g.

1.56 ×1000 = 1560 g

Formula:

D=m/v

D= density

m=mass

V=volume

v = m/d

v =  1560 g / 41.2 g/ml

v = 37.9 ml

7 0
3 years ago
BRAINLIESTTT ASAP!! PLEASE HELP ME :))
goldenfox [79]

Answer:

Solar panels and solar cells.

Explanation:

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5 0
3 years ago
Read 2 more answers
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boyakko [2]

Answer:

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8 0
3 years ago
Let’s assume that you put a balloon into the freezer. Initially,the balloon had 3.0liters of gas at a pressure of 400kPa and was
EastWind [94]
Here, we should use combined gas law which can be derived from combined gas law, “PV=nRT”. Rearranging, we can get PV/T=nR. Then we can set the two states in the problem together to get

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Then just plug in and solve algebraically.

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6 0
3 years ago
Free-energy change, ΔG∘, is related to cell potential, E∘, by the equation ΔG∘=−nFE∘ where n is the number of moles of electrons
elena-s [515]

Answer:

-372000 J or -372 KJ

Explanation:

We have the electrochemical reaction as;

Mg(s)  +  Fe^2+(aq)→  Mg^2+(aq)  +   Fe(s)

We must first calculate the E∘cell from;

E∘cathode -  E∘anode

E∘cathode = -0.44 V

E∘anode = -2.37 V

Hence;

E∘cell = -0.44 V -(-2.37 V)

E∘cell = 1.93 V

n= 2 since two electrons were transferred

F=96,500C/(mol e−)

ΔG∘=−nFE∘

ΔG∘= -( 2 * 96,500 * 1.93)

ΔG∘= -372000 J or -372 KJ

4 0
2 years ago
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