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Elis [28]
3 years ago
10

A strip of copper metal is placed in an aqueous sodium chloride solution. According to the standard reduction potential below, w

hich of the following would be observed A. No reaction occurs B. Bubbles of gas would be observed C. A blue color would form in the solution D. Crystals of copper (II) chloride would form on the strip of copper
Physics
1 answer:
oee [108]3 years ago
3 0

Answer:

The answer to A strip of copper metal is placed in an aqueous sodium chloride solution. According to the standard reduction potential below, is Option A) No reaction occurs

Explanation:

Aqueous sodium chloride solution is similar to salt and water mixed together.

Copper doesn’t dissolve in salty water. However, exposure of the metal to air and then to water results in an oxide layer on the surface of copper which is referred: the dull reddish brown copper(I) oxide and this exists in an equilibrium with the outer oxide layer of black copper(II) oxide.

If you put a strip of copper metal is placed in an aqueous sodium chloride solution, you will not observe any reaction because the oxide layer on the surface of copper is too thin for to be noticed other than by a dulling or darkening of the surface, not thick enough to be obvious.

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1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.
marta [7]

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB

<h3>Speed of sound at the given temperature</h3>

v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )

where;

  • -v₀ is velocity of the observer moving away from the source
  • -vs is the velocity of the source moving towards the observer
  • fs is the source frequency
  • fo is the observed frequency
  • v is speed of sound

f_0 = f_s(\frac{v-v_0}{v- v_s} )

f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz

Learn more about intensity of sound here: brainly.com/question/17062836

3 0
1 year ago
2. A person applies a force of 66 N to a fridge as they push it across the length of a standard tennis court. So far today, the
Lubov Fominskaja [6]

Answer:

P=39.2205\, watt

E=374.948 \,cal

Explanation:

Given that:

  • force applied, F=66\,N
  • displacement, s=23.77\,m (length of a tennis court)
  • time taken for pushing, t = 40 s

Since, work is given by:

W=F.s

W=66\times 23.77

W=1568.82\,J

Now, power is given as:

P=\frac{W}{t}

P=\frac{1568.82}{40}

P=39.2205 \,watt

Calories consumed is:

E= 1568.82\times 0.239

E=374.948\, cal

3 0
3 years ago
Two tuning forks produce sounds of wavelengths of 3.4 meters and 3.3 meters.
KengaRu [80]

Answer:

Explanation:

wavelength, λ = 3.4 m

wavelength, λ' = 3.3 m

Speed, v = 340 m/s

f = v / λ = 340 / 3.4 = 100 Hz

f' = v / λ' = 340 / 3.3 = 103.03 Hz

Frequency of beat, n = f' - f = 103.03 - 100 = 3.03 Hz

5 0
3 years ago
What factors determine the presence of electrical energy?
luda_lava [24]
One of them would be power source.
8 0
3 years ago
A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is:_________.A) 0.00245 m3
Kamila [148]

Answer: B) 0.00337 m3.

Explanation:

Given data:

Mass of the ball = 10kg

Weight of the ball in air = 98N

Weight of the ball in water = 65N

Solution:

To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.

= 98 - 65 / 9.8

= 33 / 9.8

= 3.37kg

The volume of the ball is 3.37kg

The density of water is 1kg per Liter.

So 3.37 kg of water would have a volume of 3.37 Liters.

Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).

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