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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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tester [92]
To solve these problems first draw the free body diagram:

8 0
4 years ago
Resting energy expenditure is a. slightly higher than basal energy expenditure. b. the same as basal energy expenditure. c. slig
kvv77 [185]

Answer:

B. Resting energy expenditure is the same with basal energy expenditure.

Explanation:

Basal Energy Expenditure can be explained as the energy required to execute essential metabolic functions e.g. coordination of enzymatic reactions in the body system.

On the other hand, Resting Energy Expenditure can be simply explained as the amount of energy expended or burnt when the body is resting.

Hence, in the context of definitions, and relating both definitions, it can be argued that Basal energy expenditure is simply the energy needed to execute essential metabolic functions e.g. coordination of enzymatic reactions in the body, with special emphassy on the body being at rest. Thus, in this context, Basal energy can be looked at through the prism of Resting energy expenditure. Consequently, this two definitions can be used interchangeably, with a special emphassy on perspective.

6 0
4 years ago
Two a.c V1=100sin(wt) and V2 = 150cos(wt) are fed into one circuit, determine the combined output of the two as a single a.c
Dominik [7]

Answer:

Explanation:

V = 100sin(ωt) + 150cos(ωt)

let x = ωt

V = 100sin(x) + 150cos(x)

a maximum or minimum will occur when the derivative is zero

V' = 100cos(x) - 150sin(x)

0 = 100cos(x) - 150sin(x)

100cos(x) = 150sin(x)

100/150 = sin(x)/cos(x)

0.6667 = tan(x)

x = 0.588 rad

V = 100sin(0.588) + 150cos(0.588)

V = 180.27756

as the maximum will not occur until ωt = 0.588 radians, for a cosine function we subtract that amount as a phase angle φ

V = 180.3 cos(ωt - 0.588)

or as a sine function, the phase angle lags the cosine by a difference of π/2

V = 180.3sin(ωt - (0.588 - π/2)

V = 180.3sin(ωt + 0.983)

7 0
3 years ago
A uniform 4.4 t magnetic field points north. if an electron moves vertically downward (toward the ground) with a speed of 2.5 ×
Reil [10]

Answer:

1.8\cdot 10^{-11} N

Explanation:

The force acting on the electron due to the magnetic field is

F=qvB sin \theta

where

q=1.6\cdot 10^{-19} C is the charge of the electron

v=2.5\cdot 10^7 m/s is its speed

B=4.4 T is the intensity of the magnetic field

\theta=90^{\circ} is the angle between the direction of the field and the velocity of the electron

Substituting all the numbers into the equation, we find

F=(1.6\cdot 10^{-19}C)(2.5\cdot 10^7 m/s)(4.4 T) sin 90^{\circ}=1.8\cdot 10^{-11} N

8 0
4 years ago
A 1000-kg car traveling north at 15 m/s collides with a2000-kg truck traveling east at 10 m/s. The occupants, wearing seat belts
Aleksandr-060686 [28]

Answer:

8.33 m/s, 36.87° North of East

Explanation:

m_n = Mass of car = 1000 kg

v_n = Velocity of car = 15 m/s

m_e = Mass of truck = 2000 kg

v_e = Velocity of truck = 10 m/s

M = Combined mass = 1000+2000 = 3000 kg

Momentum

p_n=m_nv_n\\\Rightarrow p_n=1000\times 15\\\Rightarrow p_n=15000\ kgm/s

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p_e=m_ev_e\\\Rightarrow p_n=2000\times 10\\\Rightarrow p_n=20000\ kgm/s

Momentum of truck traveling North is 20000 kgm/s

Angle

\theta=tan^{-1}\frac{p_n}{p_e}\\\Rightarrow \theta=tan^{-1}\frac{15000}{20000}\\\Rightarrow \theta=36.87^{\circ}

As the two vehicles are vectors, the resultant velocity is

(Mv)^2=p_n^2+p_e^2\\\Rightarrow v=\sqrt{\frac{p_n^2+p_e^2}{M^2}}\\\Rightarrow v=\sqrt{\frac{15000^2+20000^2}{3000^2}}\\\Rightarrow v=8.33\ m/s

Velocity of the two vehicles when they are locked together is 8.33 m/s and direction is 36.87° North of East

5 0
3 years ago
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