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Darina [25.2K]
2 years ago
14

Which of the following actions cannot induce voltage in a wire?

Chemistry
2 answers:
MariettaO [177]2 years ago
7 0
Your answer is D. Since there is little to no magnetic field to  wire, if it is copper which most wires are, there will be no voltage in a wire.
andrey2020 [161]2 years ago
5 0

Answer : The correct option is, E.

Explanation :

Induced voltage : According to Faraday's Law the induced voltage in a coil is directly proportional to the number of turns in the coil and to the rate at which magnetic field is changing.

Formula for induced voltage :

e=N\frac{d\phi}{dt}

where,

e = induced voltage

N = number of turns in a coil

\phi = magnetic flux

t = time

Magnetic flux : Magnetic flux is measure the magnetic field within a closed area.

Formula used for magnetic flux :

\phi=BAcos\theta

where,

B = magnetic field

A = area

\theta = angle between the magnetic field and the area of loop

Only option (E) can not induced voltage in a wire because when moving a wire parallel to a magnetic field then the \theta will be zero, so cos\theta will become 1. Hence, the flux becomes constant.

As, the flux is a constant value for this, so, the induced voltage becomes 0 because \frac{d\phi }{dt}=0. So, it will not induced voltage.

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Answer:

B, D

Explanation:

The strategy here is to realize that the ice will be taken from -6.5 ºC to OºC where it will melt.

Lets call q₁ the heat required to bring the ice to 0ºc, q₂ the heat required to bring the phase change from solid to liquid.

q₁ is calculated from the expression

q₁ = s x m x ΔT where m is the mass, s the specific heat of ice ( 2.09 J/gºC ) and ΔT  the change in temperature.

q₂ the fusion enthalpy change   is calculated from the expression:

q₂ = C x ΔT

where C is the specific heat for the phase change , in this case named AH  given in kJ/mol.

We are given all the data needed to calculate q₁, q₂ and qtotal ( q₁ + q₂ )

q₁ = 25.0 g x ( 2.09 J/gºC) x ( 0 - ( -6.5 ºC ) )

q₁ = 339.6 J = 0.339 kJ

q₂ = (25 g/ 18 g/mol) x 6.02 kJ/mol = 1.39 x 6.02 kJ = 8.36 kJ

qtotal = 0.339 kJ + 8.36 kJ = 8.70 kJ

with these calculations, we can now proceed to answer the question:

(a) False AH is theheat capacity for the melting.

(b) True as we determined above

(c) False we only have one phase change, from solid (ice) to liquid

(d) True as calculated above

(e) False as determined in our calculations

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2 years ago
Consider the elements in Group 14. Which element has the highest electron affinity? A) Carbon B) Germanium C) Lead D) Silicon
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Electron affinity is a measure of the tendency of a neutral atom  to gain electrons and form a negative ion. It can be represented by a general equation:

X + e- → X-

The value of electron affinity decreases on moving down a group. This is because on moving down a group the atomic size increases as a result the added electron feels less pull or attraction towards the nucleus.

In group 14, Carbon 'C' is the first member and therefore will have the highest electron affinity.

Ans : A) Carbon

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Consider the chemical reaction: N2 3H2 yields 2NH3. If the concentration of the reactant H2 was increased from 1.0 x 10-2 M to 2
Brilliant_brown [7]

The equilibrium constant of a reaction is defined as:

"The ratio between equilibrium concentrations of products powered to their reaction quotient and  equilibrium concentration of reactants powered to thier reaction quotient".

The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.

To solve this question we need this additional information:

<em>For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:</em>

<em>[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M</em>

<em />

Thus, for the reaction:

N₂ + 3H₂ ⇄ 2NH₃

The equilibrium constant, K, of this reaction, is defined as:

K = 6.0x10^{-2} = \frac{[NH_3]^2}{[N_2][H_2]^3}

And Q, is:

Q = \frac{[NH_3]^2}{[N_2][H_2]^3}

Where actual concentrations are:

[NH₃] = 1.0x10⁻⁴M

[N₂] = 4.0M

[H₂] = 2.5x10⁻¹M

Replacing:

Q = \frac{[1.0x10^{-4}]^2}{[4.0][2.5x10^{-1}]^3}

<h3>Q = 1.6x10⁻⁷</h3>

As Q < K,

<h3>The chemical system will shift to the right in order to produce more NH₃</h3>

Learn more about chemical equililbrium in:

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