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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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The answer to this question is D! The ball and stick model! Hope this helps :)
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1 year ago
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HELP! Ethanol has a density of 0.8 g/cm3. a. What is the mass of 225 cm3 of ethanol? b. What is the volume of 75.0 g of ethanol?
IRINA_888 [86]

Answer:

The correct answers are:

a) 180 g

b) 93.7 cm³

Explanation:

The density of a substance is the mass of the substance per unit of volume. So, it is calculated as follows:

density= mass/volume

From the data provided in the problem:

density = 0.8 g/cm³

a) Given: volume= 225 cm³

mass= density x volume = 0.8 g/cm³ x 225 cm³ = 180 g

b) Given: mass= 75.0 g

volume = mass/density = 75.0 g/(0.8 g/cm³)= 93.75 cm³≅ 93.7 cm³

5 0
1 year ago
Which equation illustrates conservation of mass?
pochemuha
Answer = c


Conservation of mass (mass is never lost or gained in chemical reactions), during chemical reaction no particles are created or destroyed, the atoms are rearranged from the reactants to the products.
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1 year ago
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The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all
Brilliant_brown [7]

<u>Full Question:</u>

The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all of the boxes that apply.

denser than water

burns readily in air  

boiling point of –1.1°C  

odorless

does not react with water

burns readily in air

does not react with water

<h3><u>Explanation:</u></h3>

The type of alkane that is used in many products includes Butane. It is found as a natural gas in the environment. It is found on the deeper part of ground. It can be obtained by drilling process and gets used up in many of the products that is used for commercial purposes.

Molecular mass that is associated with butane is 58.12 g/mol. The boiling point of butane is -1 degree Celsius and -140 degree Celsius is its melting point. It is a liquefied gas and does not react with water. It will burn in air more readily.

6 0
2 years ago
Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f
Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0
2 years ago
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