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Leno4ka [110]
3 years ago
9

How long does it take electrons to get from the car battery to the starting motor? Assume the current is 137 A and the electrons

travel through copper wire with cross sectional area 44.6 mm2 and length 75.7 cm . The mass density of copper is 8960 kg/m3 and the molar mass is 63.5 g/mol . Avogadro’s number is 6.022 × 1023 . Assume that each copper atom contributes one conduction electron. Answer in units of min.
Chemistry
1 answer:
WARRIOR [948]3 years ago
4 0

Answer:

t = 55.79 min

Explanation:

First, the problem is asking for calculate the time that it takes electrons from the battery to the motor.

The general formula to calculate time is:

<em>t = d/V (1)</em>

Where:

d: distance or length

V: speed

Now, we don't have data of speed, but we can know an expression of current density in function of the distance which is the following:

<em>J = n*q*V (2)</em>

Where:

q: charge of the particle (1.6x10^-19 C)

n: number of charge carriers per unit of volume

Current density (J) is actually current per Area so:

<em>J = I/A (3)</em>

Replacing (3) in (2) we have:

I/A = nqV

Solving for V:

<em>V = I/Anq (4)</em>

Finally, if we replace this expression in (1) we have:

<em>t = nqAd / I (5)</em>

Now, the value of n, it's not given but it can be calculated because we have mass density, molar mass and avogadro's number, so this value of "n" can be calculated using the following expression:

<em>n = D * Av / MM (6)</em>

Where:

D: mass density (kg/m³)

Av: avogadro number (6.02x10^23 atom/mol)

MM: molar mass (kg/mol)

Putting the data that we know to calculate n we have:

n = 8960 * 6.02x10^23 / 0.0635

n = 8.49x10^28 atom/m³

Now with the value of n, we can finally calculate the time:

<em>t = nqAd / I </em>

A is the area and it should be in m²: 44.6 mm² / 1x10^6 m = 4.46x10^-5 m²

d is the length in meter: 75.7 cm / 100 cm/m = 0.757 m

so replacing these data in (5):

t = 8.49x10^28 * 1.6x10^-19 * 4.46x10^-5 * 0.757 / 137

t = 3,347.63 s

But the answer is in minute so:

t = 3,347.63 / 60

<em>t = 55.79 min</em>

so the electrons takes 56 min aprox. to go from the car battery to the starting motor.

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aliya0001 [1]

Answer:

a) 56 protons

b) 54 electrons

c) 81 neutrons

d) The sum of protons and neutrons is shown. The number of protons is always the same. So we can calculate the number of neutrons ( and also the isotopes)

e)137Ba (with 56 protons and 81 neutrons)

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Explanation:

Step 1: Data given

137 Ba2+ is an isotope of barium. The atomic number of barium( and its isotopes) is 56. This shows the number of protons.

For a neutral atom, the number of protons is equal to the number of electrons.

The different isotopes of an element have the same number of protons but a different number of neutrons.

137Ba2+ has 56 protons (this is the same as the atomic number)

137Ba2+ has 54 electrons ( since it's Ba2+, this means it has 2 electrons less than protons, that's why it's charged +2)

137Ba2+ has 81 neutrons ( 137 - 56 = 81)

In the symbol, the atomic number is not shown. The sum of the protons and neutrons is shown. (Since the number of protons is the same for every isotope, we can calculate the number of neutrons that way. By knowing the neutrons, we also know the isotope.

This isotope is 137Ba

Atomic mass is also known as atomic weight. The atomic mass is the weighted average mass of an atom of an element based on the relative natural abundance of that element's isotopes.

The atomic mass of 137Ba2+ is 136.9 u

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The mass number of 137Ba2+ is 137

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Which of the following is true of a heterogeneous mixture?
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A, a heterogeneous mixture is a mixture where all the particles are not made up of one uniform composition. you can distinguish the different particles
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Given:
<span> 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)
Required:
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A scientist measures the standard enthalpy change for the following reaction to be 67.9 kJ:
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Answer: The standard enthalpy of formation  of H_2O(g) is  -252.1 kJ/mol.

Explanation:

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The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}]

Putting the values we get :

\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]

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\bold{\huge{\orange{\underline{ Solution}}}}

\bold{\underline{ Given :- }}

  • <u>We </u><u>have </u><u>250g </u><u>of </u><u>liquid </u><u>water </u><u>and </u><u>it </u><u>needs </u><u>to </u><u>be </u><u>cool </u><u>at </u><u>temperature </u><u>from </u><u>1</u><u>0</u><u>0</u><u>°</u><u> </u><u>C </u><u>to </u><u>0</u><u>°</u><u> </u><u>C</u>
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\bold{\underline{ To \: Find :- }}

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the</u><u> </u><u>total</u><u> </u><u>number </u><u>of </u><u>joules </u><u>released</u><u>. </u>

\bold{\underline{ Let's \:Begin:- }}

<u>We </u><u>know </u><u>that</u><u>, </u>

Amount of heat energy = mass * specific heat * change in temperature

<u>That </u><u>is, </u>

\sf{\red{ Q = mcΔT }}

<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u><u>:</u><u>-</u>

\sf{ Q = 250 × 4.180 ×(0 - 100 )}

\sf{ Q = 250 × 4.180 × - 100 }

\sf{ Q = 250 × - 418}

\sf{\pink{ Q = - 104,500 J }}

Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

\bold{\underline{ Now :- }}

<u>We </u><u>have </u><u>to </u><u>tell </u><u>whether </u><u>the </u><u>above </u><u>process </u><u>is </u><u>endothermic </u><u>or </u><u>exothermic </u><u>:</u><u>-</u>

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