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garri49 [273]
2 years ago
13

Pennies today are copper-covered zinc, but older pennies are 3.1 g of solid copper. What are the total positive charge and total

negative charge in a solid copper penny that is electrically neutral?
Physics
1 answer:
k0ka [10]2 years ago
7 0

Answer:

Q_+=1.36*10^5C

Q_+=-1.36*10^5C

Explanation:

From the question we are told that

Mass of old penny M=3.1g

Generally the number of moles in a penny is given mathematically as

n=\frac{M}{A}

A=atomic\ number\ of\ copper=>63.5g/mol

Therefore

n=\frac{3.1}{63.5}

n=0.049mol

Avogadro\ Number=(6.023*10^2^3 atom/mol

Therefore

N=(6.023*10^2^3 atom/mol *0.049mol

N=2.939*10^2^2atoms

Generally the Total positive charge of the copper is given by

Since its the 29th atom of the periodic table

29 protons

29 electrons

Q_+=29(2.939*10^22)(1.6*10^1^9)

Q_+=1.36*10^5C

Generally the Total negative charge of the copper is given by

Q_+=29(2.939*10^22)(-1.6*10^1^9)

Q_+=-1.36*10^5C

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Answer

given,

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change in activation energy = 109 kJ/mole

when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.

where as activation energy of the product and the reactant decreases.

example:

ΔH = 51 kJ/mole

E_a= 83 kJ/mole

here activation energy decrease whereas change in enthalpy remains same.

5 0
3 years ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to
katovenus [111]
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If the temperature drops that means that the molecules are coming together. If the temperature rises then it means that the molecules are spreading. If the kinetic energy falls down that means that they are slower which means that they are cooler.
4 0
3 years ago
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Ignoring air resistance, if a 20 kg ball and a 400 kg crate were both dropped from the
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8 0
3 years ago
A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has
lara [203]

Answer:

The phase constant is 7.25 degree  

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = \sqrt{(\frac{v_x}{\omega })^2+y^2 }  = \sqrt{(\frac{35}{2\times \pi  \times y})^2+6.2^2 }  =  6.25 cm

put value in equation 1

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cosФ  = 0.992

Ф = 7.25 degree  

so the phase constant is 7.25 degree  

5 0
3 years ago
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