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MaRussiya [10]
3 years ago
12

If an electric current of 8.50 A flows for 3.75 hours through an electrolytic cell containing copper-sulfate (CuSO4) solution, t

hen how much copper is deposited on the cathode (the negative electrode) of the cell? (Copper ions carry two units of positive elementary charge, and the atomic mass of copper is 63.5 g/mol.)
Physics
2 answers:
muminat3 years ago
7 0

Answer:

75.5g

Explanation:

From the ionic equation, we can write

CU^{2+}+SO^{2-}_{4}\\

next we find the number of charge

Note Q=it

for i=8.5A, t=3.75 to secs 3.75*60*60=13500secs

hence

Q=8.5*13500\\Q=114750C

Since one faraday represent one mole of electron which equal 96500C

Hence the number of mole produced by 114750C is

114750/96500=1.2mol

The mass of copper produced is

mol=\frac{mass}{molar mass} \\mass=mole*molar mass\\mass=1.2*63.5\\mass=75.5g

Hence the amount of copper produced is 75.5g

Ann [662]3 years ago
5 0

Answer:

75.5 g.

Explanation:

Dissociation equation:

Cu2+ + SO4^2- --> CuSO4

Q = I * t

Where,

I = current

= 8.5 A

t = time

= 3.75 hours

= 13500 s

Q = 8.5 * 13500

= 114750 C

1 faraday represent one mole of electron which equal 96500C

Number of mole of Cu2+

= 114750/96500

= 1.19 mol.

Mass = number of moles * molar mass

= 1.19 * 63.5

= 75.5 g.

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5 0
3 years ago
A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the
Elena L [17]

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

8 0
3 years ago
2. Explain whether the size of an object's displacement could be greater than the distance the
vampirchik [111]

Answer:learn vocabulary, terms,and more with flashcards, games,and other study tools.size of an objects displacement could be greater than the distance the object travels.

Explanation:

4 0
3 years ago
A car accelerates at a rate of 3.0 m/s2. If its original speed is 8.0 m/s, how many seconds will it take the car to reach a fina
V125BC [204]

Answer:

5.7 s

Explanation:

Given:

a = 3.0 m/s²

v₀ = 8.0 m/s

v = 25.0 m/s

Find: t

v = at + v₀

25.0 m/s = (3.0 m/s²) t + 8.0 m/s

t ≈ 5.7 s

5 0
3 years ago
someone help me please haha Calculate the braking force of a lorry decelerating at a rate of 8 m/s² with a mass of 5000 kg.
ICE Princess25 [194]

Answer:

40000 N

Explanation:

Force: This cam be defined as the product of mass and acceleration.

From the question,

The braking force of the lorry will be the same in magnitude as the force at which the lorry is moving.

F = ma..................... Equation 1

Where m = mass of the lorry, a = acceleration/deceleration of the lorry.

Given: m = 5000 kg, a = 8 m/s²

Substitute these values into equation 1

F = 5000(8)

F = 40000 N

6 0
3 years ago
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