PHYSICS COLLEGE

If 3.00 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.59 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.

Answers

Answer 1

Answer:

0.158 A.

Explanation:

Mass of gold deposited = 3 x 10^-3 kg

= 3 g

Molar mass = 196 g/mol

Number of moles = 3/196

= 0.0153 mol.

Faraday's constant,

1 coloumb = 96500 C/mol

Quantity of charge, Q = 96500 * 0.0153

= 1477.04 C.

Remember,

Q = I * t

t = 2.59 hr

= 2.59 * 3600 s

= 9324 s

Current, I = 1477.04/9324

= 0.158 A.

Answer 2

Answer:

0.158A

Explanation:

Using Faraday's first law of electrolysis which states that the mass(m) of a substance deposited or liberated at any electrode is directly proportional to the quantity of charge or electricity (Q) passed. i.e

m ∝ Q

m = Z Q

Where;

Z is the proportionality constant called electrochemical equivalent.

Faraday also observed that when 1 Faraday of electricity is equivalent to 96500C of charge.

Also,

Quantity of charge (Q), which is the product of current (I) passing through and the time taken (t) for the electrolysis, is given by;

Q = I x t; ----------------------(i)

With all of these in place, now let's go answer the question.

Since the gold ions carry one elementary unit of positive charge, now let's write the cathode-half reaction for gold (Au) as follows;

Au⁺ + e⁻ = Au ---------------------(ii)

From equation (ii) it can be deduced that when;

1 Faraday (96500C) of electricity is passed, 1 mole of Au forms ( = 197 grams of Au) [molar mass of Au = 197g]

Then, 3.00 x 10⁻³ kg (= 3 g of Au) will be formed by 3g x 96500C / 197g = 1469.5C

Therefore, the quantity of charge (Q) deposited is 1469.5C

Substitute this value (Q = 1469.5C) and time t = 2.59h (= 2.59 x 3600 s) into equation (i);

Q = I x t

1469.5 = I x 2.59 x 3600

1469.5 = I x 9324

Solve for I;

I = 1469.5 / 9324

I = 0.158A

Therefore, the current in the cell during that period is 0.158A

Note:

1 mole of gold atoms = 176g

i.e the molar mass of gold (Au) is 176g

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Answers

Answer:

8m/s

Explanation:

Vavg= 16-0/2=8m/s

Which statement would most likely be found in an advertisement from a
cell phone provider

Answers

Answer:B

Explanation:

The atmosphere on Venus consists mostly of CO2. The density of the atmosphere is 65.0 kg/m3 and the bulk modulus is 1.09 x 107 N/m2. A pipe on a lander is 75.0 cm long and closed at one end. When the wind blows across the open end, standing waves are caused in the pipe (like blowing across the top of a bottle). a) What is the speed of sound on Venus? b) What are the first three frequencies of standing waves in the pipe?

Answers

Answer:

a. 409.5 m/s b. f₁ = 136.5 Hz, f₂ = 409.5 Hz, f₃ = 682.5 Hz

Explanation:

a. The speed of sound v in a gas is v = √(B/ρ) where B = bulk modulus and ρ = density. Given that on Venus, B = 1.09 × 10⁷ N/m² and ρ = 65.0 kg/m³

So, v = √(B/ρ)

= √(1.09 × 10⁷ N/m²/65.0 kg/m³)

= √(0.01677 × 10⁷ Nm/kg)

= √(0.1677 × 10⁶ Nm/kg)

= 0.4095 × 10³ m/s

= 409.5 m/s

b. For a pipe open at one end, the frequency f = nv/4L where n = mode of wave = 1,3,5,.., v = speed of wave = 409.5 m/s and L = length of pipe = 75.0 cm = 0.75 m

Now, for the first mode or frequency, n = 1

f₁ = v/4L

= 409.5 m/s ÷ (4 × 0.75 m)

= 409,5 m/s ÷ 3 m

= 136.5 Hz

Now, for the second mode or frequency, n = 2

f₂ = 3v/4L

= 3 ×409.5 m/s ÷ (4 × 0.75 m)

= 3 × 409,5 m/s ÷ 3 m

= 3 × 136.5 Hz

= 409.5 Hz

Now, for the third mode or frequency, n = 5

f₃ = 5v/4L

= 5 × 409.5 m/s ÷ (4 × 0.75 m)

= 5 × 409,5 m/s ÷ 3 m

= 682.5 Hz

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1. If a net force of 412 N is required to accelerate an object at 5.82 m/s2, what must theobject's mass be?

Answers

Answer:

The mass of the object is approximately 70.79 kilograms

Explanation:

We use Newton's second law to solve this problem. This law states that the net force on an object equals the product of its mass times the acceleration:

$F_(net)=m\,a$

Therefore, for this case, since the net force on the object and its acceleration are given, we can use the equation above to solve for the unknown mass:

$F_(net)=m\,a\n412\,N=m\,(5.82\,(m)/(s^2) )\nm=(412\,N)/(5.82\,(m)/(s^2) ) \nm=70.79\.kg$

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Answers

Answer:

Trevor's taxi would cause higher levels of air pollution

Explanation:

Trevor's taxi use diesel oil.

Diesel is less cleaner than LPG.

Compared to automotive pollution from petrol and diesel, pollutants from LPG-driven cars include lower amounts of petroleum hydrocarbons, nitrogen oxides , sulphur oxides, ozone contamination and particulate matter.

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