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3 years ago

What is produced at each electrode in the electrolysis of an aqueous solution of both nabr and agf?

Physics

1 answer:

Vitek1552 [10]3 years ago

7 0

Answer:

Explanation:

reduction potential of Ag+/Ag = 0.80 V

the oxidation potential of Br-/Br2 = -1.066 V

reduction potential of Na+/Na = 2.71 V

reduction potential of 2H+/H2 = 0.00 V

oxidation potential of 2F-/F2 = -2.87 V

oxidation potential of 2H2O↔ O2 + 4H+ + 4e- = -1.23 V

So, at cathode reduction occurs so, here Ag(s) will form because it has high reduction potential.

However, at anode oxidation occurs and , here the substance with high oxidation potential can oxidises easily so bromine will form.

You might be interested in

Why are isotopes not shown on the periodic table

Fofino [41]

Hey there!

The periodic table is arranged by number of protons/electrons which is the same for all isotopes of an element. A different isotope will have a different number of neutrons only, which does not qualify it for a separate space on the periodic table.

I hope this helps!

3 0

3 years ago

A 3.90 kg block is in equilibrium on an incline of 31.0◦. The acceleration of gravity is 9.81 m/s2 . What is Fn of the incline o

storchak [24]

Answer:

Explanation:

The sum of the pore along the plane is expressed according to Newton's law

Fn-Ff = ma

Fn is the moving force

Ff = nR = frictional force

m is the Mass

a is the acceleration

Substitute the given values

Fn - nR = ma

Fn - tan31°(mgcostheta) =3.9(9.8)

Fn - tan31(3.9(9.8)cos31) = 3.9(9.8)

Fn - tan31(38.22cos31)= 38.22

Fn - 32.76tan31 = 38.22

Fn-19.68 = 38.22

Fn = 38.22+19.68.

Fn = 57.90N

Hence Fn (moving force) of the inclined block is 57.90

4 0

3 years ago

A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre

pantera1 [17]

Answer:

Part a)

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

$T = \frac{m}{L}xg + Mg$

so the speed of the wave at that position is given as

$v = \sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{m}{L}$

now we have

$v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}$

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

time taken by the wave to reach the top is given as

$t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}$

$t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})$

$t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})$

$t = 12 s$

4 0

3 years ago

If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of

kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass m₁ and m₂ a distance r apart is

F = G m₁ m₂ / r²

where G = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

F = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

F ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)g, where g = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

F = G m₁ m₂ / (2r)² = 1/4 G m₁ m₂ / r²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0

3 years ago

What would happen to a loop in a metal tube when it is heated

Galina-37 [17]

I think it most likely burn since th metal tube is going to transfer so much heat

8 0

3 years ago