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marissa [1.9K]
3 years ago
8

What is Keq for an electrochemical reaction involving the transfer of 3 electrons, if E°cell = 0.652 V at

Physics
1 answer:
madam [21]3 years ago
6 0

Answer:

2.49 V.

Explanation:

The formula which relates the equilibrium condition in the electrochemical cell is,

logk_{eq}=\frac{n}{0.059}  E_{cell} ^{0}

Given that the number of electron is 3 and the value of E^{0} _{cell} is  0.652 V.

Now put these values in above equation.

logk_{eq}=\frac{3}{0.059}  (0.652 V)\\logk_{eq}=33.15 V\\k_{eq}=e^{33.15}V\\k_{eq}=2.49V

Therefore, the value of Keq for an electrochemical reaction is 2.49 V.

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Two 20kg spheres are placed with their
Maslowich

Answer:

F = 1.07 x 10⁻⁷ N

Explanation:

The gravitational force of attraction between two objects can be found by the use of Newton's Gravitational Law:

F = \frac{Gm_{1}m_{2}}{r^2}\\\\

where,

F = Gravitational Force of attraction = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = m₂ = mass of spheres = 20 kg

r = distance between the objects = 50 cm = 0.5 m

Therefore,

F = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(20\ kg)(20\ kg)}{(0.5\ m)^2}\\\\

<u>F = 1.07 x 10⁻⁷ N</u>

5 0
3 years ago
14 km equals how much meters
MrRa [10]
14 km equals 14,000 metres
6 0
3 years ago
Read 2 more answers
You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

3 0
4 years ago
A strong-armed physics student throws a tennis ball vertically. The ball stays in the air for 5.5 seconds. Assuming the ball lef
Mila [183]

Answer:

27.5 m/s

Explanation:

applying motion equations we can find the answer,

v = u + a*t

Let assume ,

u = starting speed(velocity)

v = Final speed (velocity)

t =  time taken for the motion

a = acceleration

by the time of reaching the highest point subjected to the gravity , the speed should be equal to zero  (only a vertical speed component is there)

for the complete motion it takes 5.5 s. that means to reach the highest point it will take 5.5/2 =2.75 seconds

we consider the motion upwards , in this case the gravitational  acceleration should be negative in upwards (assume g=10 m/s2)

that is,

v = 0 , a = -10ms^{-2}    , t =2.75

v = u + at

0 = u -10*2.75

u = 27.5 ms^{-1}

4 0
4 years ago
3. A rocket is launched at an angle of 53 degrees above the 1 point
irina1246 [14]

Answer:

24,000 m

Explanation:

First find the rocket's final position and velocity during the first phase in the y direction.

Given:

v₀ = 75 sin 53° m/s

t = 25 s

a = 25 sin 53° m/s²

Find: Δy and v

Δy = v₀ t + ½ at²

Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²

Δy = 7736.8 m

v = at + v₀

v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)

v = 559.0 m/s

Next, find the final position of the rocket during the second phase (as a projectile).

Given:

v₀ = 559.0 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 15945.5 m

The total displacement is:

7736.8 m + 15945.5 m

23682.2 m

Rounded to two significant figures, the maximum altitude reached is 24,000 m.

3 0
3 years ago
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