If a machine has an efficiency of 50% and an input of 3 J, what is the output?

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

madreJ [45]

Complete Question

A voltaic cell utilizes the following reaction and operates at 298 K:

3Ce4+(aq)+Cr(s)→3Ce3+(aq)+Cr3+(aq).

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

Answer:

The value is $E^o_{cell} = 2.35 V$

Explanation:

From the question we are told that

The ionic equation is

$3 Ce^{4 +} _{(aq)} + Cr _{(s)} \to 3 Ce^{3+} _{(aq)} + Cr^{3r} _{(aq)}$

Now under standard conditions the reduction half reaction is

$Ce^{4+} + e \to Ce^{3+} ; \ \ E^o_r = 1.61 V$

And the oxidation half reaction is

$Cr^{3+} + 3e^{-} \to Cr ; \ \ \ E^o_o = - 0.74 V$

The emf of this cell under standard conditions is mathematically represented as

$E^o_{cell} = E^o _r - E^o _o$

substituting values

$E^o_{cell} = 1.61 - (- 0.74)$

$E^o_{cell} = 2.35 V$

3 0

3 years ago

On a visit to a science lab, Madison observes a blob of shiny material that is floating in the air.

AfilCa [17]

Answer:

Each force acting on the blob has another one to cancel it out

Explanation:

7 0

3 years ago

Read 2 more answers

A light spring of constant 179 N/m rests vertically on the bottom of a large beaker of water. A 5.32 kg block of wood of density

Digiron [165]

Answer:

Compression of the spring: 0.18 m (downward)

Explanation:

The forces acting on the block of wood are:

- The force of gravity, acting downward, of magnitude $mg$ , where m = 5.32 kg is the mass of the block and $g=9.8 m/s^2$ is the acceleration due to gravity

- The force exerted by the spring, downward, of magnitude $kx$ , where $k=179N/m$ is the spring constant and $x$ is the elongation of the spring

- The buoyant force, upward, of magnitude $\rho V g$ , where $\rho=1000 kg/m^3$ is the water density and V the volume of the block

Since the block is in equilibrium, the net force is zero, so we can write

$mg+kx-\rho V g=0$ (1)

We have to find the volume of the block first. We have:

m = 5.32 kg (mass)

$\rho_w = 622 kg/m^3$ (wood density)

So, the volume is

$V=\frac{m}{\rho_w}=\frac{5.32}{622}=0.0086 m^3$

So now we can re-arrange eq.(1) to find the elongation of the spring, x:

$x=\frac{-mg+\rho Vg}{k}=\frac{-(5.32)(9.8)+(1000)(0.0086)(9.8)}{179}=0.18 m$

So, the spring is compressed by 0.18 m.

7 0

3 years ago

A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the t

Alexandra [31]

Answer:

the coin does not slide off

Explanation:

mass (m) = 5 g = 0.005 kg

distance (r) = 15 cm = 0.15 m

static coefficient of friction (μs) = 0.8

kinetic coefficient of friction (μk) = 0.5

speed (f) = 60 rpm

acceleration due to gravity (g) = 9.8 m/s^{2}

lets first find the angular speed of the table

ω = 2πf

ω = 2 x π x 60 x $\frac{1}{60}$

ω = 6.3 s^{-1]

Now lets find the maximum static force between the coin and the table so we can get the maximum velocity the coin can handle without sliding

static force (Fs) = ma

static force (Fs) = μs x Fn = μs x m x g

Fs = 0.8 x 0.005 x 9.8 = 0.0392 N

Fs = ma

0.0392 = 0.005 x a

a = 7.84 m/s^{2}

$(Vmax)^{2}$ = a x r

$(Vmax)^{2}$ = 7.84 x 0.15

Vmax = 1.08 m/s

ωmax = $\frac{Vmax}{r}$

ωmax = $\frac{1.08}{0.15}$ = 7.2 s^{-1}

now that we have the maximum angular acceleration of the table, we can calculate its maximum speed in rpm

Fmax = $\frac{ωmax}{2π}$

Fmax = $\frac{7.2}{2 x π}$ = 68.7 rpm

since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.

4 0

4 years ago

You are an employee of the city of Chicago. A city official declares that all cables holding traffic lights in the city should b

Irina18 [472]

If you have ever lived in Chicago, then the answer might very well be C.
Newton's Second law states F = ma. This problem has nothing to do with acceleration.
Newton's Third law is the familiar action/reaction law. If the poles are anchored well enough and have a flexural strength greater than the tension exerted on them by the wire; then D is just.
That leaves B. There's no problem all we have to do is to increase the horizontal tension in the cable.

5 0

3 years ago