What's the job of a resistor in an electric circuit?

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial

elena55 [62]

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution : Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}$

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm

6 0

3 years ago

Calculate the wavelength λ1 for gamma rays of frequency f1 = 7.20×1021 hz .

bagirrra123 [75]

The relationship between wavelength $\lambda$ , frequency f and speed of light c for an electromagnetic wave is
$\lambda= \frac{c}{f}$
Using the data of the problem, we find
$\lambda= \frac{3\cdot 10^8 m/s }{7.20 \cdot 10^{21} Hz}=4.17 \cdot 10^{-14} m$

5 0

3 years ago

You are launching a 2 kg potato out of a potato cannon. The cannon is 2.0 m long and is aimed 70 degrees above the horizontal. I

DochEvi [55]

Answer:

Explanation:

The net force on the potatoes is given by:

F= 52 - mgSintheta

F= 52- (2×9.8× Sin70°)

F = 52 -18.4

F= 33.58N

Using Newton's 2nd law

F = ma

a=F/m = 33.58/ 2 = 16.79m/s^2

Using the equation of motion:

V^2= u^2 + 2as

V^2 = 0 + 2× 16.79 x2

V^2 = 67.16

V=sqrt(68.16)

V= 8.195m/s This is the exit velocity of the potatoes

Kinetic energy, K.E = 1/2mv^2

KE= 1/2 × 2 × 8.195^2

KE = 67.16J

8 0

3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

3 years ago

What causes breakers?

nata0808 [166]

Answer:

nreaker

Explanation:

A switch that automatically interrupts or shuts off an electric current at the first indication of a overload

3 0

3 years ago