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2 years ago

Iron(II) carbonate (FeCO3) has a solubility product constant of 3.13 x 10-11 . Calculate the molar solubility of FeCO3 in water

1 answer:

fomenos2 years ago

7 0

The molar solubility of a substance is the number of moles that dissolve per liter of solution. The dissociation equation FeCO3 in water would be as follows:

FeCO3 → Fe2+ + CO3 2-

We calculate as follows:

Ksp = [Fe2+][CO3 2-]

Let X = [Fe2+] - then X = [CO3 2-]

Ksp = X² 
(3.13*10^-11) = X² 
X = √( 3.13*10^-11) 
X = 5.59*10^-6 M

Hope this answers the question. Have a nice day.

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Dane is standing on the moon holding an 8 kilogram brick 2 metres above the ground. How much energy is in the brick's gravitatio

Nadya [2.5K]

The gravitational potential energy of the brick is 25.6 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.

Near the surface of a planet, the gravitational potential energy is given by

$PE=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the brick in this problem, we have:

m = 8 kg is its mass

g = 1.6 N/kg is the strenght of the gravitational field on the moon

h = 2 m is the height above the ground

Substituting, we find:

$PE=(8)(1.6)(2)=25.6 J$

Learn more about potential energy:

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3 0

2 years ago

Read 2 more answers

Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. Wha

MArishka [77]

Answer:

$P = 1.64 \times 10^4 Watt$

Explanation:

Here we know that the glider is accelerated uniformly from rest to final speed of 25.7 m/s in total distance of d = 46.9 m

so we will have

$v_f = 25.7 m/s$

$v_i = 0$

d = 46.9

so for uniformly accelerated motion we have

$d = \frac{v_f + v_i}{2} t$

$46.9 = \frac{25.7 + 0}{2}t$

$t = 3.65 s$

now we will find the total work done given as change in kinetic energy

$W = \frac{1}{2}mv^2$

$W = \frac{1}{2}(181)(25.7^2)$

$W = 5.97 \times 10^4 J$

now power is given as

$P = \frac{W}{t}$

$P = \frac{5.97 \times 10^4}{3.65}$

$P = 1.64 \times 10^4 Watt$

6 0

2 years ago

A woman is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follow

Ymorist [56]

Answer:

Explanation:

Given

Force P is acting upward

C is vertical contact Force

W is the weight of the crate

As P is unable to move the Block therefore Normal reaction keeps on acting on block

thus we can say that

P-W+C=0

P=W-C

7 0

2 years ago

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a 0.0215m diameter coin rolls up a 20 degree inclined plane. the coin starts with an initial angular speed of 55.2rad/s and roll

marissa [1.9K]

Answer:

$h = 0.0362\,m$

Explanation:

Given the absence of non-conservative force, the motion of the coin is modelled after the Principle of Energy Conservation solely.

$U_{g,A} + K_{A} = U_{g,B} + K_{B}$

$U_{g,B} - U_{g,A} = K_{A} - K_{B}$

$m\cdot g \cdot h = \frac{1}{2}\cdot I \cdot \omega_{o}^{2}$

The moment of inertia of the coin is:

$I = \frac{1}{2}\cdot m \cdot r^{2}$

After some algebraic handling, an expression for the maximum vertical height is derived:

$m\cdot g \cdot h = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega_{o}^{2}$

$h = \frac{r^{2}\cdot \omega_{o}^{2}}{g}$

$h = \frac{(0.0108\,m)^{2}\cdot (55.2\,\frac{rad}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }$

$h = 0.0362\,m$

3 0

2 years ago

An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

4 0

2 years ago