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lesya692 [45]
3 years ago
11

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

Physics
1 answer:
fomenos3 years ago
7 0
<span>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:

</span><span>FeCO3 → Fe2+ + CO3 2-
</span>
We calculate as follows:

Ksp = [Fe2+][CO3 2-] 

<span>Let X = [Fe2+] - then X = [CO3 2-] 
</span>
<span>Ksp = X² </span>
<span>(3.13*10^-11) = X² </span>
<span>X = √( 3.13*10^-11) </span>
<span>X = 5.59*10^-6 M
</span>
Hope this answers the question. Have a nice day.
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A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be  a = 0.003* 6 =0.018\  m/s^2    

Explanation:

The objective of this solution is to obtain the acceleration of the particle

       Now looking at Newton law which is mathematically represented as

                       F = ma

  Where F is the force experience by a particle

              m is the mass of the particle

              a is the acceleration of the particle

  And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

                    F = qvB

Where q is the charge of the particle

             v is the velocity of the  charge

             B is the magnetic field the charge is under it influence

  Now equating this two formulas

                   ma = qvB

 Making a the subject we have

                  a = \frac{qvB}{m}

In the question the direction of the is in the positive x-axis which is i hence the direction would be in the i direction

      So substituting  (2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T for B

                    a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}

      Substituting       3.0 Km /s = 3.0*10^{3}\ m/s  for v  and -6.0 \muC = -6.0*10^{-6} C for q

                     a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}

                       a = 0.003 * 3i(2i+3j+4k)

                      a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j  \ + (3*4)i \ \cdot \ k)

According to vector multiplication

                                             i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k  = 0

     So

               a = 0.003* 6 =0.018\  m/s^2          

     

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Answer:

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Explanation:

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Answer:

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Explanation:

hope this helps :)

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