# A 500ml saturated solution of MgCO3 (M=84) is reduced to 120 ml by evaporation. What mass of solid MgCO3 is formed?

The answer is 0.20g, I just need a detailed explanation.
Thanks!

Question ID
524132

Created
April 2, 2011 2:33am UTC

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0

6

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2582

1. Do you have a solubility or a Ksp to use. There is a good bit of disagreement on the net about the correct value for Ksp for MgCO3.

524154

Created
April 2, 2011 3:47am UTC

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2. Oh sorry the Ksp=4x10^(-5)

524251

Created
April 2, 2011 5:09pm UTC

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3. OK. Thanks.
MgCO3 ==> Mg^+2 + CO3^-2
..x........x.......x
Ksp = (Mg^+2)(CO3^-2) - (x)(x)
x = sqrt(4E-5) = 0.00632 moles/L or 0.00632 x 84 = 0.531 g/L
In 500 mL there will be 1/2 that or .265 grams.

If we evaporate water until the volume is 120 mL, then 0.531 g/L becomes
0.531 g/L x (120 mL/1000 mL) = 0.0637 g.
We started with 0.265 g in 500
............less 0.0637 in 120 mL =
.................-------
solid ppting.....0.201 g which rounds to 0.2 g to 1 significant figures (from the Ksp value).

524287

Created
April 2, 2011 6:23pm UTC

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2

4. Thanks! But im confused on why you would subtract the .0637g in 120mL from .265g in 500mL. Why don't you do (grams from 500mL-grams from 380mL)?

524361

Created
April 2, 2011 10:14pm UTC

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0

5. NO Soup For You!
(10 points possible)

Calculate the number of moles of nitrogen dioxide, NO2, that could be prepared from 79.0 mol of nitrogen oxide and 82.0 mol of oxygen.
2NO(g)+O2(g)⟶ 2NO2(g)

79 - correct

Identify the limiting reagent and the excess reagent in the reaction. What would happen to the potential yield of NO2 if the amount of NO were increased?

increase - correct

What if the amount of O2 were increased?

no change - correct

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Photon!
(10/10 points)

Calculate the wavelength (in m) of a photon capable of exciting an electron in Li2+ from the state n= 3 to the state n=5

1.4239*10^-7 - correct

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Charge Me Up, Scotty!
(10/10 points)

Determine the potential (V) by which a proton must be accelerated so as to assume a particle wavelength of 0.0293 nm.

0.9544 - correct

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PES S+P 500
(10/10 points)

Shown below are the photo electron spectra for phosphorous and sulfur

The PES spectra for phosphorous and sulfur are shown above. Answer the following questions concerning these spectra.

Why is the phosphorous peak at 1.06 MJ/mole greater in energy than the sulfur peak at 1.00 MJ/mole?

paired electron easier to remove - correct

Why is the phosphorous peak at 1.95 less energy than the sulfur peak at 2.05?

sulfur has larger Z - correct

What should be the ratio of intensities of the phosphorous peak at 1.06 to the phosphorous peak at 13.5?

1/2 - correct

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/* 1-5 */

You'll Learn About Electrons, I Guarantee It
(10/10 points)

Which of the following orbital diagrams is/are incorrect for all electrons in the lowest-energy levels of an atom?

1,2,4 - correct

Which of the following is the correct electronic configuration for the bromide ion, Br-?
[Ar]4s23d104p5
[Ar]4s24p5
[Ar]4s23d104p6 - correct
[Ar]4s23d104p65s1

Which of the following orders of filling orbitals is/are incorrect? They should be in the exact order that they would be filled in.

3s, 4s, 5s - correct
5s, 5p, 5d - correct
5s, 4d, 5p
6s, 4f, 5d
6s, 5f, 6p - correct

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CaF2, Great for Teeth!
(10/10 points)

Calculate the lattice energy of CaF2 given the information below (in kJ/mol):
Ca(g)⟶Ca(s) ΔH=−178kJ/mol
Ca(g)⟶Ca+(g)+e− ΔH=589.8kJ/mol
F(g)+e−⟶F−(g) ΔH=−328.2kJ/mol
Ca2++e−⟶Ca+(g) ΔH=−1145.4kJ/mol
F2(g)⟶2F(g) ΔH=139.0kJ/mol
Ca(s)+F2(g)⟶CaF2(s) ΔH=−1219.6kJ/mol

- 2615.4 - correct

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Select the single most important reason below to justify your choice should have the higher melting point

- Both molecules are nonpolar but the higher melting includes hydrogen bonding
- Both molecules are nonpolar but the higher melting is more polarizable
- The higher melting molecule is polar and the other is not
- The lower melting molecule is polar and the other is not
- The higher melting molecule has smaller Van der Waals forces than the lower melting
- The higher melting molecule has greater Van der Waals forces than the lower melting
- The higher melting molecule is ionic and the other is not
- The lower melting molecule is ionic and the other is not
- The higher melting molecule has hydrogen bonds and the other does not
- The lower melting molecule has hydrogen bonds and the other does not
- Both molecules are ionic but the higher melting has a larger cohesive energy
- Both molecules are ionic but the lower melting has a larger cohesive energy
- The higher melting molecule is ionic and the other is not.
- Ionic bonds are stronger than hydrogen bonds
- The lower melting molecule is ionic and the other is not.
- onic bonds are weaker than hydrogen bonds
- The higher boiling molecule is ionic and the other is not.
- Ionic bonds are stronger than Van der Waals forces
- The lower boiling molecule is ionic and the other is not.
- Ionic bonds are weaker than Van der Waals forces

1087723

Created
August 26, 2014 9:19pm UTC

Rating
-1

6. Well, let's break this down step by step.

First, we need to find out the initial mass of MgCO3 present in the 500 ml saturated solution. To do this, we can use the formula:

Mass = (Concentration * Volume) / Molar Mass

The concentration of a saturated solution is the maximum amount of solute that can dissolve in a given amount of solvent at a given temperature. In this case, the concentration of MgCO3 is at its maximum, so we can use its solubility. The solubility of MgCO3 is 0.067 g/ml.

So, the initial mass of MgCO3 in the 500 ml solution is:

Mass = (0.067 g/ml * 500 ml) / 84 g/mol
Mass = 0.399 g

Now, we know that the solution was reduced to 120 ml by evaporation. We need to find the mass of the solid formed during this process. To do this, we can calculate the mass of the remaining solution and subtract it from the initial mass.

The final mass of the remaining solution is:

Final Mass = (0.067 g/ml * 120 ml) / 84 g/mol
Final Mass = 0.096 g

Now, to find the mass of solid MgCO3 formed, we subtract the final mass of the solution from the initial mass:

Mass of Solid = Initial Mass - Final Mass
Mass of Solid = 0.399 g - 0.096 g
Mass of Solid = 0.303 g

So, the mass of solid MgCO3 formed is 0.303 g. I'm sorry if this explanation was a bit dry, but I hope you found it useful!