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Bogdan [553]
3 years ago
6

A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is

then mixed with 75.0 mL of a 0.800 M aqueous solution of AlBr3. Calculate the concentration (M) of Na+ and Br− in the final solution.
Chemistry
1 answer:
Colt1911 [192]3 years ago
7 0

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

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The<em> effective nuclear charge</em> is that portion of the total nuclear charge that a given electron in an atom feels.

Since, the inner electrons repel the outer electrons, t<em>he effective nuclear charg</em>e of a determined electron is the sum of the positive charge (number of protons or atomic number) that it feels from the nucleus less the number of electrons that are in the shells that are are closer to the nucleus than the own shell of such (determined) electron.

Mathematically, <em>the effective nuclear charge (Zeff)</em> is equal to the atomic number (Z) minus the amount (S) that other electrons in the atom shield the given (determined) atom from the nucleus.

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Let's look at the group to which Si belongs, which is the group 14. This table summarizes the relevant data:

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C              6      14                      4                     6 - 4 = 2             6 -  2 = +4

Si             14     14                      4                     14 - 4 = 10         14 - 10 = +4

Ge           32     14                     4                     32 - 4 = 28       32 -28 = +4

Sn           50     14                     4                     50 - 4 = 46       50 - 46 = +4

Pb           82     14                     4                     82 - 4 = 78        82 - 78 = +4  

With that, you have shown that the valence electrons of the unknown substance's atoms feel an effective nuclear charge of +4 and you have a short list of 4 elements which can be the unknown element: C, Ge, Sn or Pb.

The second known characteristic of the unknown substance's atoms is that it has a <em>higher electronegativity than silicon (Si)</em><em>.</em>

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Other two periodic trends attending the group number are the <em>atomic radii and the ionization energy</em>.

The atomic radii generally increases as you go from top to bottom in a group. This is because you are adding electrons to new higher main energy levels. So, you can conclude that the originally unknwon substance (carbon) has a smaller atomic radii, than Si.

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Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

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2.3  × 0.523 J/g.°C

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