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fomenos
2 years ago
7

Which has the largest atomic radius Pb or O

Chemistry
1 answer:
Sergeeva-Olga [200]2 years ago
4 0
<h2>LARGEST ATOMIC</h2>

\large\mathcal{QUESTION:}

  • Which has the largest atomic radius Pb or O?

\large\mathcal{ANSWER:}

  • Pb

______

  • The atomic radius of Pb, <u>also known as "Lead"</u> is <u>180 pm</u> while the atomic radius of O, <u>also known as "Oxygen</u>" is <u>152 pm</u>. <u>The largest atomic radius</u> between Pb and O <u>is Pb</u>.

<h3>Atomic radius definition:</h3>

  • A <u>chemical element's</u> <u>atomic radius</u> is a <u>measurement of the size of its atoms</u>, usually the mean or typical distance between the nucleus's center and the boundary of the surrounding electron shells.

<h3>#ProvideUniqueAnswers</h3>

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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
Colt1911 [192]

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.

7 0
3 years ago
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