Based on the trend of electronegativities values of the elements <em>Be, Mg, Ca, </em>and,<em> Sr</em> within their group (increasing from bottom to top), the atoms of the element Sr will have the <u>least attraction</u> for an electron.
We need to remember that electronegativity indicates the affinity or attraction of an element for an electron. Hence, the <u>higher</u> the <em>electronegativity,</em> the<u> higher</u> the <em>attraction </em>of that element for an <em>electron</em>.
The values of <em>electronegativities </em>of the given <em>elements </em>are the following:
Be = 1.57
Mg = 1.31
Ca = 1.0
Sr = 0.95
We can see that the increasing trend of <em>electronegativities </em>in this group is from the <u>bottom to the top</u>, having the Sr with the lowest electronegativity value and the <em>Be</em> with the <em>highest</em>.
This trend is related to atomic size, the <u>larger</u> the <em>atomic size</em>, the <u>lower</u> the <em>electronegativity</em> because the <em>electron </em>in the outermost shell will feel <u>less attraction</u> towards the nucleus of an atom. In a group of the periodic table, the <em>size </em>of an <em>atom </em>will <em>increase </em>from <u>top to bottom</u>.
Therefore, according to the said above, the atoms of the Sr will have the least attraction for an electron (it has the lower electronegativity value).
Find more about electronegativity here:
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Answer:
<h3>the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
Explanation:
Amount of HBr dissociated
2HBr(g) ⇆ H2(g) + Br2(g)
Initial Changes 2.15 0 0 (mol)
- 0.789 + 0.395 + 0.395 (mol)
At equilibrium 1.361 0.395 0.395 (mole)
Concentration 1.361 / 1 0.395 / 1 0.395 / 1
at equilibrium (mole/L)
<h3>Therefore, the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
n(O₂)=1.407 mol
M(C₃H₈)=44.1 g/mol
m(C₃H₈)/M(C₃H₈)=n(O₂)/5
m(C₃H₈)=M(C₃H₈)n(O₂)/5
m(C₃H₈)=44.1*1.407/5=12.410 g
the mass of propane is 12.410 g