To get this it helps to know the electronegativity numbers of the elements but it isn't required. You just need to know that Fluorine is the most electronegative element and that the farther away from Fluorine you are on the periodic table, the less electronegative you get. The one exception to this rule is hydrogen with actually has an electronegativity of 2.1 while lithium has one of 1.0. Also the higher difference in electronegativity between two atoms the more polar the bond is.
Now to start the question. H-Br could be a contender since H has an electronegativity number of 2.1 and Br is relatively close to Fluorine so we'll put that one aside for now. H-Cl knocks out A because both bonds have H but one bond has Br and the other has Cl. Cl is closer to Fluorine than Br so answer B is the contender now. For answer C, I and Br are too close to have a higher electronegativity difference than H-Cl so that one isn't it. Finally for answer D, I is much closer to Cl than H is so the electronegativity difference is much less, making your answer B.
<span>c. energy needed to remove an electron from an atom or ion in the gas phase</span>
Answer:
The formula of the original halide is SrCl₂.
Explanation:
- The balanced equation of this reaction is:
SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X is the halide.
- From the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of SrSO₄.
- The number of moles of SrSO₄ <em>(n = mass/molar mass) </em>= (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
- The number of moles of SrX are 4.11 x 10⁻³ moles from the stichiometry of the balanced equation.
- n = mass / molar mass, n = 4.11 x 10⁻³ moles and mass = 0.652 g.
- The molar mass of SrX₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
- The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
- The atomic mass of halide X = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.
- This is the atomic mass of Cl.
- <em>So, the formula of the original halide is SrCl₂</em>.
