Give the number of lone pairs around the central atom and the geometry of the ion IBr2.
1 answer:
Answer:
Option E!
Explanation:
If we were to draw the lewis dot structure for IBr2 -, we would first count the total number of valence electrons ( " available electrons " ). Iodine has 7 valence electrons, and so does Bromine, but as Bromine exists in 2, the total number of valence electrons would be demonstrated below;
Don't forget the negative on the Bromine!
Now go through the procedure below;
1 ) Place Iodine in the middle and draw single bonds to each of the bromine.
2 ) Add three lone pairs on each of the Bromine's
3 ) Now we have 6 electrons left, if we were to exclude the electrons shared in the " single bonds. " This can be placed as three lone pairs on Iodine ( central atom )!
The molecular geometry can't be linear, as there are lone pairs on the atoms. This makes it bent.
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Answer:
A. 0.143 M
B. 0.0523 M
Explanation:
A.
Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).
KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄
The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:
1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol
The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.
5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:
M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M
B.
Let's consider the neutralization of potassium hydroxide and perchloric acid.
KOH + HClO₄ → KClO₄ + H₂O
When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.
The equation that scientists could use to find the wavelength of the emission lines of the hydrogen atom would be that of Balmer.
The wavelength of the emission lines of the hydrogen atom can be derived using the Balmer series:
1/λ
Where λ = wavelength, = Rydberg constant, and n = level of the original orbital.
The equation becomes applicable in getting the wavelength of emitted light when electrons in hydrogen atoms transition from higher (n) orbital to lower orbital (2) levels.
More on the Balmer series can be found here: brainly.com/question/5295294