Answer is: Both a fluorine atom and a bromine atom gain one electron, and both atoms become stable.
Fluorine and bromine are in group 17 in Periodic table of elements. Group 17 (halogens) elements are in group 17: fluorine (F), chlorine (Cl), bromine (Br) and iodine (I). They are very reactive and easily form many compounds.
Halogens need to gain one electron to have electron cofiguration like next to it noble gas.
Fluorine has atomic number 9, it means it has 9 protons and 9 electrons.
Fluorine tends to have eight electrons in outer shell like neon (noble gas) and gains one electron in chemical reaction.
Electron configuration of fluorine: ₉F 1s² 2s² 2p⁵.
Electron configuration of neon: ₁₀Ne 1s² 2s² 2p⁶.
the answer is the nature of science
Answer:
The value of the carbon bond angles are 109.5 °
Explanation:
CH3CH2CH2OH = propanol . This is an alcohol.
All bonds here are single bonds.
Single bonds are sp³- hybdridization type. To be sp3 hybridized, it has an s orbital and three p orbitals : sp³. This refers to the mixing character of one 2s-orbital and three 2p-orbitals. This will create four hybrid orbitals with similar characteristics.
Sp3- types have angles of 109.5 ° between the carbon - atoms.
This means that the value of the carbon bond angles are 109.5 °
Answer:
pH = 12.22
Explanation:
<em>... To make up 170mL of solution... The temperature is 25°C...</em>
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The dissolution of Barium Hydroxide, Ba(OH)₂ occurs as follows:
Ba(OH)₂ ⇄ Ba²⁺(aq) + 2OH⁻(aq)
<em>Where 1 mole of barium hydroxide produce 2 moles of hydroxide ion.</em>
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To solve this question we need to convert mass of the hydroxide to moles with its molar mass. Twice these moles are moles of hydroxide ion (Based on the chemical equation). With moles of OH⁻ and the volume we can find [OH⁻] and [H⁺] using Kw. As pH = -log[H⁺], we can solve this problem:
<em>Moles Ba(OH)₂ molar mass: 171.34g/mol</em>
0.240g * (1mol / 171.34g) = 1.4x10⁻³ moles * 2 =
2.80x10⁻³ moles of OH⁻
<em>Molarity [OH⁻] and [H⁺]</em>
2.80x10⁻³ moles of OH⁻ / 0.170L = 0.01648M
As Kw at 25°C is 1x10⁻¹⁴:
Kw = 1x10⁻¹⁴ = [OH⁻] [H⁺]
[H⁺] = Kw / [OH⁻] = 1x10⁻¹⁴/0.01648M = 6.068x10⁻¹³M
<em>pH:</em>
pH = -log [H⁺]
pH = -log [6.068x10⁻¹³M]
<h3>pH = 12.22</h3>
Answer:
Explanation:
The actual boiling point is probably between 34C and 40C.
