The electronic configuration of a chlorine ion in BeCl2 compound is
[2.8.8]^- (answer B)
chlorine atom gain on electron form Be to form chloride ions
chlorine atom has a electronic configuration of 2.8.7 and it gains one electron to form chloride ion with 2.8.8 electronic configuration
Answer:
The resultant structure is shown below. This structure contains four shared pairs of electrons, which are located on all four "sides" of carbon's electron dot structure. Each of these shared pairs was created by pairing one of carbon's unpaired electrons with an unpaired electron from chlorine.
Explanation:
Answer:
a. 0.5 mol
b. 1.5 mol
c. 0.67
Explanation:
Fe3+ + SCN- -----> [FeSCN]2+
a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium
b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium
c. KC = 0.5/(0.5*1.5) = 0.67
Answer:
Kc = 1.09x10⁻⁴
Explanation:
<em>HF = 1.62g</em>
<em>H₂O = 516g</em>
<em>F⁻ = 0.163g</em>
<em>H₃O⁺ = 0.110g</em>
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To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:
Kc = [H₃O⁺] [F⁻] / [HF]
<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>
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[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M
[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M
[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M
Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]
<h3>Kc = 1.09x10⁻⁴</h3>