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>
<em />
[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>
Answer:
condensation is the answer
CH3 is the empirical formula for the compound.
A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.
The number of atom or moles in the compound is
1.17 g C X 1 mol of C / 12.011 g C = 0.097411 mol of C.
0.287 g H x 1 mol of H / 1 g H = 0.28474 mol H.
This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.
So we can represent the compound with the formula C0.974H0.284.
Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.
we can divide 0.284 by 0.0974.
0.284 / 0.0974 = 3.
So here, Carbon is one and hydrogen is 3.
We can write the above formula as a CH3.
Hence the empirical formula for the sample compound is CH3.
For a detailed study of the empirical formula refer given link brainly.com/question/13058832.
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Magnesium (Mg)
The reason for this is the reactivity of the listed metals. Gold and silver are extremely unreactive metals. It is because of this unreactive nature that they remain in good condition for long periods of time, and are preferred in jewelry. Copper, although more reactive than gold and silver, is still not reactive enough to react with HCl.
The only metal that will react is magnesium.
For the absorbance of the solution in a 1.00 cm cell at 500 nm is mathematically given as
A’ = 0.16138
<h3>What is the absorbance of the solution in a 1.00 cm cell at 500 nm?</h3>
Absorbance (A) 2 – log (%T) = 2 – log (15.6) = 0.8069
Generally, the equation for the Beer’s law is mathematically given as
A = ε*c*l
0.8069 = ε*c*(5.00 )
ε*c = 0.16138 cm-1
then for when ε*c is constant
l’ = 1.00
A’ = (0.16138 cm-1)*(1.00 cm)
A’ = 0.16138
In conclusion, the absorbance of the solution in a 1.00 cm cell at 500 nm is
A’ = 0.16138
Read more about Wavelength
brainly.com/question/3004869
