Answer:
the equilibrium concentration of HF is 2.85 M
Option a) 2.85 M is the correct answer.
Explanation:
Given the data in the question;
H₂ + F₂ ⇄ 2HF
I 1.69 M 1.69 M 0
C -x -x +2x
E 1.69-x 1.69-x +2x
given that Kc = 115
Kc = [ HF ]² / [H₂][F₂]
we substitute
115 = [ 2x ]² / [ 1.69-x ][ 1.69-x ]
lets find the square root of both sides
10.7238 = 2x / [ 1.69-x ]
10.7238[ 1.69-x ] = 2x
18.123222 - 10.7238x = 2x
2x + 10.7238x = 18.123222
12.7238x = 18.123222
x = 18.123222 / 12.7238
x = 1.424356
Hence, equilibrium concentration of HF = 2x
that is;
HF = 2 × 1.424356
HF = 2.8487 ≈ 2.85 M
Therefore, the equilibrium concentration of HF is 2.85 M
Option a) 2.85 M is the correct answer.
Answer:
Ionic bonds - an intramolecular force exist in a nonmetal and metal compounds such as NaCl. The Na donate 1 electon to Cl to complete its octet rule.
Covalent bond - an intramolecular force exist in a nonmetal and nonmetal compounds such as bonds O2, Cl2, CO2, sugar, proteins and most of organic compounds and biomolecules by sharing electrons to bond.
There are two types of covalent bonds: polar and nonpolar. Polar bond is a bond between two different nonmetal atoms of different electronegativities. While nonpolar bond is a bond between the same atom or two differenct atoms of the same electronegativities (if there is). Their electronegativities pull will cancel so that their overall polarity is zero.
Q1)
molarity is defined as the number of moles of solute in 1 L of solution.
the NaCl solution volume is 1.00 L
number of moles NaCl = NaCl mass present / molar mass of NaCl
NaCl moles = 112 g / 58.5 g/mol = 1.91 mol
the number of moles of NaCl in 1.00 L of solution is - 1.91 mol
therefore molarity of NaCl is 1.91 M
Q2)
molality is defined as the number of moles of solute in 1 kg of solvent.
density is mass per volume.
density of the solution is 1.08 g/mL.
therefore mass of the solution is = density x volume
mass = 1.08 g/mL x 1000 mL = 1080 g
since we have to find the moles in 1 kg of solvent
mass of solvent = 1080 g - 112 g = 968 g
number of moles of NaCl in 968 g of solvent - 1.91 mol
therefore number of NaCl moles in 1000 g - (1.91 mol / 968 g) x 1000 g/kg = 1.97 mol/kg
molality of NaCl solution is 1.97 mol/kg
Q3)
mass percentage is the percentage of mass of solute by total mass of the solution
mass percentage of solution = mass of solute / total mass of the solution
mass of solute = 112 g
total mass of solution = 1080 g
mass % of NaCl = 112 g / 1080 g x 100%
therefore mass % of NaCl = 10.4 %
answer is 10.4 %
Answer:
118 elements
Explanation:
Of these 118 elements, 94 occur naturally on Earth.
