The term for a push or pull is called Force.
Equilibrium expression is ![Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BH3O%2B%5D%5BHCO3%5E-%5D%7D%7B%5BH2CO3%5D%7D%5C%5C)
<u>Explanation:</u>
Equilibrium expression is denoted by Keq.
Keq is the equilibrium constant that is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.
Example -
aA + bB = cC + dD
So, Keq = conc of product/ conc of reactant
![Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BC%5D%5Ec%20%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%20%5BB%5D%5Eb%7D)
So from the equation, H₂CO₃+H₂O = H₃O+HCO₃⁻¹
![Keq = \frac{[H3O^+]^1 [HCO3^-]^1}{[H2CO3]^1 [H2O]^1}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BH3O%5E%2B%5D%5E1%20%5BHCO3%5E-%5D%5E1%7D%7B%5BH2CO3%5D%5E1%20%5BH2O%5D%5E1%7D)
The concentration of pure solid and liquid is considered as 1. Therefore, concentration of H2O is 1.
Thus,
Therefore, Equilibrium expression is
Answer:
Oil is extracted by three general methods: rendering, used with animal products and oleaginous fruits; mechanical pressing, for oil-bearing seeds and nuts; and extracting with volatile solvents, employed in large-scale operations for a more complete extraction than is possible with pressing.
Explanation:
The atomic number increases moving left to right across a period and subsequently so does the effective nuclear charge. Therefore, moving left to right across a period the nucleus has a greater pull on the outer electrons and the atomic radii decreases.
Answer:
127.3° C, (This is not a choice)
Explanation:
This is about the colligative property of boiling point.
ΔT = Kb . m . i
Where:
ΔT = T° boling of solution - T° boiling of pure solvent
Kb = Boiling constant
m = molal (mol/kg)
i = Van't Hoff factor (number of particles dissolved in solution)
Water is not a ionic compound, but we assume that i = 2
H₂O → H⁺ + OH⁻
T° boling of solution - 118.1°C = 0.52°C . m . 2
Mass of solvent = Solvent volume / Solvent density
Mass of solvent = 500 mL / 1.049g/mL → 476.6 g
Mol of water are mass / molar mass
76 g / 18g/m = 4.22 moles
These moles are in 476.6 g
Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m
T° boling of solution = 0.52°C . 8.85 m . 2 + 118.1°C = 127.3°C
