Answer:
HOAc is stronger acid than HClO
ClO⁻ is stronger conjugate base than OAc⁻
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
Explanation:
Assume 0.10M HOAc => H⁺ + OAc⁻ with Ka = 1.8 x 10⁻⁵
=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺
Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸
=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺
HOAc delivers more H⁺ than HClO and is more acidic.
Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
The answer is B because isotopes have the same number of protons and neutrons.
Ideal gas law:
PV=nRT ⇒ V=nRT / P
P=pressure=1 atm
V=volume
n=number moles=2.10 moles
R=0,082 Atm l/ºK mol
T=temperature=273 K
V=(2.10 moles*0.082 (atm l)/º(K mol)*237ºK) / 1 atm=47.01 litres
47.1 L
mol of Na2CO3 = 2.36 x 10⁻⁴
<h3>Further explanation</h3>
Given
Mass : 0.025 g of Na2CO3
Required
moles
Solution
The mole is the number of particles contained in a substance
1 mol = 6.02.10²³
Moles can also be determined from the amount of substance mass and its molar mass :
mol = mass : molar mass
mass = mol x molar mass
Input the value :
mol = mass : MW Na2CO3
mol = 0.025 g : 106 g/mol
mol = 2.36 x 10⁻⁴
Answer:
The ratio of staggered to eclipsed conformers is 134
Explanation:
It is possible to determine the ratio of staggered to eclipsed conformers of a reactant, using the equilibrium:
Staggered ⇄ Eclipsed
Keq = [Eclipsed] / [Staggered]
That means Keq is equal to the ratio we need to find:
Using:
G˚= -RTln Keq
<em>Where G° = -12133.6J/mol</em>
<em>R is gas constant: 8.314J/molK</em>
<em>T is absolute temperature: 298K</em>
<em />
-12133.6J/mol= -8.314J/molK*298K ln Keq
4.8974 = ln Keq
134 = Keq = [Eclipsed] / [Staggered]
<h3>The ratio of staggered to eclipsed conformers is 134</h3>
