Answer:
The equilibrium constant Ksp of the generic salt AB2 = 6.4777 *10^-8 M
Explanation:
Step 1: The balanced equation
AB2 ⇒ A2+ + 2B-
Step 2: Given data
Concentration of A2+ = 0.00253 M
Concentration of B- = 0.00506 M
Step 3: Calculate the equilibrium constant
Equilibrium constant Ksp of [AB2] = [A2+][B-]²
Ksp = 0.00253 * 0.00506² = 6.4777 *10^-8 M
The equilibrium constant Ksp of the generic salt AB2 = 6.4777 *10^-8 M
Answer:
A. The amount of energy needed to remove 1 mole of electrons from 1 mole of ground-state atoms or ions in the gas phase.
Explanation:
Ionization energy is the quantity of energy required to remove an electron in ground electronic state from an isolated gaseous atom or ion, resulting in a cation. kJ/mol is the expresion we use for this energy, it refers to the amount of energy it takes for all the atoms in a mole to lose one electron each.
Ionization energy can be used as an indicator of reactivity and can be used to help predict the strength of chemical bonds because the more electrons are lost, the more positive the ion will be and the harder it will be to separate the electrons from the atom.
I hope you find this information useful and interesting! Good luck!
Answer:
Option C = 30 j
Explanation:
Given data:
mass of snowboard = 5 Kg
Initial speed = 2 m/s
final speed = 4 m/s
work done = ΔE= ?
ΔE= change in kinetic energy
Solution:
Formula:
K.E (initial) = 1/2 × mv²
K.E (initial) = 1/2 × 5 Kg . (2m/s)²
K.E (initial) = 1/2 × 20 Kg.m²/s²
K.E (initial) = 10 Kg.m²/s² or 10 J
Kg.m²/s² = J
K.E (finial) = 1/2 × mv²
K.E (finial) = 1/2 × 5 Kg . (4m/s)²
K.E (finial) = 1/2 × 5 Kg . 16 m²/s²
K.E (finial) = 1/2 × 80 Kg.m²/s²
K.E (finial) = 40 Kg.m²/s² or 40 J
work done = ΔE = K.E (finial) - K.E (initial)
work done = ΔE = 40 J - 10 J
work done = ΔE = 30 J
